Conflict Exam 1 Solutions
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics
Conflict Exam 1 Solutions Problem 1: (25 points) Concept Questions: Circle the correct answer for the following five questions. A. (5 points): Suppose a spring is attached at one end to a cart of mass m that lies on a frictionless plane. The other end of the spring is attached to a force sensor. If you pull the cart so that the spring stretches, the direction of the force on the cart due to the interaction between the spring and the cart
a) points towards the force sensor. b) points towards the motion sensor. c) does not point in any direction because the cart exerts an equal an opposite force on the spring. d) points in some other direction. Solution (a). When the spring is extended the spring exerts a force on the cart that pulls the cart towards the force sensor.
1
B. (5 points): B. (5 points): During a race on a straight track, a person moves with constant acceleration, then continues at constant velocity, then experiences at a constant deceleration until coming to a stop. Which of the following graphs best represents the position as a function of time?
(1)
(2)
(3)
(4)
(5)
None of the above. (6)
Solution: (5).
2
C. (5 points): A skier of mass M slides down a ramp shaped as a circle of radius R . At the end point of the ramp just before the skier is in the air, the magnitude of the normal force exerted by the ramp on the skier is N . The gravitational constant is g .
1) The magnitude of the normal force N greater than Mg . 2) The magnitude of the normal force N equal to Mg . 3) The magnitude of the normal force N less than Mg . 4) The magnitude of the normal force N can be greater than, equal to, or less than Mg depending on the speed of the skier. Solution (1): The forces on the skier on shown in the figure below. Since the skier is still undergoing circular motion at the instant just before leaving the track, the skier is accelerating upwards. If the skier has speed v at that instant then applying Newton’s Second Law to the skier yields N − Mg = Mv 2 / R
(1.1)
We can solve this for the magnitude of the normal force N = Mg + Mv 2 / R . (1.2) Therefore the magnitude of the normal force is greater than the magnitude of the gravitational force at the instant just before the skier leaves the track.
3
D. (5 points): A puck of mass m is moving in a circle at constant speed on a frictionless table as shown above. The puck is connected by a string to a suspended bob, also of mass m , which is at rest below the table. Half of the length of the string is above the tabletop and half below. What is the magnitude of the centripetal acceleration of the moving puck? Let g be the gravitational constant.
a) The magnitude of the centripetal acceleration of the moving puck is less than g . b) The magnitude of the centripetal acceleration of the moving puck is equal to g . c) The magnitude of the centripetal acceleration of the moving puck is greater than g . d) The magnitude of the centripetal acceleration of the moving puck is zero. e) There is not enough information given to determine how the magnitude of the centripetal acceleration of the moving puck compares to g . Solution (b): Since the suspended object is not accelerating the tension force upwards exactly balances the gravitational force downwards, so T = mg . The inward force on the puck is due to the tension in the string which is equal in to the mass times the centripetal acceleration, T = macentripetal . Therefore mg = macentripetal or
g = acentripetal .
4
E. (5 points): A bead is given a small push at the top of a hoop (position A) and is constrained to slide around a frictionless circular wire (in a vertical plane). Circle the arrow that best describes the direction of the acceleration when the bead is at the position B.
Solution (4): The bead is speeding up at position B therefore it has a tangential component of the acceleration (pointing downward) and it is traveling in a circular trajectory so it has a radial component of the acceleration pointing towards the center of the circle. Direction 4 best describes the sum of these two components.
5
Problem 2 (25 points): A person is riding a bicycle at a constant speed v0 (which is unknown) along a straight road and then gradually applies the brakes during a time interval 0 < t < t f until the bicycle comes to a stop. Assume that the magnitude of the braking force increases linearly in time according to
F = bt ,
0 < t < tf
where b is a constant. The rider and bicycle have a total mass m . At the instant the person applies the brakes, an iPod drops from a tree branch from a height h above the ground. The rider catches the iPod at the instant she has come to a stop.. You may assume that she catches it at a height s above the ground. The goal of this problem is to determine how far the cyclist traveled during the time interval that the brakes were applied. Express your answer in terms of the variables, b , m , h , s , and g but not v0 .
a) Write a brief description of the strategy that you plan to use and any diagrams or graphs that you have chosen for solving this problem. Make sure you clearly state which concepts you plan to use to calculate any relevant physical quantities. Solution: There are two objects moving, the bicyclist and the iPod. We shall choose a coordinate system for the bicyclist and the iPod as shown in the figure below.
