Physics 2211 A Quiz #3 Solutions Summer 2007 G ...
Physics 2211 A Summer 2007
Quiz #3
Solutions
G = 6.673 × 10−11 N·m2 /kg2 gEarth = 9.8 m/s2 Unless otherwise directed, all springs, cords, and pulleys are ideal, and drag should be neglected. I. (16 points) Astronaut Arlene stands at the equator of the (fictional) planet Planteen and measures her apparent weight to be 220 N. Arlene knows her own mass to be 82 kg, while Planteen has a mass of 3.0×1023 kg and a radius of 2.5 × 106 m. With what angular speed does Planteen rotate on its axis? . . . . . . . . . . . . . . . . . . . . . . .
This is a Newton’s Second Law problem with Uniform Circular Motion. Sketch a Free Body Diagram of Arlene. Choose the positive direction to be in the direction of the centripetal acceleration, toward the center of the planet. The only forces acting on her are the force of gravity, and a normal force. The normal force is her apparent weight. The force of gravity on Arlene will be calculated using the Law of Universal Gravitation—it can’t be calculated from w = mg because the acceleration of gravity on Planteen isn’t known. Letting mP be the mass of Planteen, and mA be the mass of Arlene, X
Fr = FG − n = mA ar
⇒
G
mP mA v2 − n = mA = mA rω 2 2 r r
Arlene moves in a circle whose radius is that of Planteen. Solving for the angular speed ω, r ω= s =
GmP n − r3 mA r (6.673 × 10−11 N·m2 /kg2 ) (3.0 × 1023 kg) (2.5 ×
= 4.6 × 10
−4
106
3
m)
−
220 N (82 kg) (2.5 × 106 m)
rad/s
Quiz #3 Solutions Page 1 of 5
II. (16 points) The upper block in the illustration, with mass m1 , is pulled horizontally by a rope with a tension of magnitude A. The lower block has mass m2 , and rests on a horizontal frictionless surface. The coefficient of kinetic friction between the two blocks is µk . What is the acceleration magnitude of the lower block, in terms of any or all of m1 , m2 , A, µk , and physical or mathematical constants? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . .
This is a Newton’s Second Law problem. Sketch a Free Body Diagram of each block. Choose a coordinate system for each block in which one axis points in the direction of that block’s acceleration. For the top block, the normal force and the weight act in the vertical direction. X
Fy = n1 − w1 = m1 ay = 0
⇒
n1 = m1 g
The applied force, the tension, and the force of kinetic friction act in the horizontal direction. X
F x = A − T − fk = m 1 a x
⇒
A − T − µk n1 = m1 ax
⇒
A − T − µk m1 g = m1 ax
The acceleration is the answer to the question, but the tension is unknown. For the bottom block, the tension and the force of kinetic friction act in the horizontal direction. Note that the tension has the same magnitude as the tension on the bottom block (ideal cord and pulley), and the force of kinetic friction is equal and opposite to the force of kinetic friction on the top block (Newton’s Third Law). X
F u = T − fk = m 2 a u
⇒
T = m2 au + µk m1 g
Note that, by wise choice of coordinate systems, ax = au . Let them both be a. Substituting this expression for T into the horizontal equation for the top block, A − (m2 a + µk m1 g) − µk m1 g = m1 a and solving for the acceleration A − µk m1 g − µk m1 g = m1 a + m2 a
⇒
A − 2µk m1 g = (m1 + m2 ) a
⇒
a=
A − 2µk m1 g m1 + m2
1. (6 points) In the problem above, how does the frictional force on the bottom block, f~2 , compare to that on the top block, f~1 ? . . . . . . . . . . . . . . . . . . . . . . . These two forces constitute a force pair. From Newton’s Third Law f~2 is in the opposite direction to f~1 , and f2 = f1 .
Quiz #3 Solutions Page 2 of 5
III. (11 points) In an amusement park ride called “The Roundup”, passengers stand inside a rotating ring of radius R. After the ring has acquired sufficient speed, it tilts into a vertical plane, as illustrated. At the very bottom of the ring, a person of mass m finds that their apparent weight is three times their actual weight. What is the tangential speed vt of a person in the ring, in terms of any or all of m, R, and physical or mathematical constants? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . .
