How to do these STP gas and mass stoichiometry problems in general.
All of the problems in this set are stoichiometry problems with at
least one equation participant as a gas at STP. (a) Write and
balance the chemical equation. (2) Do the math in DA style using
1 mole gas at STP = 22.4 liters as a factor. In the following
problems ALL GASES ARE AT STP.
In this set of problems for the first time you will be
using the idea of stoichiometry. These problems always refer to a chemical
reaction. The chemical reaction must be first written and balanced. You will
be given an amount of one of the materials and be expected to find out how
much that corresponds to another one of the materials in that chemical reaction.
Because we are beginning with an amount of a material and we are looking for
another amount of (another) material, the DA method
is standardly used for this type of problem.
Begin each problem with what you know, the
GIVEN material and the GIVEN amount in whatever units it is given.
Change the GIVEN amount of material to units of
mols. In the case of gases at STP, the conversion factor is: 1 mol of gas =
22.4 liters at STP. In the case of masses, you must use the formula weight of
the material to change from mass to mols.
Use the mol ratio to change from one material
to another. The mol ratio is the name (or symbol) of the material and the
coefficient of that material in the balanced chemical reaction.
Change the mols of the new material to
whatever units are asked for.
Back to gas @ STP problems.
ANSWER AND DISCUSSION
PROBLEM #1, SET 2 - STP GAS AND MASS STOICHIOMETRY
1. How many moles of nitrogen gas is needed to react with 44.8
liters of hydrogen gas to produce ammonia gas?
3H2 +
N2
2NH3
GIVEN: 44.8 L of H2 at STP.
FIND: mols of N2.
Here the sequence is: GIVEN liters of H2 at STP,
CHANGE liters of H2 at STP to mols of
H2, MOL RATIO to change from H2
to N2. There is no need to go any further
to change the N2 into mols, because the
mol ratio leaves the material in that unit anyway.
The MVG is the "molar volume of gas at STP." The
MR is the mol ratio. Notice the numbers in
blue in the mol ratio. Those numbers come from the
coefficients in the balanced chemical equation.
The math is: 44.8 ÷ 22.4 ÷ 3 =
calculator
Back to gas @ STP problems.
ANSWER AND DISCUSSION
PROBLEM #2, SET 2 - STP GAS AND MASS Stoichiometry
2. How many liters of ammonia are produced when 89.6 liters of
hydrogen are used in the above reaction?
The "above reaction" from problem #1 is: N2 +
3H2
2NH3
GIVEN: 89.6 L of H2 at STP.
FIND: Volume of ammonia (in liters at STP)
Take the GIVEN quantity, use the Molar Volume of Gas at
STP (MVG) to change it to mols, change the material with the mol ratio (MR),
and change the mols of new material to the requested liters at STP using the
MVG again.
The math is: 89.6 ÷ 22.4 ÷ 3 x 2 x 22.4 =
or, if the MVG's cancel, 89.6 ÷ 3 x 2 =
calculator
Back to gas @ STP problems.
ANSWER AND DISCUSSION
PROBLEM #3, SET 2 - STP GAS AND MASS STOICHIOMETRY
3. Ten grams of calcium carbonate was produced when carbon
dioxide was added to lime water (calcium hydroxide in solution).
What volume of carbon dioxide at STP was needed?
CO2 +
Ca(OH)2
CaCO3 +
H2O
GIVEN: 10.0 g = mass of calcium carbonate
FIND: Volume of carbon dioxide (in liters at STP)
Take the GIVEN quantity, a mass, use the Formula Weight of the given
quantity to change it to mols, change the material with the mol ratio (MR),
and change the mols of new material to the requested liters at STP using the
MVG. Find this pathway on the Stoichiometry Roadmap
The math is: 10 ÷ 100.1 x 22.4 =
calculator
Back to gas @ STP problems.
ANSWER AND DISCUSSION
PROBLEM #4, SET 2 - STP GAS AND MASS STOICHIOMETRY
4. When 11.2 liters of hydrogen gas is made by adding zinc to
sulfuric acid, what mass of zinc is needed?
Zn + H2
SO4
H2 + ZnSO4
GIVEN: 11.2 L of H2 at STP
FIND: mass of Zn
Take the GIVEN quantity, a volume at STP, use the MVG to change it to mols, change the material with the mol ratio (MR),
and change the mols of new material to the mass using the formula weight of the
new material. Find this pathway on the Stoichiometry Roadmap
The math is: 11.2 ÷ 22.4 x 65.4 =
calculator
Back to gas @ STP problems.
ANSWER AND DISCUSSION
PROBLEM #5, SET 2 - STP GAS AND MASS STOICHIOMETRY
5. What volume of ammonia at STP is needed to add to water to
produce 11 moles of ammonia water?
The balanced chemical equation is:
NH
3
+ H2
O
NH4
OH
GIVEN: 11.0 mols of ammonia water (NH4OH)
FIND: Volume of ammonia gas (NH3) in liters at STP.
Take the GIVEN quantity, a number of mols, and directly use the mol ratio
(MR) to change to the other material. The mols of FIND material can be changed
to the requested liters at STP using the MVG.
Find this pathway on the Stoichiometry Roadmap
ANSWER AND DISCUSSION
PROBLEM #6, SET 2 - STP GAS AND MASS STOICHIOMETRY
6. How many grams of carbonic acid is produced when 55 liters
of carbon dioxide is pressed into water?
There is no such thing as solid carbonic acid. It only exists in ionic form in
solution, but we can consider this exercise anyway.
The balanced chemical equation is:
CO
3
+ H2
O H2
CO3
GIVEN: 55.0 mols of carbon dioxide (CO2)
FIND: mass of carbonic acid (H2CO3).
Tuesday, 23 August 2016
Stoichiometry : Sample Problems
1. A sample contains 27.1 g of calcium oxide. How many moles of calcium oxide are in the sample?
- NOTE: Use the Periodic Table to find the molecular mass (grams per
mole)
- NOTE: A mole is by definition 6.0220 x 1023 particles which
can generally be rounded to 6.02 x 1023.
- NOTE: At STP, 1 mole of any gas = 22.4L. STP is 273K (0C)
and 1 atm.
- NOTE: Molarity is MOLES per LITER. The volume in
milliliters must be converted to Liters.
6. How many grams of carbon dioxide are there in a container with a volume of 4.50L at STP?
7. How many moles of nitrogen are there in a 7.45 mol sample of ammonium phosphate?
8. If 120. g of propane, C3H8, is burned in excess oxygen, how many grams of water are formed?
9. 90.0 g of FeCl3 reacts with 52.0 g of H2S. What is the limiting reactant? What is the mass of HCl produced? What mass of excess reactant remains after the reaction?
- NOTE: The limiting reactant is the reactant that limits the amount
of product that can be formed and is completely consumed during the reaction.
Practice Problems: Stoichiometry
- Balance the following chemical reactions:
a. CO + O2 CO2
b. KNO3 KNO2 + O2
c. O3 O2
d. NH4NO3 N2O + H2O
e. CH3NH2 + O2 CO2 + H2O + N2 Hint
f. Cr(OH)3 + HClO4 Cr(ClO4)3 + H2O - Write the balanced chemical equations of each reaction:
a. Calcium carbide (CaC2) reacts with water to form calcium hydroxide (Ca(OH)2) and acetylene gas (C2H2).
b. When potassium chlorate (KClO3) is heated, it decomposes to form KCl and oxygen gas (O2).
c. C6H6 combusts in air. Hint
d. C5H12O combusts in air. - Given the following reaction: Na2S2O3 + AgBr
NaBr + Na3[Ag(S2O3)2]
a. How many moles of Na2S2O3 are needed to react completely with 42.7 g of AgBr?
b. What is the mass of NaBr that will be produced from 42.7 g of AgBr?