6
We can use Newton’s Second Law to calculate the acceleration of the bicyclist. We can then integrate the acceleration to find the x-component of the velocity of the bicyclist. We then use the equations for free fall to determine the time it takes the iPod to fall a distance h − s . The x-component of the velocity of the bicyclist is zero at this instant so we can solve for the initial x-component of the velocity of the bicyclist. We can now integrate the x-component of the velocity of the bicyclist to find the position and with substitution of the time of travel, we can find the distance traveled. b) How far did the cyclist travel during the time interval that the brakes were applied? Express your answer in terms of the variables, b , m , h , s , and g but not v0 . You may or may not need all of these quantities. G G From Newton’s Second Law, F = ma , are in the x-direction:
ˆi : −bt = ma . x Thus the x-component of the acceleration is
ax = −
b t m
(1.3)
(1.4)
We can now integrate the x-component of the acceleration of the bicyclist to get the ycomponent of the velocity of the bicyclist
b 2 ⎛ b ⎞ vx (t ) − vx 0 = ∫ ax dt = ∫ ⎜ − t ⎟ dt = − t . m ⎠ 2m 0 0⎝ t
t
(1.5)
The y-component of the position of the iPod is given by y (t ) = (h − s ) −
g 2 t 2
(1.6)
The bicyclist catches the iPod at the instant when
7
y (t = t f ) = 0 = (h − s ) −
g 2 tf 2
(1.7)
We can solve eq. (1.7) for the flight time: 2(h − s ) g
tf =
(1.8)
The bicyclist comes to a full stop at this instant so we can substitute Eq. (1.8) into Eq. (1.5) and solve for the initial x-component of the velocity of the bicyclist vx 0 =
b 2 b(h − s ) tf = 2m mg
(1.9)
We can now integrate Eq. (1.5) using the value for vx 0 that we found in Eq. (1.9) to find that the x-component of the position the bicyclist when it just came to a stop tf
t
f ⎛ b( h − s ) b 2 ⎞ b 2⎞ b( h − s ) b 3 ⎛ − x(t f ) − x0 = ∫ vx dt = ∫ ⎜ vx 0 − t ⎟ dt = ∫ ⎜ t ⎟ dt = tf − t f .(1.10) mg mg 2m ⎠ 2m ⎠ 6m 0 0⎝ 0⎝
t
We chose x0 = 0 and so substituting x0 = 0 and Eq. (1.8) into Eq. (1.10) yields b ⎛ 2(h − s ) ⎞ x(t f ) = ⎜ ⎟⎟ 3m ⎜⎝ g ⎠
3
(1.11)
8
Problem 3 (25 points):
An object with mass m slides down a roof a distance d that is inclined at an angle φ with respect to the horizontal. The contact surface between the roof and object has a coefficient of kinetic friction μ . The edge of the bottom of the roof is at a height h above the ground. What is the horizontal distance from the edge of the roof to the point where the object hits the ground?
Include with your answer a plan for solving the problem. Make sure you clearly state which concepts you plan to use to calculate any relevant physical quantities. Also clearly state any assumptions you make. Be sure you include any diagrams or sketches that you plan to use. Solution: It is clear that the distance the object travels will depend on how fast it is moving when it leaves the roof, and that this speed will be larger for larger values of d and φ .
Consider then the two stages of the motion, the first when the object is sliding down the roof and the second where it is in free fall. We shall use Newton’s second Law to find the acceleration for the sliding down the roof and then use the kinematic equations for position and velocity along the roof to find the speed of the object just when it reaches the end of the roof. We can then use that speed as an initial condition for the projectile motion trajectory. We can then solve for the time that it takes to reach the ground and hence solve for the horizontal distance the object traveled from edge of the roof. For the first stage, choose a coordinate system with the positive ˆi -direction down the roof (the steepest downward direction) and the positive ˆj -direction to be perpendicular to the roof (with positive upward component).
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The forces on the object are gravity
(
G mg = mg ˆi cos φ − ˆj sin φ
and the contact force
)
G G G C = N + fk = N ˆj − f k ˆi
(1.12)
(1.13)
G G . The components of the vectors in Newton’s Second Law, F = ma , are
mg sin φ − f k = m ax N − mg cos φ = m a y .
(1.14) (1.15)
The condition that the object remains on the roof is expressed as a y = 0 and the model for kinetic friction is
f k = μk N .