This is a Newton’s Second Law problem with Uniform Circular Motion. Sketch a Free Body Diagram of the person at the bottom. Choose the positive direction to be in the direction of the centripetal acceleration, upward toward the center of the ride. The only forces acting on the person are the force of gravity (or weight), and a normal force. The normal force is the apparent weight. X
Fr = n − w = mar
⇒
3mg − mg = m
v2 R
⇒
v=
p 2gR
2. (11 points) In the problem above, what is the magnitude of the total force the ring exerts on the person, when they are in the side position, halfway up, indicated with an asterisk? Again, express your answer in terms of any or all of m, R, and physical or mathematical constants. Hint: Don’t forget friction! . . . . . . . . . . . . . . . . . . . . . . . This is still a Newton’s Second Law problem with Uniform Circular Motion. Sketch a Free Body Diagram of the person at the side. Choose the positive direction to be in the direction of the centripetal acceleration, sideways toward the center of the ride. The forces acting on the person are the force of gravity (or weight), a normal force, and friction which keeps them from sliding down the ring. In the vertical, or tangential direction, X Ft = f − w = mat = 0
⇒
f = mg
Toward the center, and substituting the tangential speed found above X
v2 Fr = n = mar = m = m R
¡√
2gR R
¢2 = 2mg
The normal and frictional forces both are from the ring on the person. They are necessarily at right angles, so the total force magnitude of the ring on the person can be found from the Pythagorean Theorem. Fring =
p
q f 2 + n2 =
√ 2 2 (mg) + (2mg) = mg 5
Quiz #3 Solutions Page 3 of 5
3. (10 points) It is the beginning of the dog sled race, and the illustrated dog accelerates toward the right as it leaves the starting line over level ground, pulling two sleds of equal mass. The runners of one sled have been waxed to reduce the coefficient of kinetic friction between that sled and the ground. How is the tension in rope 2 related to the tension in rope 1? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . Rope 2 must pull the mass of both sleds, while rope 1 must only pull the rear sled. The waxing of the runners is irrelevant. Tension in rope 2 is greater than tension in rope 1, regardless of which sled has waxed runners.
4. (10 points) A horizontal turntable rotates about a vertical axis with constant angular speed. A coin lying on the turntable rotates with it, without sliding. At the instant illustrated, which diagram correctly shows the forces acting on the coin? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . .
The forces acting on the coin are its weight downward, a normal force perpendicular to the surface of the turntable (upward) and a frictional force toward the center (there must be a force toward the center if the coin is to move in a circle). Diagram i shows these forces.
Quiz #3 Solutions Page 4 of 5
5. (10 points) Two satellites, A and B, of the same mass are going around the Earth in concentric orbits. The distance of satellite B from the Earth’s center is four times that of satellite A. What is the ratio of the tangential speed of B to that of A? vB /vA = . . . . . . . . . . . . . . . . . . . . . . . . . . These satellites are in circular orbits. That is, they have uniform circular motion, and the radial force is the force of gravity. From Newton’s Second Law X
Fr = Fg = mar
Mm v2 G 2 =m r r
⇒
⇒
M G = v2 r
r ⇒
v=
GM r
where M is the mass of the Earth, and m is the mass of each satellite. Let r = R for satellite A. Then r = 4R for satellite B. So r vA =
r
GM R
and
vB =
1 GM = 4R 2
r
GM = 12 vA R
⇒
vB /vA = 1/2
6. (10 points) Rank the gravitational acceleration at the surface of the four illustrated planets, from greatest to least. Planet i has mass M and radius R. . . . . . . . . . . . . . . . . . . . . . . . Weight is the force of gravity. FG = G
Mm = mg r2
⇒
g=G
M r2
For planet i this is GM/R2 , for planet ii this is 2GM/R2 , for iii it’s 14 GM/R2 , and for planet iv it is 12 GM/R2 . So ii > i > iv > iii
Quiz #3 Solutions Page 5 of 5
Quiz #3
Solutions
G = 6.673 × 10−11 N·m2 /kg2 gEarth = 9.8 m/s2 Unless otherwise directed, all springs, cords, and pulleys are ideal, and drag should be neglected. I. (16 points) Astronaut Arlene stands at the equator of the (fictional) planet Planteen and measures her apparent weight to be 220 N. Arlene knows her own mass to be 82 kg, while Planteen has a mass of 3.0×1023 kg and a radius of 2.5 × 106 m. With what angular speed does Planteen rotate on its axis? . . . . . . . . . . . . . . . . . . . . . . .