Hint - From the reaction: B2H6 + O2 HBO2 + H2O
a. What mass of O2 will be needed to burn 36.1 g of B2H6?
b. How many moles of water are produced from 19.2 g of B2H6? - Calculate the mass (in kg) of water produced from the combustion of 1.0 gallon (3.8 L) of gasoline (C8H18). The density of gasoline is 0.79 g/mL.
- One mole of aspartame (C14H18N2O5) reacts with two moles of water to produce
one mole of aspartic acid (C4H7NO4), one mole of methanol (CH3OH) and one
mole of phenylalanine.
a. What is the molecular formula of phenylalanine?
b. What mass of phenylalanine is produced from 378 g of aspartame? - KO2 is used in a closed-system breathing apparatus. It removes carbon dioxide
and water from exhaled air. The reaction for the removal of water is: KO2
+ H2O
O2 + KOH. The KOH produced is used to remove carbon dioxide by the following
reaction: KOH + CO2
KHCO3.
a. What mass of KO2 produces 235 g of O2?
b. What mass of CO2 can be removed by 123 g of KO2?
Sunday, 14 August 2016
SAT / GMSAT / MCAT / ECAT physics help
1) Two forces have magnitudes of 11 newtons and 5 newtons. The magnitude
of their sum could be equal to all of the following values EXCEPT one.
Which one is the EXCEPTION?
A. 10 N
B. 7 N
C. 5 N
D. 16 N
2) If the sum of all the forces acting on a moving object is zero, the object will:
A. slow down and stop.
B. decelerate uniformly.
C. increase velocity only if there is no friction.
D. continue moving with constant velocity.
3) If the direction of a moving car changes and its speed remains constant, which quantity must remain the same?
A. Momentum
B. Displacement
C. Velocity
D. Kinetic energy
4) Two cylindrical wires are drawn from the same material, one having a radius twice that of the other but both wires being of the same length. What is the ratio of the resistances in the two wires?
A. 1:1
B. 2:1
C. 4:1
D. 8:1
5) An object weighs 60 newtons on Earth's surface. When the same object is moved to a point one Earth radius above Earth's surface, what would be its weight?
A. 120 N
B. 15 N
C. 30 N
D. 60 N
6) A spring scale reads 10 newtons as it pulls a 25 kilogram mass across a flat surface. What is the magnitude of the force exerted by the mass on the spring scale?
A. 250 N
B. 50 N
C. 25 N
D. 10 N
7) A 20 kg child is playing on a swing of negligible mass. It is attached to its pivot point with two 3 m ropes. The maximum speed of the child during a swing cycle is found to be 3 m/s. What is the maximum tension in each rope?
A. 200 N
B. 260 N
C. 60 N
D. 130 N
8) A raft is constructed from wood that has a density of 400 kg/m3. Ten identical sections of wood (width, 25 cm; depth, 10 cm; length, 4 m) are tied together side by side to build the raft. The density of water is 1000 kg/m3. How much of the unloaded raft, in depth, is underwater?
A. 2 cm
B. 4 cm
C. 6 cm
D. 10 cm
9) A copper rod which is 5 centimeters in diameter and infinitely long carries a current of 2 amps. The current is distributed uniformly throughout the rod. The magnetic field half way between the axis of the rod and its outside edge is:
A. zero.
B. pointing radially inward.
C. pointing radially outward.
D. encircling the axis of the rod.
10) A ball is thrown up in the air with a velocity of 10 m/s. Which of the following is true?
A. Time in air is 1 s; maximum height is 5 m.
B. Time in air is 1 s; maximum height is 10 m.
C. Time in air is 2 s; maximum height is 5 m.
D. Time in air is 2 s; maximum height is 10 m.
11) An electron moving at constant velocity enters the area between two charged plates, as shown in the diagram below. Which of the paths correctly indicates the electron's trajectory after leaving the area between the charged plates?
12) A person stands on a scale in an elevator. He notices that the scale reading is lower than his usual weight. Which of the following could possibly describe the motion of the elevator?
A. It is moving down and slowing down.
B. It is moving down at constant speed.
C. It is moving up and slowing down.
D. It is moving up with constant speed.
13) The cruising altitude of most planes is about 40,000 feet above sea level where there is less air above the plane pushing down, so the air pressure is lower (about 20 kPa outside compared to about 100 kPa atmospheric pressure at sea level). To keep everyone comfortable inside the plane, the cabin is pressurized to about 75 kPa. The passenger doors on a 747 are about 1 meter wide by 2 meters tall.
If a metric ton is 1000 kg, to pull the door open at 40,000 feet is equivalent to moving how many metric tons?
A. 5 x 10-2 tons
B. 0.8 tons
C. 2 tons
D. 11 tons
14) A box sits on an inclined plane without sliding. As the angle of the plane (measured from the horizontal) increases, the normal force:
A. decreases nonlinearly.
B. increases nonlinearly.
C. increases linearly.
D. does not change.
15) What is the voltage drop across R3 in the circuit diagram below?
A. 15 V
B. 30 V
C. 40 V
D. 100 V
Answer Key:
1. C
2. D
3. D
4. C
5. B
6. D
7. D
8. B
9. D
10. C
11. B
12. C
13. D
14. A
15. B
SOLUTIONS / EXPLANATIONS:
*Gold Standard GAMSAT textbook Reference
1) If you are adding these 2 vectors, let's consider the 2 extreme situations (i.e. the highest possible value and the lowest possible value): (1) they are pointing in exactly the same direction so the sum is 11 + 5 = 16 N; (2) they are pointing in exactly the opposite directions so the sum is 11 + (-5) = 6 N. Anything from 6 N to 16 N is possible but nothing above or below.
*PHY 1.1
2) Newton's 1st law states that an object in motion will stay in motion at constant velocity unless acted upon by an unbalanced force. For something to change direction or accelerate there must be an unbalanced force acting on it.
*PHY 4.2
3) If the direction changes, then all vectors change: momentum, displacement and velocity. Kinetic energy is a scalar and does not depend on direction. As long as the speed remains the same, the kinetic energy remains the same.
*PHY 1.1, 5.1, 5.2
4) Imagine that you needed to run down a hallway but you had 2 choices: (1) a very narrow hallway with a certain density of students there, just milling around OR (2) a wide hallway with the SAME density of students. Think for a moment. If you run down a wide hallway with the same density, you always have more options to choose from: left, right, sideways and so you can get through faster. It's the same with the electrons going through these 2 wires, same density but one with a greater radius.
Actually, a wire is like a tiny cylinder and if you cut a wire and look at its edge, the shape is a tiny circle. So it's not just a matter of the radius that is bigger, it's actually the area of a circle that is bigger. So we can summarize by saying: as the area increases, resistance decreases. Or, in “math-speak”, resistance is inversely proportional to the cross-sectional area.
Now we have our first basic high school equation that must be memorized!
The resistance (R) is inversely proportional to the cross-sectional area (A), where
Acircle = π(radius)2
If radius1 = 2 x radius2, then A1 = (2A2)2 = 4A2 and R1 = 1/4 R2, giving us a ratio of R1:R2 --> 4:1.