(1.16)
The condition a y = 0 in Eq. (1.15) gives
N = mg cos φ ,
(1.17)
f k = μk mg cos φ .
(1.18)
and so Eq. (1.16) becomes
Using this in Eq. (1.14) allows solution for the x -component of acceleration ax = g ( sin φ − μk cos φ ) .
(1.19)
Let the time that the object takes to slide down the roof be troof and the speed of the object just as it leaves the roof be v0 . Then, we have that d=
1 1 ( ax troof ) 2 ax troof = ax 2 2
2
(1.20)
We note that v0 = ax troof .
(1.21)
So Eq. (1.20) becomes
10
v0 2 d= 2a x
(1.22)
We can now solve Eq. (1.22) for the speed of the object the instant it leaves the roof v0 = 2 ax d = 2 g d ( sin φ − μk cos φ ) .
(1.23)
For the second stage, the object is in free fall. For this stage, it should be clear that the coordinate system used for the first stage will cause great difficulty. So, take the positive ˆi -direction to be horizontal and the positive ˆj -direction to be vertically upward, so that the ˆi - ˆj plane contains the object’s motion.
We then have ax = 0 , ay = − g .
(1.24) (1.25)
Additionally, if we now “reset our clock” so that t = 0 when the object leaves the roof, vx ,0 = v0 cos φ , (1.26) v y ,0 = −v0 sin φ .
(1.27)
If we further take the origin to be the point on the ground directly below the point where the object leaves the roof, x0 = 0 , y0 = h . The equations describing the object’s motion as a function of time t are then
1 x(t ) = x0 + vx ,0 t + ax t 2 = v0 cos φ t , 2 1 1 y (t ) = y0 + v y ,0 t + a y t 2 = h − v0 sin φ t − gt 2 . 2 2
(1.28) (1.29)
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When the object hits the ground, y = 0 , the x -coordinate may be denoted x f and the time when the object hits may be denoted t f . Eq. (1.29) then becomes 0 = h − v0 sin φ t f −
1 2 gt f 2
(1.30)
This is a quadratic equation in t f , and the quadratic formula gives tf =
(
1 −v0 sin φ ± v02 sin 2 φ + 2 gh g
)
(1.31)
and substitution of the positive solutions into Eq. (1.28) gives xf =
(
v0 cos φ −v0 sin φ + v02 sin 2 φ + 2 gh g
⎛ v02 2 gh ⎞ = cos φ ⎜ − sin φ + sin 2 φ + 2 ⎟ ⎜ g v0 ⎟⎠ ⎝
) .
(1.32)
We can rewrite Eq. (1.22) as v02 / g = d ( sin φ − μk cos φ )
(1.33)
Then substitute Eq. (1.33) into Eq. (1.32) which yields ⎛ ⎞ 2h x f = d ( sin φ − μk cos φ ) cos φ ⎜ − sin φ + sin 2 φ + ⎟ ⎜ d ( sin φ − μk cos φ ) ⎟⎠ ⎝
Note that the gravitational constant g does not appear in this expression. simplification is of questionable value.
(1.34)
Further
12
Problem 4: (25 points)
A coin of mass m (which you may treat as a point object) lies on a turntable, exactly at the rim, a distance R from the center. The turntable turns at constant angular speed ω and the coin rides without slipping. Suppose the coefficient of static friction between the turntable and the coin is given by μ . Let g be the gravitational constant.
a) What is the maximum angular speed ω max such that the coin does not slip? Explain your plan for solving this problem. Include all graphs or diagrams that you intend to use. Solution: The coin undergoes circular motion at constant speed so it is accelerating inward. The force inward is static friction and at the just slipping point it has reached its maximum value. We can use Newton’s Second Law to find the maximum angular speed ω max . Our coordinate system and free body force diagram is shown in the figure below. We choose the inward direction as positive.
The contact force is given by
G G G C = N + fs = N ˆj + fs ˆi .
The gravitational force is given by
G Fdisc = − mgˆj .
(1.35)
(1.36)
Newton’s Second Law, Fx = max , in the x-direction (inward direction), noting that the centripetal acceleration has magnitude ax = Rω 2 , is given by fs = m Rω 2 .
(1.37)
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Newton’s Second Law, Fy = ma y , in the y-direction (inward direction), noting that the disc is static hence a y = 0 , is given by
N − mg = 0 .
(1.38)
N = mg .
