# LON-CAPA HOMEWORK 10

First: Calculate the acceleration of the ball using ∆x = ½at².
Second: Use the acceleration to find the force acting on the ball, and then remember that the change in Kinetic Energy equals the net work done on it (W = F∆x). (Don’t worry about initial because it starts at rest)

ANS: 864 J

A ball of mass 3 kg is uniformly accelerated from rest and travels 48 m in 4 s. What is its final kinetic energy after 4 s?
Remember that the initial momentum pi = mvo = [ mvxo ] i + [ mvyo ] j.

ANS: i + [-30] j kg/m*s

A 1-kg ball in the air has an initial velocity vi = [ 20 ] i + [ -30 ] j m/s. It falls for a total time of 1 s. Assume that positive x-values are to the right and positive y-values are upward.

What is the initial momentum pi of the ball?

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Remember that the impulse I = ∆p = F∆t = [-mg∆t] j. (no i component). The force is gravity in the negative y-direction.

ANS: i + [-10] j kg/m*s

A 1-kg ball in the air has an initial velocity vi = [ 20 ] i + [ -30 ] j m/s. It falls for a total time of 1 s. Assume that positive x-values are to the right and positive y-values are upward.

What is the impulse or change in momentum ∆p of the ball?

Remember that the final momentum pf = pi + ∆p = [ pxi ] i + [ pyi + ∆py ] j. Note that ∆px = 0 in this problem since no force acts in the x-direction.

ANS: i + [-40] j kg/m*s

A 1-kg ball in the air has an initial velocity vi = [ 20 ] i + [ -30 ] j m/s. It falls for a total time of 1 s. Assume that positive x-values are to the right and positive y-values are upward.

What is the final momentum pf of the ball?

Remember that the initial kinetic energy Ki = (pxi² + pyi²)/2m

ANS: (400 + 900)/2*1 ==> 650 J

A 1-kg ball in the air has an initial velocity vi = [ 20 ] i + [ -30 ] j m/s. It falls for a total time of 1 s. Assume that positive x-values are to the right and positive y-values are upward.

What is the initial kinetic energy Ki of the ball?

Remember that the final kinetic energy Kf = (pxf2 + pyf2)/2m.

ANS: (400 + 1600)/2*1 ==> 1000 J

A 1-kg ball in the air has an initial velocity vi = [ 20 ] i + [ -30 ] j m/s. It falls for a total time of 1 s. Assume that positive x-values are to the right and positive y-values are upward.

What is the final kinetic energy Kf of the ball?

Remember that the change in kinetic energy ∆K = Kf − Ki.

ANS: 1000 J – 650 J = 350 J

A 1-kg ball in the air has an initial velocity vi = [ 20 ] i + [ -30 ] j m/s. It falls for a total time of 1 s. Assume that positive x-values are to the right and positive y-values are upward.

What is the change in kinetic energy ∆K of the ball?

Remember that the net work on the ball Wnet = ∆K = W1E * ∆y. Do not forget that the ball is falling, so its change in height is negative.

ANS: 350J = (-10) * ∆y ==> ∆y = -35 m

A 1-kg ball in the air has an initial velocity vi = [ 20 ] i + [ -30 ] j m/s. It falls for a total time of 1 s. Assume that positive x-values are to the right and positive y-values are upward.

What is the ball’s change in height ∆y after it falls 1 s?

Since the total momentum of the system is conserved, the initial total momentum (= m1v1i) equals the final total momentum (=m1v1f + m2v2f). You are given the masses of both pucks and two of the three velocities, which is sufficient information to solve for v2f.

ANS: (0.5*40m/s) = (0.5*-8m/s)+(2*V₂f) ==> V₂f = 12 m/s

Puck 1 (0.5 kg) travels with velocity 40 m/s to the right when it collides with puck 2 (2 kg) which is initially at rest. After the collision, puck 1 moves with a velocity of -8 m/s. Assume that no external forces are present and therefore the momentum for the system of pucks is conserved.

What is the final velocity, V₂f, of puck 2 after the collision?

Remember to wrap your right hand in the direction of rotation and your thumb will give the direction of the angular velocity.

ANS: UP

A spinning disk is rotating at a rate of 2 rad/s in the counterclockwise direction as shown in the figure. Assume that the initial angular velocity is positive.

What is the direction of the angular velocity vector?

It will be opposite the angular velocity vector because it is slowing down.

ANS: DOWN

If the disk is slowing down at a rate of 2 rad/s², what is the direction of the angular acceleration vector?
Remember that ω = ωinit + αt.

ANS: ω = 2 + (-2 * 4s) ==> -6 rad/s

Find the wheel’s angular velocity after 4 s.
Remember that the angle is given by: θ = θo + ωinit*t + 0.5αt²

ANS: ?

Find the angle through which the wheel has turned after 4 s in radians.
Remember that the linear velocity v = rω. Do not forget to convert the given units of centimeters to meters for the answer.

ANS: (.1m * – 6 rad/s) ==> -.6 m/s

Find the velocity of a point located at r = 10 cm from the center of the wheel after 4 s. The sign of the velocity will be the same as the sign for the angular velocity at that time.
Remember that the parallel acceleration a// = rα. Do not forget that the sign of the parallel acceleration is positive if the disk is speeding up and negative if it is slowing down.