*PHY 10.2 (B: PHY 3.3)
5) You would be expected to know: Newton's Law of Gravitation F = Gm1m2/r2 (you would never perform a complete calculation on the GAMSAT using values for each of the unknowns in the equation because there is just not enough time! However, you must know the relationship so that you can make assessments as to the relative changes of the variables in the equation).
When the object is moved 1 Earth radius away from the Earth then r is now 2r (on the surface of the earth the object is 1 Earth radius away from the center of the Earth, when the object is moved 1 radius away the distance is now r + r = 2r).
When you plug (2r)2 into the denominator of the equation it becomes 4r2. Thus when you double the distance, you quarter the force (same relationship with Coulomb's Law and point charges). So if we start with a force of 60 N, then we end with 15 N. Not much to calculate!
*PHY 2.4
6) This is a common exam question testing your understanding of Newton's 3rd Law: for every action, there is an equal and opposite reaction. When you push a couch, the couch is pushing back against your body with an equal force. The mass in the question exerts a 10 N force back on the spring because action/reaction forces are equal.
*PHY 2.3
7) The tension in the ropes is largest when the ropes are vertical because at that moment, the centripetal force is maximum and in the same direction as gravity. Clearly, the speed of the child is maximum at the point of lowest (gravitational) potential energy; therefore, the centripetal force is also maximum at that point.
Centripetal force = mv2/r = (20 x 3 x 3)/3 = 60 N
Weight = 200 NMaximum force = 200 + 60 = 260 N
Maximum tension in each rope = 260/2 = 130 N*PHY 3.3
8) First let's calculate the specific gravity (SG) of wood which is simply the density of wood divided by the density of water which is 400/1000 = 0.4. SG is equal to the fraction of the height of a buoyant (= floating) object below the surface. So we know the fraction is 0.4 and now we need to apply that to the total “height” which is the depth of the wood 10 cm. So 0.4 x 10 = 4 cm.
*PHY 6.1.1, 6.1.2
9) The right-hand rule describes the direction of the magnetic field: as you "grab" the wire with your thumb in the direction of the flow of current, the tips of your fingers trace the direction of the magnetic field which encircles the wire. See PHY 9.2.3 of the Gold Standard GAMSAT book.
*PHY 9.2.2, 9.2.3
10) v = vo + gt
t = (v-vo)/g = 10/10 = 1 s
total time in air (to go up and back down) = 2t = 2 s
And, s = vot + (g/2)(t2)
= (10 x 1) - (10/2)(1)
= 5 m (max. height)
Yes, there are some very basic physics equations that you should know for the GAMSAT and we have listed them all here: www.gamsat-prep.com/section3-physics-formula-list/
*PHY 1.6, 2.6
11) While in the area between the plates, the negatively charged electron is attracted to the positive plate and so it would bend “upwards”. But after leaving the plates, there is no more net force acting on the electron. Therefore, the electron would continue in motion in a straight line according to Newton's first law (inertia).
Oh, btw, the real GAMSAT: only 4 answer choices for all multiple choice questions. We just had to make this question 20% better but we promise not to do it again!! : )
*PHY 9.1.3, 4.2
12) You can draw a free-body diagram of the person:
If upwards is positive, and we know that the sum of forces is equal to ma from Newton's 2nd Law, we get: Fscale - weight = ma The scale reading is less than his normal weight, so the acceleration is negative; i.e., down. Downward acceleration means either something speeds up while moving downward, or slows down while moving upward.
*PHY 2.2, 3.4
13) Key points: (1) pressure is defined as force per unit area (just apply some pressure to the table in front of you and you'll would have to say that you are applying a certain force over a particular area); (2) dimensional analysis: kilo means 1000 in the SI system and pascals or Pa, being force/area, has the equivalent units of N/m2.
The difference in pressure between inside and outside the cabin at 40,000 feet is: 75 – 20 = 55 kPa = 55 000 Pa or N/m2 of pressure pushing on the door.
Pressure: P = F/A
The area of the door is 2 x 1 = 2m2
So F = PA = (55 000)(2) = 110 000 N
Thus weight mg = 110 000 meaning m = 11 000 kg or 11 metric tons.
*PHY 6.1.2
14) Because no forces act perpendicular to the incline except for the normal force and the perpendicular component of weight, and there is no acceleration perpendicular to the incline, the normal force is equal to the perpendicular component of weight, which is mg cosθ. As the angle increases, the cosine of the angle decreases. This decrease is nonlinear because a graph of the normal force N vs. θ would show a curve, not a line.
*PHY 3.2, 3.2.1
15) The 10 A of current splits into two paths. The path through R2 takes 7 A, so the path through R3 takes 3 A (this is Kirchoff's First Law, the junctional theorem). Now use Ohm's law, where the voltage V is equal to the current I times the resistance R (Ohm's law is a frequent visitor in the real GAMSAT). V = IR V = (3 A) (10 Ω) = 30 V.
*PHY 10.1, 10.2, 10.3
A. 10 N
B. 7 N
C. 5 N
D. 16 N
2) If the sum of all the forces acting on a moving object is zero, the object will:
A. slow down and stop.
B. decelerate uniformly.
C. increase velocity only if there is no friction.
D. continue moving with constant velocity.
3) If the direction of a moving car changes and its speed remains constant, which quantity must remain the same?
A. Momentum
B. Displacement
C. Velocity
D. Kinetic energy
4) Two cylindrical wires are drawn from the same material, one having a radius twice that of the other but both wires being of the same length. What is the ratio of the resistances in the two wires?
A. 1:1
B. 2:1
C. 4:1
D. 8:1
5) An object weighs 60 newtons on Earth's surface. When the same object is moved to a point one Earth radius above Earth's surface, what would be its weight?
A. 120 N
B. 15 N
C. 30 N
D. 60 N
6) A spring scale reads 10 newtons as it pulls a 25 kilogram mass across a flat surface. What is the magnitude of the force exerted by the mass on the spring scale?
A. 250 N
B. 50 N
C. 25 N
D. 10 N
7) A 20 kg child is playing on a swing of negligible mass. It is attached to its pivot point with two 3 m ropes. The maximum speed of the child during a swing cycle is found to be 3 m/s. What is the maximum tension in each rope?
A. 200 N
B. 260 N
C. 60 N
D. 130 N
8) A raft is constructed from wood that has a density of 400 kg/m3. Ten identical sections of wood (width, 25 cm; depth, 10 cm; length, 4 m) are tied together side by side to build the raft. The density of water is 1000 kg/m3. How much of the unloaded raft, in depth, is underwater?
A. 2 cm
B. 4 cm
C. 6 cm
D. 10 cm
9) A copper rod which is 5 centimeters in diameter and infinitely long carries a current of 2 amps. The current is distributed uniformly throughout the rod. The magnetic field half way between the axis of the rod and its outside edge is:
A. zero.
B. pointing radially inward.
C. pointing radially outward.
D. encircling the axis of the rod.
10) A ball is thrown up in the air with a velocity of 10 m/s. Which of the following is true?
A. Time in air is 1 s; maximum height is 5 m.
B. Time in air is 1 s; maximum height is 10 m.
C. Time in air is 2 s; maximum height is 5 m.
D. Time in air is 2 s; maximum height is 10 m.
11) An electron moving at constant velocity enters the area between two charged plates, as shown in the diagram below. Which of the paths correctly indicates the electron's trajectory after leaving the area between the charged plates?