(1.39)
Thus the normal force is
As ω increases, the static friction increases in magnitude until at ω = ωmax and static friction reaches its maximum value (noting Eq. (1.39)). ( f s ) max = μ N = μ mg .
(1.40)
At this value the disc slips. Thus substituting this value for the maximum static friction into Eq. (1.37) yields 2 μ mg = mRωmax . (1.41) We can now solve Eq. (1.41) for maximum angular speed ω max such that the coin does not slip
ωmax =
μg R
.
(1.42)
Suppose an identical second coin is now stacked on top of the first coin and the turntable turns at constant angular speed ω such that the coins ride without slipping. The coefficient of static friction between the two coins is also μ .
b) What is the magnitude of the radial force exerted by the turntable on the bottom coin? Does this force point inward or outward? Explain your plan for solving this problem. Include all graphs or diagrams that you intend to use.
Solution: We will solve this problem in a similar fashion with a coordinate system and free body diagram on each coin shown in the figures below.
14
We will now apply Newton’s Second Law to each coin and determine the magnitude of the radial force exerted by the turntable on the bottom coin, ( fs )bG . The key point is that the static friction between the coins form an action-reaction pair, (neither coin is slipping). So the static friction that makes the top coin accelerate inward also acts on the bottom coin in the opposite direction (radially outward or the negative x-direction). Newton’s Second Law on the bottom coin, Fx = max , in the x-direction (inward direction), noting that the centripetal acceleration has magnitude ax = Rω 2 , is given by ( f s )bG − ( fs )bt = m Rω 2 .
(1.43)
Newton’s Second Law on the top coin, Fx = max , in the x-direction (inward direction), noting that the centripetal acceleration has magnitude ax = Rω 2 , is given by ( f s )tb = m Rω 2 .
(1.44)
( f s )bt = ( f s )tb .
(1.45)
Newton’s Third Law, requires that
So substituting Eq. (1.44) into (1.44) yields ( fs )bt = mRω 2 .
(1.46)
Substituting Eq. (1.46) into Eq. (1.43) then yields ( f s )bG − m Rω 2 = m Rω 2 .
(1.47)
Hence the magnitude of the radial inward force exerted on the bottom coin due to the turntable is then ( f s )bG = 2m Rω 2
(1.48)
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Conflict Exam 1 Solutions Problem 1: (25 points) Concept Questions: Circle the correct answer for the following five questions. A. (5 points): Suppose a spring is attached at one end to a cart of mass m that lies on a frictionless plane. The other end of the spring is attached to a force sensor. If you pull the cart so that the spring stretches, the direction of the force on the cart due to the interaction between the spring and the cart
a) points towards the force sensor. b) points towards the motion sensor. c) does not point in any direction because the cart exerts an equal an opposite force on the spring. d) points in some other direction. Solution (a). When the spring is extended the spring exerts a force on the cart that pulls the cart towards the force sensor.
1
B. (5 points): B. (5 points): During a race on a straight track, a person moves with constant acceleration, then continues at constant velocity, then experiences at a constant deceleration until coming to a stop. Which of the following graphs best represents the position as a function of time?
(1)
(2)
(3)
(4)
(5)
None of the above. (6)
Solution: (5).
2
C. (5 points): A skier of mass M slides down a ramp shaped as a circle of radius R . At the end point of the ramp just before the skier is in the air, the magnitude of the normal force exerted by the ramp on the skier is N . The gravitational constant is g .
1) The magnitude of the normal force N greater than Mg . 2) The magnitude of the normal force N equal to Mg . 3) The magnitude of the normal force N less than Mg . 4) The magnitude of the normal force N can be greater than, equal to, or less than Mg depending on the speed of the skier. Solution (1): The forces on the skier on shown in the figure below. Since the skier is still undergoing circular motion at the instant just before leaving the track, the skier is accelerating upwards. If the skier has speed v at that instant then applying Newton’s Second Law to the skier yields N − Mg = Mv 2 / R
(1.1)
We can solve this for the magnitude of the normal force N = Mg + Mv 2 / R . (1.2) Therefore the magnitude of the normal force is greater than the magnitude of the gravitational force at the instant just before the skier leaves the track.