This is a Newton’s Second Law problem with Uniform Circular Motion. Sketch a Free Body Diagram of Arlene. Choose the positive direction to be in the direction of the centripetal acceleration, toward the center of the planet. The only forces acting on her are the force of gravity, and a normal force. The normal force is her apparent weight. The force of gravity on Arlene will be calculated using the Law of Universal Gravitation—it can’t be calculated from w = mg because the acceleration of gravity on Planteen isn’t known. Letting mP be the mass of Planteen, and mA be the mass of Arlene, X
Fr = FG − n = mA ar
⇒
G
mP mA v2 − n = mA = mA rω 2 2 r r
Arlene moves in a circle whose radius is that of Planteen. Solving for the angular speed ω, r ω= s =
GmP n − r3 mA r (6.673 × 10−11 N·m2 /kg2 ) (3.0 × 1023 kg) (2.5 ×
= 4.6 × 10
−4
106
3
m)
−
220 N (82 kg) (2.5 × 106 m)
rad/s
Quiz #3 Solutions Page 1 of 5
II. (16 points) The upper block in the illustration, with mass m1 , is pulled horizontally by a rope with a tension of magnitude A. The lower block has mass m2 , and rests on a horizontal frictionless surface. The coefficient of kinetic friction between the two blocks is µk . What is the acceleration magnitude of the lower block, in terms of any or all of m1 , m2 , A, µk , and physical or mathematical constants? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . .
This is a Newton’s Second Law problem. Sketch a Free Body Diagram of each block. Choose a coordinate system for each block in which one axis points in the direction of that block’s acceleration. For the top block, the normal force and the weight act in the vertical direction. X
Fy = n1 − w1 = m1 ay = 0
⇒
n1 = m1 g
The applied force, the tension, and the force of kinetic friction act in the horizontal direction. X
F x = A − T − fk = m 1 a x
⇒
A − T − µk n1 = m1 ax
⇒
A − T − µk m1 g = m1 ax
The acceleration is the answer to the question, but the tension is unknown. For the bottom block, the tension and the force of kinetic friction act in the horizontal direction. Note that the tension has the same magnitude as the tension on the bottom block (ideal cord and pulley), and the force of kinetic friction is equal and opposite to the force of kinetic friction on the top block (Newton’s Third Law). X
F u = T − fk = m 2 a u
⇒
T = m2 au + µk m1 g
Note that, by wise choice of coordinate systems, ax = au . Let them both be a. Substituting this expression for T into the horizontal equation for the top block, A − (m2 a + µk m1 g) − µk m1 g = m1 a and solving for the acceleration A − µk m1 g − µk m1 g = m1 a + m2 a
⇒
A − 2µk m1 g = (m1 + m2 ) a
⇒
a=
A − 2µk m1 g m1 + m2
1. (6 points) In the problem above, how does the frictional force on the bottom block, f~2 , compare to that on the top block, f~1 ? . . . . . . . . . . . . . . . . . . . . . . . These two forces constitute a force pair. From Newton’s Third Law f~2 is in the opposite direction to f~1 , and f2 = f1 .
Quiz #3 Solutions Page 2 of 5
III. (11 points) In an amusement park ride called “The Roundup”, passengers stand inside a rotating ring of radius R. After the ring has acquired sufficient speed, it tilts into a vertical plane, as illustrated. At the very bottom of the ring, a person of mass m finds that their apparent weight is three times their actual weight. What is the tangential speed vt of a person in the ring, in terms of any or all of m, R, and physical or mathematical constants? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . .
This is a Newton’s Second Law problem with Uniform Circular Motion. Sketch a Free Body Diagram of the person at the bottom. Choose the positive direction to be in the direction of the centripetal acceleration, upward toward the center of the ride. The only forces acting on the person are the force of gravity (or weight), and a normal force. The normal force is the apparent weight. X
Fr = n − w = mar
⇒
3mg − mg = m
v2 R
⇒
v=
p 2gR
2. (11 points) In the problem above, what is the magnitude of the total force the ring exerts on the person, when they are in the side position, halfway up, indicated with an asterisk? Again, express your answer in terms of any or all of m, R, and physical or mathematical constants. Hint: Don’t forget friction! . . . . . . . . . . . . . . . . . . . . . . . This is still a Newton’s Second Law problem with Uniform Circular Motion. Sketch a Free Body Diagram of the person at the side. Choose the positive direction to be in the direction of the centripetal acceleration, sideways toward the center of the ride. The forces acting on the person are the force of gravity (or weight), a normal force, and friction which keeps them from sliding down the ring. In the vertical, or tangential direction, X Ft = f − w = mat = 0
⇒
f = mg
Toward the center, and substituting the tangential speed found above X
v2 Fr = n = mar = m = m R
¡√
2gR R
¢2 = 2mg
The normal and frictional forces both are from the ring on the person. They are necessarily at right angles, so the total force magnitude of the ring on the person can be found from the Pythagorean Theorem. Fring =
p
q f 2 + n2 =
√ 2 2 (mg) + (2mg) = mg 5
Quiz #3 Solutions Page 3 of 5
3. (10 points) It is the beginning of the dog sled race, and the illustrated dog accelerates toward the right as it leaves the starting line over level ground, pulling two sleds of equal mass. The runners of one sled have been waxed to reduce the coefficient of kinetic friction between that sled and the ground. How is the tension in rope 2 related to the tension in rope 1? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . . Rope 2 must pull the mass of both sleds, while rope 1 must only pull the rear sled. The waxing of the runners is irrelevant. Tension in rope 2 is greater than tension in rope 1, regardless of which sled has waxed runners.
4. (10 points) A horizontal turntable rotates about a vertical axis with constant angular speed. A coin lying on the turntable rotates with it, without sliding. At the instant illustrated, which diagram correctly shows the forces acting on the coin? (On Earth.) . . . . . . . . . . . . . . . . . . . . . . .
The forces acting on the coin are its weight downward, a normal force perpendicular to the surface of the turntable (upward) and a frictional force toward the center (there must be a force toward the center if the coin is to move in a circle). Diagram i shows these forces.
Quiz #3 Solutions Page 4 of 5
5. (10 points) Two satellites, A and B, of the same mass are going around the Earth in concentric orbits. The distance of satellite B from the Earth’s center is four times that of satellite A. What is the ratio of the tangential speed of B to that of A? vB /vA = . . . . . . . . . . . . . . . . . . . . . . . . . . These satellites are in circular orbits. That is, they have uniform circular motion, and the radial force is the force of gravity. From Newton’s Second Law X
Fr = Fg = mar
Mm v2 G 2 =m r r
⇒
⇒
M G = v2 r
r ⇒
v=
GM r
where M is the mass of the Earth, and m is the mass of each satellite. Let r = R for satellite A. Then r = 4R for satellite B. So r vA =
r
GM R
and
vB =
1 GM = 4R 2
r
GM = 12 vA R
⇒
vB /vA = 1/2
6. (10 points) Rank the gravitational acceleration at the surface of the four illustrated planets, from greatest to least. Planet i has mass M and radius R. . . . . . . . . . . . . . . . . . . . . . . . Weight is the force of gravity. FG = G
Mm = mg r2
⇒
g=G
M r2
For planet i this is GM/R2 , for planet ii this is 2GM/R2 , for iii it’s 14 GM/R2 , and for planet iv it is 12 GM/R2 . So ii > i > iv > iii
Quiz #3 Solutions Page 5 of 5