12) A person stands on a scale in an elevator. He notices that the scale reading is lower than his usual weight. Which of the following could possibly describe the motion of the elevator?
A. It is moving down and slowing down.
B. It is moving down at constant speed.
C. It is moving up and slowing down.
D. It is moving up with constant speed.
13) The cruising altitude of most planes is about 40,000 feet above sea level where there is less air above the plane pushing down, so the air pressure is lower (about 20 kPa outside compared to about 100 kPa atmospheric pressure at sea level). To keep everyone comfortable inside the plane, the cabin is pressurized to about 75 kPa. The passenger doors on a 747 are about 1 meter wide by 2 meters tall.
If a metric ton is 1000 kg, to pull the door open at 40,000 feet is equivalent to moving how many metric tons?
A. 5 x 10-2 tons
B. 0.8 tons
C. 2 tons
D. 11 tons
14) A box sits on an inclined plane without sliding. As the angle of the plane (measured from the horizontal) increases, the normal force:
A. decreases nonlinearly.
B. increases nonlinearly.
C. increases linearly.
D. does not change.
15) What is the voltage drop across R3 in the circuit diagram below?
A. 15 V
B. 30 V
C. 40 V
D. 100 V
1. C
2. D
3. D
4. C
5. B
6. D
7. D
8. B
9. D
10. C
11. B
12. C
13. D
14. A
15. B
*Gold Standard GAMSAT textbook Reference
1) If you are adding these 2 vectors, let's consider the 2 extreme situations (i.e. the highest possible value and the lowest possible value): (1) they are pointing in exactly the same direction so the sum is 11 + 5 = 16 N; (2) they are pointing in exactly the opposite directions so the sum is 11 + (-5) = 6 N. Anything from 6 N to 16 N is possible but nothing above or below.
*PHY 1.1
2) Newton's 1st law states that an object in motion will stay in motion at constant velocity unless acted upon by an unbalanced force. For something to change direction or accelerate there must be an unbalanced force acting on it.
*PHY 4.2
3) If the direction changes, then all vectors change: momentum, displacement and velocity. Kinetic energy is a scalar and does not depend on direction. As long as the speed remains the same, the kinetic energy remains the same.
*PHY 1.1, 5.1, 5.2
4) Imagine that you needed to run down a hallway but you had 2 choices: (1) a very narrow hallway with a certain density of students there, just milling around OR (2) a wide hallway with the SAME density of students. Think for a moment. If you run down a wide hallway with the same density, you always have more options to choose from: left, right, sideways and so you can get through faster. It's the same with the electrons going through these 2 wires, same density but one with a greater radius.
Actually, a wire is like a tiny cylinder and if you cut a wire and look at its edge, the shape is a tiny circle. So it's not just a matter of the radius that is bigger, it's actually the area of a circle that is bigger. So we can summarize by saying: as the area increases, resistance decreases. Or, in “math-speak”, resistance is inversely proportional to the cross-sectional area.
Now we have our first basic high school equation that must be memorized!
The resistance (R) is inversely proportional to the cross-sectional area (A), where
Acircle = π(radius)2
If radius1 = 2 x radius2, then A1 = (2A2)2 = 4A2 and R1 = 1/4 R2, giving us a ratio of R1:R2 --> 4:1.
*PHY 10.2 (B: PHY 3.3)
5) You would be expected to know: Newton's Law of Gravitation F = Gm1m2/r2 (you would never perform a complete calculation on the GAMSAT using values for each of the unknowns in the equation because there is just not enough time! However, you must know the relationship so that you can make assessments as to the relative changes of the variables in the equation).
When the object is moved 1 Earth radius away from the Earth then r is now 2r (on the surface of the earth the object is 1 Earth radius away from the center of the Earth, when the object is moved 1 radius away the distance is now r + r = 2r).
When you plug (2r)2 into the denominator of the equation it becomes 4r2. Thus when you double the distance, you quarter the force (same relationship with Coulomb's Law and point charges). So if we start with a force of 60 N, then we end with 15 N. Not much to calculate!
*PHY 2.4
6) This is a common exam question testing your understanding of Newton's 3rd Law: for every action, there is an equal and opposite reaction. When you push a couch, the couch is pushing back against your body with an equal force. The mass in the question exerts a 10 N force back on the spring because action/reaction forces are equal.
*PHY 2.3
7) The tension in the ropes is largest when the ropes are vertical because at that moment, the centripetal force is maximum and in the same direction as gravity. Clearly, the speed of the child is maximum at the point of lowest (gravitational) potential energy; therefore, the centripetal force is also maximum at that point.
Centripetal force = mv2/r = (20 x 3 x 3)/3 = 60 N
Weight = 200 NMaximum force = 200 + 60 = 260 N
Maximum tension in each rope = 260/2 = 130 N*PHY 3.3
8) First let's calculate the specific gravity (SG) of wood which is simply the density of wood divided by the density of water which is 400/1000 = 0.4. SG is equal to the fraction of the height of a buoyant (= floating) object below the surface. So we know the fraction is 0.4 and now we need to apply that to the total “height” which is the depth of the wood 10 cm. So 0.4 x 10 = 4 cm.
*PHY 6.1.1, 6.1.2
9) The right-hand rule describes the direction of the magnetic field: as you "grab" the wire with your thumb in the direction of the flow of current, the tips of your fingers trace the direction of the magnetic field which encircles the wire. See PHY 9.2.3 of the Gold Standard GAMSAT book.
*PHY 9.2.2, 9.2.3
10) v = vo + gt
t = (v-vo)/g = 10/10 = 1 s
total time in air (to go up and back down) = 2t = 2 s
And, s = vot + (g/2)(t2)
= (10 x 1) - (10/2)(1)
= 5 m (max. height)
Yes, there are some very basic physics equations that you should know for the GAMSAT and we have listed them all here: www.gamsat-prep.com/section3-physics-formula-list/
*PHY 1.6, 2.6
11) While in the area between the plates, the negatively charged electron is attracted to the positive plate and so it would bend “upwards”. But after leaving the plates, there is no more net force acting on the electron. Therefore, the electron would continue in motion in a straight line according to Newton's first law (inertia).
Oh, btw, the real GAMSAT: only 4 answer choices for all multiple choice questions. We just had to make this question 20% better but we promise not to do it again!! : )
*PHY 9.1.3, 4.2
12) You can draw a free-body diagram of the person:
If upwards is positive, and we know that the sum of forces is equal to ma from Newton's 2nd Law, we get: Fscale - weight = ma The scale reading is less than his normal weight, so the acceleration is negative; i.e., down. Downward acceleration means either something speeds up while moving downward, or slows down while moving upward.
*PHY 2.2, 3.4
13) Key points: (1) pressure is defined as force per unit area (just apply some pressure to the table in front of you and you'll would have to say that you are applying a certain force over a particular area); (2) dimensional analysis: kilo means 1000 in the SI system and pascals or Pa, being force/area, has the equivalent units of N/m2.
The difference in pressure between inside and outside the cabin at 40,000 feet is: 75 – 20 = 55 kPa = 55 000 Pa or N/m2 of pressure pushing on the door.
Pressure: P = F/A
The area of the door is 2 x 1 = 2m2
So F = PA = (55 000)(2) = 110 000 N
Thus weight mg = 110 000 meaning m = 11 000 kg or 11 metric tons.