3
D. (5 points): A puck of mass m is moving in a circle at constant speed on a frictionless table as shown above. The puck is connected by a string to a suspended bob, also of mass m , which is at rest below the table. Half of the length of the string is above the tabletop and half below. What is the magnitude of the centripetal acceleration of the moving puck? Let g be the gravitational constant.
a) The magnitude of the centripetal acceleration of the moving puck is less than g . b) The magnitude of the centripetal acceleration of the moving puck is equal to g . c) The magnitude of the centripetal acceleration of the moving puck is greater than g . d) The magnitude of the centripetal acceleration of the moving puck is zero. e) There is not enough information given to determine how the magnitude of the centripetal acceleration of the moving puck compares to g . Solution (b): Since the suspended object is not accelerating the tension force upwards exactly balances the gravitational force downwards, so T = mg . The inward force on the puck is due to the tension in the string which is equal in to the mass times the centripetal acceleration, T = macentripetal . Therefore mg = macentripetal or
g = acentripetal .
4
E. (5 points): A bead is given a small push at the top of a hoop (position A) and is constrained to slide around a frictionless circular wire (in a vertical plane). Circle the arrow that best describes the direction of the acceleration when the bead is at the position B.
Solution (4): The bead is speeding up at position B therefore it has a tangential component of the acceleration (pointing downward) and it is traveling in a circular trajectory so it has a radial component of the acceleration pointing towards the center of the circle. Direction 4 best describes the sum of these two components.
5
Problem 2 (25 points): A person is riding a bicycle at a constant speed v0 (which is unknown) along a straight road and then gradually applies the brakes during a time interval 0 < t < t f until the bicycle comes to a stop. Assume that the magnitude of the braking force increases linearly in time according to
F = bt ,
0 < t < tf
where b is a constant. The rider and bicycle have a total mass m . At the instant the person applies the brakes, an iPod drops from a tree branch from a height h above the ground. The rider catches the iPod at the instant she has come to a stop.. You may assume that she catches it at a height s above the ground. The goal of this problem is to determine how far the cyclist traveled during the time interval that the brakes were applied. Express your answer in terms of the variables, b , m , h , s , and g but not v0 .
a) Write a brief description of the strategy that you plan to use and any diagrams or graphs that you have chosen for solving this problem. Make sure you clearly state which concepts you plan to use to calculate any relevant physical quantities. Solution: There are two objects moving, the bicyclist and the iPod. We shall choose a coordinate system for the bicyclist and the iPod as shown in the figure below.
6
We can use Newton’s Second Law to calculate the acceleration of the bicyclist. We can then integrate the acceleration to find the x-component of the velocity of the bicyclist. We then use the equations for free fall to determine the time it takes the iPod to fall a distance h − s . The x-component of the velocity of the bicyclist is zero at this instant so we can solve for the initial x-component of the velocity of the bicyclist. We can now integrate the x-component of the velocity of the bicyclist to find the position and with substitution of the time of travel, we can find the distance traveled. b) How far did the cyclist travel during the time interval that the brakes were applied? Express your answer in terms of the variables, b , m , h , s , and g but not v0 . You may or may not need all of these quantities. G G From Newton’s Second Law, F = ma , are in the x-direction:
ˆi : −bt = ma . x Thus the x-component of the acceleration is
ax = −
b t m
(1.3)
(1.4)
We can now integrate the x-component of the acceleration of the bicyclist to get the ycomponent of the velocity of the bicyclist
b 2 ⎛ b ⎞ vx (t ) − vx 0 = ∫ ax dt = ∫ ⎜ − t ⎟ dt = − t . m ⎠ 2m 0 0⎝ t
t
(1.5)
The y-component of the position of the iPod is given by y (t ) = (h − s ) −
g 2 t 2
(1.6)
The bicyclist catches the iPod at the instant when
7
y (t = t f ) = 0 = (h − s ) −
g 2 tf 2
(1.7)
We can solve eq. (1.7) for the flight time: 2(h − s ) g
tf =
(1.8)
The bicyclist comes to a full stop at this instant so we can substitute Eq. (1.8) into Eq. (1.5) and solve for the initial x-component of the velocity of the bicyclist vx 0 =
b 2 b(h − s ) tf = 2m mg
(1.9)
We can now integrate Eq. (1.5) using the value for vx 0 that we found in Eq. (1.9) to find that the x-component of the position the bicyclist when it just came to a stop tf
t
f ⎛ b( h − s ) b 2 ⎞ b 2⎞ b( h − s ) b 3 ⎛ − x(t f ) − x0 = ∫ vx dt = ∫ ⎜ vx 0 − t ⎟ dt = ∫ ⎜ t ⎟ dt = tf − t f .(1.10) mg mg 2m ⎠ 2m ⎠ 6m 0 0⎝ 0⎝
t
We chose x0 = 0 and so substituting x0 = 0 and Eq. (1.8) into Eq. (1.10) yields b ⎛ 2(h − s ) ⎞ x(t f ) = ⎜ ⎟⎟ 3m ⎜⎝ g ⎠
3
(1.11)
8
Problem 3 (25 points):
An object with mass m slides down a roof a distance d that is inclined at an angle φ with respect to the horizontal. The contact surface between the roof and object has a coefficient of kinetic friction μ . The edge of the bottom of the roof is at a height h above the ground. What is the horizontal distance from the edge of the roof to the point where the object hits the ground?
Include with your answer a plan for solving the problem. Make sure you clearly state which concepts you plan to use to calculate any relevant physical quantities. Also clearly state any assumptions you make. Be sure you include any diagrams or sketches that you plan to use. Solution: It is clear that the distance the object travels will depend on how fast it is moving when it leaves the roof, and that this speed will be larger for larger values of d and φ .
Consider then the two stages of the motion, the first when the object is sliding down the roof and the second where it is in free fall. We shall use Newton’s second Law to find the acceleration for the sliding down the roof and then use the kinematic equations for position and velocity along the roof to find the speed of the object just when it reaches the end of the roof. We can then use that speed as an initial condition for the projectile motion trajectory. We can then solve for the time that it takes to reach the ground and hence solve for the horizontal distance the object traveled from edge of the roof. For the first stage, choose a coordinate system with the positive ˆi -direction down the roof (the steepest downward direction) and the positive ˆj -direction to be perpendicular to the roof (with positive upward component).
9
The forces on the object are gravity
(
G mg = mg ˆi cos φ − ˆj sin φ
and the contact force
)
G G G C = N + fk = N ˆj − f k ˆi
(1.12)
(1.13)
G G . The components of the vectors in Newton’s Second Law, F = ma , are
mg sin φ − f k = m ax N − mg cos φ = m a y .
(1.14) (1.15)
The condition that the object remains on the roof is expressed as a y = 0 and the model for kinetic friction is
f k = μk N .
(1.16)
The condition a y = 0 in Eq. (1.15) gives
N = mg cos φ ,
(1.17)
f k = μk mg cos φ .
(1.18)
and so Eq. (1.16) becomes
Using this in Eq. (1.14) allows solution for the x -component of acceleration ax = g ( sin φ − μk cos φ ) .
(1.19)
Let the time that the object takes to slide down the roof be troof and the speed of the object just as it leaves the roof be v0 . Then, we have that d=
1 1 ( ax troof ) 2 ax troof = ax 2 2
2
(1.20)
We note that v0 = ax troof .
(1.21)
So Eq. (1.20) becomes
10
v0 2 d= 2a x
(1.22)
We can now solve Eq. (1.22) for the speed of the object the instant it leaves the roof v0 = 2 ax d = 2 g d ( sin φ − μk cos φ ) .
(1.23)
For the second stage, the object is in free fall. For this stage, it should be clear that the coordinate system used for the first stage will cause great difficulty. So, take the positive ˆi -direction to be horizontal and the positive ˆj -direction to be vertically upward, so that the ˆi - ˆj plane contains the object’s motion.
We then have ax = 0 , ay = − g .
(1.24) (1.25)
Additionally, if we now “reset our clock” so that t = 0 when the object leaves the roof, vx ,0 = v0 cos φ , (1.26) v y ,0 = −v0 sin φ .
(1.27)
If we further take the origin to be the point on the ground directly below the point where the object leaves the roof, x0 = 0 , y0 = h . The equations describing the object’s motion as a function of time t are then
1 x(t ) = x0 + vx ,0 t + ax t 2 = v0 cos φ t , 2 1 1 y (t ) = y0 + v y ,0 t + a y t 2 = h − v0 sin φ t − gt 2 . 2 2
(1.28) (1.29)
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When the object hits the ground, y = 0 , the x -coordinate may be denoted x f and the time when the object hits may be denoted t f . Eq. (1.29) then becomes 0 = h − v0 sin φ t f −
1 2 gt f 2
(1.30)
This is a quadratic equation in t f , and the quadratic formula gives tf =
(
1 −v0 sin φ ± v02 sin 2 φ + 2 gh g
)
(1.31)
and substitution of the positive solutions into Eq. (1.28) gives xf =
(
v0 cos φ −v0 sin φ + v02 sin 2 φ + 2 gh g
⎛ v02 2 gh ⎞ = cos φ ⎜ − sin φ + sin 2 φ + 2 ⎟ ⎜ g v0 ⎟⎠ ⎝
) .