*PHY 6.1.2
14) Because no forces act perpendicular to the incline except for the normal force and the perpendicular component of weight, and there is no acceleration perpendicular to the incline, the normal force is equal to the perpendicular component of weight, which is mg cosθ. As the angle increases, the cosine of the angle decreases. This decrease is nonlinear because a graph of the normal force N vs. θ would show a curve, not a line.
*PHY 3.2, 3.2.1
15) The 10 A of current splits into two paths. The path through R2 takes 7 A, so the path through R3 takes 3 A (this is Kirchoff's First Law, the junctional theorem). Now use Ohm's law, where the voltage V is equal to the current I times the resistance R (Ohm's law is a frequent visitor in the real GAMSAT). V = IR V = (3 A) (10 Ω) = 30 V.
*PHY 10.1, 10.2, 10.3
Physics fact
10 Physics Questions to Ponder for a Millennium or Two
By GEORGE JOHNSON
ho of us would not be glad to lift the veil behind which the future lies hidden; to cast a glance at the next advances of our science and at the secrets of its development during future centuries?"One hundred years ago, with those inviting thoughts, the German mathematician David Hilbert opened his landmark address to the International Congress of Mathematicians in Paris, laying out 23 of the great unsolved problems of the day. "For the close of a great epoch," Hilbert declared, "not only invites us to look back into the past but also directs our thoughts to the unknown future."
With another century ending -- a whole millennium in fact -- the pressure is all the greater to tabulate human ignorance with lists of the most enticing cosmic mysteries.
In May, the Clay Mathematics Institute of Cambridge, Mass., emulated Hilbert, announcing (in Paris, for full effect) seven "Millennium Prize Problems," each with a bounty of $1 million.
The list is at: www.claymath.org/prize_problems/.
And last month physicists, with a typically lighter touch, ended a conference on superstring theory at the University of Michigan with a session called "Millennium Madness," choosing 10 of the most perplexing problems in their field. It was like a desert island game, involving some of science's smartest people.
"The way I thought about this challenge was to imagine what question I would ask if I woke up from a coma 100 years from now," said Dr. David Gross, a theoretical physicist at the University of California at Santa Barbara, as he unveiled the winners. He and the other judges made the selection, he noted, "in the middle and after this party in which we were sufficiently drunk."
After weeding out unanswerable questions (like "How do you get tenure?"), the judges came up with enough puzzles to occupy physicists for the next century or so. There are no monetary prizes, though solving any one of these would almost guarantee a trip to Stockholm.
1. Are all the (measurable) dimensionless parameters that characterize the physical universe calculable in principle or are some merely determined by historical or quantum mechanical accident and uncalculable? Einstein put it more crisply: did God have a choice in creating the universe? Imagine the Old One sitting at his control console, preparing to set off the Big Bang. "How fast should I set the speed of light?" "How much charge should I give this little speck called an electron?" "What value should I give to Planck's constant, the parameter that determines the size of the tiny packets -- the quanta -- in which energy shall be parceled?" Was he randomly dashing off numbers to meet a deadline? Or do the values have to be what they are because of a deep, hidden logic?
These kinds of questions come to a point with a conundrum involving a mysterious number called alpha. If you square the charge of the electron and then divide it by the speed of light times Planck's constant, all the dimensions (mass, time and distance) cancel out, yielding a so-called "pure number" -- alpha, which is just slightly over 1/137. But why is it not precisely 1/137 or some other value entirely? Physicists and even mystics have tried in vain to explain why.
2. How can quantum gravity help explain the origin of the universe? Two of the great theories of modern physics are the standard model, which uses quantum mechanics to describe the subatomic particles and the forces they obey, and general relativity, the theory of gravity. Physicists have long hoped that merging the two into a "theory of everything" -- quantum gravity -- would yield a deeper understanding of the universe, including how it spontaneously popped into existence with the Big Bang. The leading candidate for this merger is superstring theory, or M theory, as the latest, souped-up version is called (with the M standing for "magic," "mystery," or "mother of all theories").
3. What is the lifetime of the proton and how do we understand it? It used to be considered gospel that protons, unlike, say, neutrons, live forever, never decaying into smaller pieces. Then in the 1970's, theorists realized that their candidates for a grand unified theory, merging all the forces except gravity, implied that protons must be unstable. Wait long enough and, very occasionally, one should break down.
The trick is to catch it in the act. Sitting in underground laboratories, shielded from cosmic rays and other disturbances, experimenters have whiled away the years watching large tanks of water, waiting for a proton inside one of the atoms to give up the ghost. So far the fatality rate is zero, meaning that either protons are perfectly stable or their lifetime is enormous -- an estimated billion trillion trillion years or more.
4. Is nature supersymmetric, and if so, how is supersymmetry broken? Many physicists believe that unifying all the forces, including gravity, into a single theory would require showing that two very different kinds of particles are actually intimately related, a phenomenon called supersymmetry.
The first, fermions, are loosely described as the building blocks of matter, like protons, electrons and neutrons. They clump together to make stuff. The others, the bosons, are the particles that carry forces, like photons, conveyors of light. With supersymmetry, every fermion would have a boson twin, and vice versa.
Physicists, with their compulsion for coining funny names, call the so-called superpartners "sparticles": For the electron, there would be the selectron; for the photon, the photino. But since the sparticles have not been observed in nature, physicists would also have to explain why, in the jargon, the symmetry is "broken": the mathematical perfection that existed at the moment of creation was knocked out of kilter as the universe cooled and congealed into its present lopsided state.
5. Why does the universe appear to have one time and three space dimensions? "Just because" is not considered an acceptable answer. And just because people can't imagine moving in extra directions, beyond up-and-down, left-and-right, and back-and-forth, doesn't mean that the universe had to be designed that way. According to superstring theory, in fact, there must be six more spatial dimensions, each one curled up too tiny to detect. If the theory is right, then why did only three of them unfurl, leaving us with this comparatively claustrophobic dominion?
6. Why does the cosmological constant have the value that it has? Is it zero and is it really constant? Until recently cosmologists thought the universe was expanding at a steady clip. But recent observations indicate that the expansion may be getting faster and faster. This slight acceleration is described by a number called the cosmological constant. Whether the constant turns out to be zero, as earlier believed, or some very tiny number, physicists are at a loss to explain why.
According to some fundamental calculations, it should be huge -- some 10 to 122 times as big as has been observed.
The universe, in other words, should be ballooning in leaps and bounds. Since it is not, there must be some mechanism suppressing the effect. If the universe were perfectly supersymmetric, the cosmological constant would become canceled out entirely. But since the symmetry, if it exists at all, appears to be broken, the constant would still remain far too large. Things would get even more confusing if the constant turned out to vary over time.
7. What are the fundamental degrees of freedom of M-theory (the theory whose low-energy limit is eleven-dimensional supergravity and that subsumes the five consistent superstring theories) and does the theory describe nature? For years, one big strike against superstring theory was that there were five versions. Which, if any, described the universe? The rivals have been recently reconciled into an overarching 11-dimensional framework called M theory, but only by introducing complications.
Before M theory, all the subatomic particles were said to be made from tiny superstrings. M theory adds to the subatomic mix even weirder objects called "branes" -- like membranes but with as many as nine dimensions. The question now is, Which is more fundamental -- are strings made from branes or vice versa? Or is there something else even more basic that no one has thought of yet? Finally, is any of this real, or is M theory just a fascinating mind game?
8. What is the resolution of the black hole information paradox? According to quantum theory, information -- whether it describes the velocity of a particle or the precise manner in which ink marks or pixels are arranged on a document -- cannot disappear from the universe.
But the physicists Kip Thorne, John Preskill and Stephen Hawking have a standing bet: what would happen if you dropped a copy of the Encyclopaedia Britannica down a black hole? It does not matter whether there are other identical copies elsewhere in the cosmos. As defined in physics, information is not the same as meaning, but simply refers to the binary digits, or some other code, used to precisely describe an object or pattern. So it seems that the information in those particular books would be swallowed up and gone forever. And that is supposed to be impossible.
Dr. Hawking and Dr. Thorne believe the information would indeed disappear and that quantum mechanics will just have to deal with it. Dr. Preskill speculates that the information doesn't really vanish: it may be displayed somehow on the surface of the black hole, as on a cosmic movie screen.
9. What physics explains the enormous disparity between the gravitational scale and the typical mass scale of the elementary particles? In other words, why is gravity so much weaker than the other forces, like electromagnetism? A magnet can pick up a paper clip even though the gravity of the whole earth is pulling back on the other end.
According to one recent proposal, gravity is actually much stronger. It just seems weak because most of it is trapped in one of those extra dimensions. If its full force could be tapped using high-powered particle accelerators, it might be possible to create miniature black holes. Though seemingly of interest to the solid waste disposal industry, the black holes would probably evaporate almost as soon as they were formed.
10. Can we quantitatively understand quark and gluon confinement in quantum chromodynamics and the existence of a mass gap? Quantum chromodynamics, or QCD, is the theory describing the strong nuclear force. Carried by gluons, it binds quarks into particles like protons and neutrons. According to the theory, the tiny subparticles are permanently confined. You can't pull a quark or a gluon from a proton because the strong force gets stronger with distance and snaps them right back inside.
But physicists have yet to prove conclusively that quarks and gluons can never escape. When they try to do so, the calculations go haywire. And they cannot explain why all particles that feel the strong force must have at least a tiny amount of mass, why it cannot be zero. Some hope to find an answer in M theory, maybe one that would also throw more light on the nature of gravity.
11. (Question added in translation). Why is any of this important? In presenting his own list of mysteries, Hilbert put it this way: "It is by the solution of problems that the investigator tests the temper of his steel; he finds new methods and new outlooks, and gains a wider and freer horizon."
And in physics, the horizon is no less than a theory that finally makes sense of the universe.
Physics MCAT help ii
The 20 physics questions given below are both interesting and highly
challenging. You will likely have to take some time to work through
them. These questions go beyond the typical problems you can expect to
find in a physics textbook. Some of these physics questions make use of
different concepts, so (for the most part) there is no single formula or
set of equations that you can use to solve them. These questions make
use of concepts taught at the high school and college level (mostly
first year).
It is recommended that you persist through these physics questions, even if you get stuck. It's not a race, so you can work through them at your own pace. The result is that you will be rewarded with a greater understanding of physics. discuss your answers in the comments
Problem # 1
A crank drive mechanism is illustrated below. A uniform linkage BC of length L connects a flywheel of radius r (rotating about fixed point A) to a piston at C that slides back and forth in a hollow shaft. A variable torque T is applied to the flywheel such that it rotates at a constant angular velocity. Show that for one full rotation of the flywheel, energy is conserved for the entire system; consisting of flywheel, linkage, and piston (assuming no friction).
Note that gravity g is acting downwards, as shown.
Even though energy is conserved for the system, why is it a good idea to make the components of the drive mechanism as light as possible (with the exception of the flywheel)?
Problem # 2
An engine uses compression springs to open and close valves, using cams. Given a spring stiffness of 30,000 N/m, and a spring mass of 0.08 kg, what is the maximum engine speed to avoid “floating the valves”?
During the engine cycle the spring is compressed between 0.5 cm (valve fully closed) and 1.5 cm (valve fully open). Assume the camshaft rotates at the same speed as the engine.
Floating the valves occurs when the engine speed is high enough so that the spring begins to lose contact with the cam when the valve closes. In other words, the spring doesn’t extend quickly enough to maintain contact with the cam, when the valve closes.
For simplicity, you may assume that Hooke’s Law applies to the spring, where the force acting on the spring is proportional to its amount of compression (regardless of dynamic effects).
You may ignore gravity in the calculations.
Problem # 3
An object is traveling in a straight line. Its acceleration is given by
where C is a constant, n is a real number, and t is time.
Find the general equations for the position and velocity of the object as a function of time.
Problem # 4
In archery, when an arrow is released it can oscillate during flight. If we know the location of the center of mass of the arrow (G) and the shape of the arrow at an instant as it oscillates (shown below), we can determine the location of the nodes. The nodes are the “stationary” points on the arrow as it oscillates.
Using a geometric argument (no equations), determine the location of the nodes.
Assume that the arrow oscillates in the horizontal plane, so that no external forces act on the arrow in the plane of oscillation.
Problem # 5
A gyroscope wheel is spinning at a constant angular velocity ws while precessing about a vertical axis at a constant angular velocity wp. The distance from the pivot to the center of the front face of the spinning gyroscope wheel is L, and the radius of the wheel is r. The rod connecting the pivot to the wheel makes a constant angle θ with the vertical.
Determine the acceleration components normal to the wheel, at points A, B, C, D labeled as shown.
Problem # 6
When a vehicle makes a turn, the two front wheels trace out two arcs as shown in the figure below. The wheel facing towards the inside of the turn has a steering angle that is greater than that of the outer wheel. This is necessary to ensure that both front wheels smoothly trace out two arcs, which have the same center, otherwise the front wheels will skid on the ground during the turn.
During a turn, do the rear wheels necessarily trace out the same arcs as the front wheels? Based on your answer, what are the implications for making a turn close to the curb?
Problem # 7
A horizontal turntable at an industrial plant is continuously fed parts into a slot (shown on the left). It then drops these parts into a basket (shown on the right). The turntable rotates 180° between these two stages. The turntable briefly stops at each 1/8th of a turn in order to receive a new part into the slot on the left.
If the rotational speed of the turntable is w radians/second, and the outer radius of the turntable is R2, what must be the inner radius R1 so that the parts fall out of the slot and into the basket, as shown?
Assume:
• The angular speed w of the turntable can be treated as constant and continuous; which means you can ignore the brief stops the turntable makes at each 1/8th of a turn.
• The location of the basket is 180° from the feed location.
• The slots are very well lubricated so that there is no friction between the slot and part.
• The parts can be treated as particles, which means you can ignore their dimensions in the calculation.
• The slots are aligned with the radial direction of the turntable.
Problem # 8
A flywheel for a single piston engine rotates at an average speed of 1500 RPM. During half a rotation the flywheel has to absorb 1000 J of energy. If the maximum permissible speed fluctuation is ± 60 RPM, what is the minimum rotational inertia of the flywheel? Assume there is no friction.
Problem # 9
An aluminum extrusion process is simulated numerically with a computer. In this process, a punch pushes an aluminum billet of diameter D through a die of smaller diameter d. In the computer simulation, what is the maximum punch velocity Vp so that the net dynamic force (predicted by the simulation) acting on the aluminum during extrusion is at most 5% of the force due to deformation of the aluminum? Evaluate for a specific case where D = 0.10 m, d = 0.02 m, and the density of aluminum is ρ = 2700 kg/m3.