(1.32)
We can rewrite Eq. (1.22) as v02 / g = d ( sin φ − μk cos φ )
(1.33)
Then substitute Eq. (1.33) into Eq. (1.32) which yields ⎛ ⎞ 2h x f = d ( sin φ − μk cos φ ) cos φ ⎜ − sin φ + sin 2 φ + ⎟ ⎜ d ( sin φ − μk cos φ ) ⎟⎠ ⎝
Note that the gravitational constant g does not appear in this expression. simplification is of questionable value.
(1.34)
Further
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Problem 4: (25 points)
A coin of mass m (which you may treat as a point object) lies on a turntable, exactly at the rim, a distance R from the center. The turntable turns at constant angular speed ω and the coin rides without slipping. Suppose the coefficient of static friction between the turntable and the coin is given by μ . Let g be the gravitational constant.
a) What is the maximum angular speed ω max such that the coin does not slip? Explain your plan for solving this problem. Include all graphs or diagrams that you intend to use. Solution: The coin undergoes circular motion at constant speed so it is accelerating inward. The force inward is static friction and at the just slipping point it has reached its maximum value. We can use Newton’s Second Law to find the maximum angular speed ω max . Our coordinate system and free body force diagram is shown in the figure below. We choose the inward direction as positive.
The contact force is given by
G G G C = N + fs = N ˆj + fs ˆi .
The gravitational force is given by
G Fdisc = − mgˆj .
(1.35)
(1.36)
Newton’s Second Law, Fx = max , in the x-direction (inward direction), noting that the centripetal acceleration has magnitude ax = Rω 2 , is given by fs = m Rω 2 .
(1.37)
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Newton’s Second Law, Fy = ma y , in the y-direction (inward direction), noting that the disc is static hence a y = 0 , is given by
N − mg = 0 .
(1.38)
N = mg .
(1.39)
Thus the normal force is
As ω increases, the static friction increases in magnitude until at ω = ωmax and static friction reaches its maximum value (noting Eq. (1.39)). ( f s ) max = μ N = μ mg .
(1.40)
At this value the disc slips. Thus substituting this value for the maximum static friction into Eq. (1.37) yields 2 μ mg = mRωmax . (1.41) We can now solve Eq. (1.41) for maximum angular speed ω max such that the coin does not slip
ωmax =
μg R
.
(1.42)
Suppose an identical second coin is now stacked on top of the first coin and the turntable turns at constant angular speed ω such that the coins ride without slipping. The coefficient of static friction between the two coins is also μ .
b) What is the magnitude of the radial force exerted by the turntable on the bottom coin? Does this force point inward or outward? Explain your plan for solving this problem. Include all graphs or diagrams that you intend to use.
Solution: We will solve this problem in a similar fashion with a coordinate system and free body diagram on each coin shown in the figures below.
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We will now apply Newton’s Second Law to each coin and determine the magnitude of the radial force exerted by the turntable on the bottom coin, ( fs )bG . The key point is that the static friction between the coins form an action-reaction pair, (neither coin is slipping). So the static friction that makes the top coin accelerate inward also acts on the bottom coin in the opposite direction (radially outward or the negative x-direction). Newton’s Second Law on the bottom coin, Fx = max , in the x-direction (inward direction), noting that the centripetal acceleration has magnitude ax = Rω 2 , is given by ( f s )bG − ( fs )bt = m Rω 2 .
(1.43)
Newton’s Second Law on the top coin, Fx = max , in the x-direction (inward direction), noting that the centripetal acceleration has magnitude ax = Rω 2 , is given by ( f s )tb = m Rω 2 .
(1.44)
( f s )bt = ( f s )tb .
(1.45)
Newton’s Third Law, requires that
So substituting Eq. (1.44) into (1.44) yields ( fs )bt = mRω 2 .
(1.46)
Substituting Eq. (1.46) into Eq. (1.43) then yields ( f s )bG − m Rω 2 = m Rω 2 .
(1.47)
Hence the magnitude of the radial inward force exerted on the bottom coin due to the turntable is then ( f s )bG = 2m Rω 2
(1.48)
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