Hint:
The extrusion of the aluminum through the die is analogous to fluid flowing through a pipe which transitions from a larger diameter to a smaller diameter (e.g. water flowing through a fireman’s hose). The net dynamic force acting on the fluid is the net force required to accelerate the fluid, which occurs when the velocity of the fluid increases as it flows from the larger diameter section to the smaller diameter section (due to conservation of mass).
Problem # 10
A child on a horizontal merry-go-round gives an initial velocity Vrel to a ball. Find the initial direction and velocity Vrel of the ball relative to the merry-go-round so that, relative to the child, the ball goes around in a perfect circle as he’s sitting on the merry-go-round. Assume there is no friction between merry-go-round and ball.
The merry-go-round is rotating at a constant angular velocity of w radians/second, and the ball is released at a radius r from the center of the merry-go-round.
Problem # 11
A heavy pump casing with a mass m is to be lifted off the ground using a crane. For simplicity, the motion is assumed to be two-dimensional, and the pump casing is represented by a rectangle having side dimensions ab (see figure). A cable of length L1 is attached to the crane (at point P) and the pump casing (at point O). The crane pulls up vertically on the cable with a constant velocity Vp.
The center of mass G of the pump casing is assumed to lie in the center of the rectangle. It is located at a distance L2 from point O. The right side of the pump casing is located at a horizontal distance c from the vertical line passing though point P.
Find the maximum cable tension during the lift, which includes the part of the lift before the pump casing loses contact with the ground, and after the pump casing loses contact with the ground (lift off). In this stage the pump casing swings back and forth.
Evaluate for a specific case where:
a = 0.4 m
b = 0.6 m
c = 0.2 m
L1 = 3 m
m = 200 kg
IG = 9 kg-m2 (rotational inertia of pump casing about G)
Assume:
• The friction between the pump casing and ground is high enough so that the pump casing does not slide along the ground (towards the right), before lift off occurs.
• Before lift off occurs, dynamic effects are negligible.
• The velocity Vp is fast enough so that the bottom of the pump casing swings clear of the ground after lift off occurs.
• For purposes of approximating the cable tension, you can model the system as a regular pendulum during swinging (you can ignore double pendulum effects).
• The mass of the cable can be neglected.
Problem # 12
A linkage arrangement is shown below. The pin joints O1 and O2 are attached to a stationary base and are separated by a distance b. The linkages of identical color have the same length. All linkages are pin jointed and allow for rotation. Determine the path traced by the end point P as the blue linkage of length b rotates back and forth.
Why is this result interesting?
Problem # 13
A conveyor belt carrying aggregate is illustrated in the figure below. A motor turns the top roller at a constant speed, and the remaining rollers are allowed to spin freely. The belt is inclined at an angle θ. To keep the belt in tension a weight of mass m is suspended from the belt, as shown.
Find the point of maximum tension in the belt. You don’t have to calculate it, just find the location and give a reason for it.
Problem # 14
A quality test has determined that a pump impeller is too heavy on one side by an amount equal to 0.0045 kg-m. To correct this imbalance it is recommended to cut out a groove around the outer circumference of the impeller, using a milling machine, on the same side as the imbalance. This will remove material with the intent of correcting the imbalance. The dimension of the groove is 1 cm wide and 1 cm deep. The groove will be symmetric with respect to the heavy spot. How far around the outer circumference of the impeller should the groove be? Specify the answer in terms of θ. Hint: Treat the groove as a thin ring of material.
The outer radius of the impeller, at the location of the groove, is 15 cm.
The impeller material is steel, with a density of ρ = 7900 kg/m3.
Problem # 15
As part of a quality check, an axisymmetric container is placed over a very well lubricated fixed mandrel, as shown below. The container is then given an initial pure rotation w, with no initial translational motion. What do you expect to see if the center of mass of the container is offset from the geometric center O of the container?
Problem # 16
A stream of falling material hits the plate of an impact weigher and the horizontal force sensor allows the mass flow rate to be calculated from this. If the speed of the material just before it strikes the plate is equal to the speed of the material just after it strikes the plate, determine an equation for the mass flow rate of the material, based on the horizontal force readout on the sensor. Ignore friction with the plate.
Hint: This can be treated as a fluid flow problem.
Problem # 17
The SunCatcher is a Stirling engine that is powered by solar energy. It uses large parabolic mirrors to focus sunlight onto a central receiver, which powers a Stirling engine. In the parabolic mirror you can see the reflection of the landscape. Why is the reflection upside down?
Source: http://www.stirlingenergy.com
Problem # 18
On a cold, dry winter day your glasses fog up when you go indoors after being outside for a while. Why is that?
And if you go back outside with your glasses still fogged up, they quickly clear up. Why is that?
Problem # 19
In an astronaut training exercise, an airplane at high altitude travels along a circular arc in order to simulate weightlessness for its passengers. Explain how this is possible.
Problem # 20
A rope is wrapped around a pole of radius R = 3 cm. If the tension on one end of the rope is T = 1000 N, and the coefficient of static friction between the rope and pole is μ = 0.2, what is the minimum number of times the rope must be wrapped around the pole so that it doesn’t slip off?
Assume that the minimum number of times the rope must be wrapped around the pole corresponds to a tension of 1 N on the other end of the rope.
It is recommended that you persist through these physics questions, even if you get stuck. It's not a race, so you can work through them at your own pace. The result is that you will be rewarded with a greater understanding of physics. discuss your answers in the comments
Problem # 1
A crank drive mechanism is illustrated below. A uniform linkage BC of length L connects a flywheel of radius r (rotating about fixed point A) to a piston at C that slides back and forth in a hollow shaft. A variable torque T is applied to the flywheel such that it rotates at a constant angular velocity. Show that for one full rotation of the flywheel, energy is conserved for the entire system; consisting of flywheel, linkage, and piston (assuming no friction).
Note that gravity g is acting downwards, as shown.
Even though energy is conserved for the system, why is it a good idea to make the components of the drive mechanism as light as possible (with the exception of the flywheel)?
Problem # 2
An engine uses compression springs to open and close valves, using cams. Given a spring stiffness of 30,000 N/m, and a spring mass of 0.08 kg, what is the maximum engine speed to avoid “floating the valves”?
During the engine cycle the spring is compressed between 0.5 cm (valve fully closed) and 1.5 cm (valve fully open). Assume the camshaft rotates at the same speed as the engine.
Floating the valves occurs when the engine speed is high enough so that the spring begins to lose contact with the cam when the valve closes. In other words, the spring doesn’t extend quickly enough to maintain contact with the cam, when the valve closes.
For simplicity, you may assume that Hooke’s Law applies to the spring, where the force acting on the spring is proportional to its amount of compression (regardless of dynamic effects).
You may ignore gravity in the calculations.
Problem # 3
An object is traveling in a straight line. Its acceleration is given by
where C is a constant, n is a real number, and t is time.
Find the general equations for the position and velocity of the object as a function of time.
Problem # 4
In archery, when an arrow is released it can oscillate during flight. If we know the location of the center of mass of the arrow (G) and the shape of the arrow at an instant as it oscillates (shown below), we can determine the location of the nodes. The nodes are the “stationary” points on the arrow as it oscillates.
Using a geometric argument (no equations), determine the location of the nodes.
Assume that the arrow oscillates in the horizontal plane, so that no external forces act on the arrow in the plane of oscillation.
Problem # 5
A gyroscope wheel is spinning at a constant angular velocity ws while precessing about a vertical axis at a constant angular velocity wp. The distance from the pivot to the center of the front face of the spinning gyroscope wheel is L, and the radius of the wheel is r. The rod connecting the pivot to the wheel makes a constant angle θ with the vertical.
Determine the acceleration components normal to the wheel, at points A, B, C, D labeled as shown.
Problem # 6
When a vehicle makes a turn, the two front wheels trace out two arcs as shown in the figure below. The wheel facing towards the inside of the turn has a steering angle that is greater than that of the outer wheel. This is necessary to ensure that both front wheels smoothly trace out two arcs, which have the same center, otherwise the front wheels will skid on the ground during the turn.
During a turn, do the rear wheels necessarily trace out the same arcs as the front wheels? Based on your answer, what are the implications for making a turn close to the curb?
Problem # 7
A horizontal turntable at an industrial plant is continuously fed parts into a slot (shown on the left). It then drops these parts into a basket (shown on the right). The turntable rotates 180° between these two stages. The turntable briefly stops at each 1/8th of a turn in order to receive a new part into the slot on the left.
If the rotational speed of the turntable is w radians/second, and the outer radius of the turntable is R2, what must be the inner radius R1 so that the parts fall out of the slot and into the basket, as shown?
Assume:
• The angular speed w of the turntable can be treated as constant and continuous; which means you can ignore the brief stops the turntable makes at each 1/8th of a turn.
• The location of the basket is 180° from the feed location.
• The slots are very well lubricated so that there is no friction between the slot and part.
• The parts can be treated as particles, which means you can ignore their dimensions in the calculation.
• The slots are aligned with the radial direction of the turntable.
Problem # 8
A flywheel for a single piston engine rotates at an average speed of 1500 RPM. During half a rotation the flywheel has to absorb 1000 J of energy. If the maximum permissible speed fluctuation is ± 60 RPM, what is the minimum rotational inertia of the flywheel? Assume there is no friction.
Problem # 9
An aluminum extrusion process is simulated numerically with a computer. In this process, a punch pushes an aluminum billet of diameter D through a die of smaller diameter d. In the computer simulation, what is the maximum punch velocity Vp so that the net dynamic force (predicted by the simulation) acting on the aluminum during extrusion is at most 5% of the force due to deformation of the aluminum? Evaluate for a specific case where D = 0.10 m, d = 0.02 m, and the density of aluminum is ρ = 2700 kg/m3.
Hint:
The extrusion of the aluminum through the die is analogous to fluid flowing through a pipe which transitions from a larger diameter to a smaller diameter (e.g. water flowing through a fireman’s hose). The net dynamic force acting on the fluid is the net force required to accelerate the fluid, which occurs when the velocity of the fluid increases as it flows from the larger diameter section to the smaller diameter section (due to conservation of mass).
Problem # 10
A child on a horizontal merry-go-round gives an initial velocity Vrel to a ball. Find the initial direction and velocity Vrel of the ball relative to the merry-go-round so that, relative to the child, the ball goes around in a perfect circle as he’s sitting on the merry-go-round. Assume there is no friction between merry-go-round and ball.
The merry-go-round is rotating at a constant angular velocity of w radians/second, and the ball is released at a radius r from the center of the merry-go-round.
Problem # 11
A heavy pump casing with a mass m is to be lifted off the ground using a crane. For simplicity, the motion is assumed to be two-dimensional, and the pump casing is represented by a rectangle having side dimensions ab (see figure). A cable of length L1 is attached to the crane (at point P) and the pump casing (at point O). The crane pulls up vertically on the cable with a constant velocity Vp.
The center of mass G of the pump casing is assumed to lie in the center of the rectangle. It is located at a distance L2 from point O. The right side of the pump casing is located at a horizontal distance c from the vertical line passing though point P.
Find the maximum cable tension during the lift, which includes the part of the lift before the pump casing loses contact with the ground, and after the pump casing loses contact with the ground (lift off). In this stage the pump casing swings back and forth.
Evaluate for a specific case where:
a = 0.4 m
b = 0.6 m
c = 0.2 m
L1 = 3 m
m = 200 kg
IG = 9 kg-m2 (rotational inertia of pump casing about G)
Assume:
• The friction between the pump casing and ground is high enough so that the pump casing does not slide along the ground (towards the right), before lift off occurs.
• Before lift off occurs, dynamic effects are negligible.
• The velocity Vp is fast enough so that the bottom of the pump casing swings clear of the ground after lift off occurs.
• For purposes of approximating the cable tension, you can model the system as a regular pendulum during swinging (you can ignore double pendulum effects).
• The mass of the cable can be neglected.
Problem # 12
A linkage arrangement is shown below. The pin joints O1 and O2 are attached to a stationary base and are separated by a distance b. The linkages of identical color have the same length. All linkages are pin jointed and allow for rotation. Determine the path traced by the end point P as the blue linkage of length b rotates back and forth.
Why is this result interesting?
Problem # 13
A conveyor belt carrying aggregate is illustrated in the figure below. A motor turns the top roller at a constant speed, and the remaining rollers are allowed to spin freely. The belt is inclined at an angle θ. To keep the belt in tension a weight of mass m is suspended from the belt, as shown.
Find the point of maximum tension in the belt. You don’t have to calculate it, just find the location and give a reason for it.
Problem # 14
A quality test has determined that a pump impeller is too heavy on one side by an amount equal to 0.0045 kg-m. To correct this imbalance it is recommended to cut out a groove around the outer circumference of the impeller, using a milling machine, on the same side as the imbalance. This will remove material with the intent of correcting the imbalance. The dimension of the groove is 1 cm wide and 1 cm deep. The groove will be symmetric with respect to the heavy spot. How far around the outer circumference of the impeller should the groove be? Specify the answer in terms of θ. Hint: Treat the groove as a thin ring of material.
The outer radius of the impeller, at the location of the groove, is 15 cm.
The impeller material is steel, with a density of ρ = 7900 kg/m3.
Problem # 15
As part of a quality check, an axisymmetric container is placed over a very well lubricated fixed mandrel, as shown below. The container is then given an initial pure rotation w, with no initial translational motion. What do you expect to see if the center of mass of the container is offset from the geometric center O of the container?
Problem # 16
A stream of falling material hits the plate of an impact weigher and the horizontal force sensor allows the mass flow rate to be calculated from this. If the speed of the material just before it strikes the plate is equal to the speed of the material just after it strikes the plate, determine an equation for the mass flow rate of the material, based on the horizontal force readout on the sensor. Ignore friction with the plate.
Hint: This can be treated as a fluid flow problem.
Problem # 17
The SunCatcher is a Stirling engine that is powered by solar energy. It uses large parabolic mirrors to focus sunlight onto a central receiver, which powers a Stirling engine. In the parabolic mirror you can see the reflection of the landscape. Why is the reflection upside down?
Source: http://www.stirlingenergy.com
Problem # 18
On a cold, dry winter day your glasses fog up when you go indoors after being outside for a while. Why is that?
And if you go back outside with your glasses still fogged up, they quickly clear up. Why is that?
Problem # 19
In an astronaut training exercise, an airplane at high altitude travels along a circular arc in order to simulate weightlessness for its passengers. Explain how this is possible.
Problem # 20
A rope is wrapped around a pole of radius R = 3 cm. If the tension on one end of the rope is T = 1000 N, and the coefficient of static friction between the rope and pole is μ = 0.2, what is the minimum number of times the rope must be wrapped around the pole so that it doesn’t slip off?
Assume that the minimum number of times the rope must be wrapped around the pole corresponds to a tension of 1 N on the other end of the rope.
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