ANS: (.1m * -2 rad/s²) ==> -.2 m/s²

Find the parallel acceleration of the point in the above part. The sign of the parallel acceleration will be the same as the sign for the angular acceleration
Incorrect: F2 = F1
Incorrect: F2 = F1/4
Incorrect: F2 = 16 F1
Correct: None of the above.
Two rotating wheels have the same mass (m2 = m1), but their radii are such that R2 = 4 R1. If the angular accelerations of the wheels are equal, then what is the relationship between the forces F1 and F2?
Incorrect: τ₂ = τ₁
Correct: τ₂ = 16 τ₁
Incorrect: τ₂ = 4 τ ₁
Incorrect: τ₂ = τ¼
What is the relationship between the torques?
Incorrect: F₂ = 4 F₁
Correct: F₂ = 28 F₁
Incorrect: F₂ = 16 F₁
Incorrect: F₂ = 49 F₁
Two rotating wheels have masses such that m2 = 7 m1, and radii such that R2 = 4 R1. If the angular accelerations of the wheels are equal, then what is the relationship between the forces F1 and F2?
Correct: τ2 = 112 τ1
Incorrect: τ2 = 49 τ1
Incorrect: τ2 = 4 τ1
Incorrect: τ2 = 28 τ1
What is the relationship between the torques?
A rope of tension 1T applies a force at radius 1R on the spool. It is located at 12 o’clock and points left.
What is the resulting torque in terms of RT?
[Remember to check whether the torque is positive (results in CCW motion) or negative (results in CW motion). ]

ANS: 1 RT

A rope of tension 1T applies a force at radius 3R on the spool. It is located at 9 o’clock and points upward.
What is the resulting torque in terms of RT?

ANS: -3 RT

A spool rests on a horizontal, frictionless surface and you are looking at it from the top view (appears as a circle).
In the following questions, the spool is pulled by tension forces that are applied at various locations, directions, and radii.
To solve each problem, draw the spool as a circle on a piece of paper and then draw the given tension vectors.

REMEMBER that the direction UPWARD points to the TOP of the page and DOWNWARD points to the BOTTOM of the page.
The resulting TORQUE will be either OUT OF the page (CCW motion, POSITIVE torque) or INTO the page (CW motion, NEGATIVE torque).

RING = 1
DISK = ½
SOLID SPHERE = 2/5
MOMENT OF INERTIA COEFFICIENTS?
I ring = 16 MR²
I disk = 8 MR²
A ring of mass 4 M and radius 2 R rotates in a CCW direction with initial angular speed 1 ω. A disk of mass 4 M and radius 2 R rotates in a CW direction with initial angular speed 4 ω. The ring and disk “collide” and eventually rotate together. Assume that positive angular momentum and angular velocity values correspond to rotation in the CCW direction.

What are the moments of inertia for the ring and disk? Write your answer in terms of MR²

Remember to use Coefficients of Inertia

Initial Angular Momentum of Ring = 16MR²ω
A ring of mass 4 M and radius 2 R rotates in a CCW direction with initial angular speed 1 ω. A disk of mass 4 M and radius 2 R rotates in a CW direction with initial angular speed 4 ω. The ring and disk “collide” and eventually rotate together. Assume that positive angular momentum and angular velocity values correspond to rotation in the CCW direction.

What is the initial angular momentum for the ring? Write your answer in terms of MR²ω.

Remember to use Coefficients of Inertia

Initial Angular Momentum of Disk = -32 MR²ω
A ring of mass 4 M and radius 2 R rotates in a CCW direction with initial angular speed 1 ω. A disk of mass 4 M and radius 2 R rotates in a CW direction with initial angular speed 4 ω. The ring and disk “collide” and eventually rotate together. Assume that positive angular momentum and angular velocity values correspond to rotation in the CCW direction.

What is the initial angular momentum for the disk? Write your answer in terms of MR²ω.

Remember that Li = Iω, where I is the moment of inertia of the disk and ω is the angular velocity of the disk. Also, the CW motion of the disk gives a negative angular momentum value.
Remember to use Coefficients of Inertia

Initial Inertia of System (ring + disk) = -16MR²ω
A ring of mass 4 M and radius 2 R rotates in a CCW direction with initial angular speed 1 ω. A disk of mass 4 M and radius 2 R rotates in a CW direction with initial angular speed 4 ω. The ring and disk “collide” and eventually rotate together. Assume that positive angular momentum and angular velocity values correspond to rotation in the CCW direction.

What is the initial angular momentum Li(ring+disk) of the system?

Remember to use Coefficients of Inertia

ωf = -16/24 ==> -0.667 ω
A ring of mass 4 M and radius 2 R rotates in a CCW direction with initial angular speed 1 ω. A disk of mass 4 M and radius 2 R rotates in a CW direction with initial angular speed 4 ω. The ring and disk “collide” and eventually rotate together. Assume that positive angular momentum and angular velocity values correspond to rotation in the CCW direction.

What is the final angular velocity ωf of the ring+disk system? Write your answer in terms of ω.

Remember that Li(ring+disk) = Lf(ring+disk) = If(ring+disk) ωf and solve for ωf using information calculated above.

Remember to use Coefficients of Inertia

Incorrect: same net force
Incorrect: insufficient information
Incorrect: simulation 1
Correct: simulation 2
Two identical black masses, m, are hung via massless strings over two pulleys of identical mass M and radius R, but different mass distributions. The bearings in the pulleys are frictionless and the strings do not slip as they unwind from their pulleys. The masses fall with different accelerations as shown in the animation (position is in meters and time is in seconds). The mass in the second simulation hits the floor first.

1) In which simulation is the net force on the falling mass larger?

Incorrect: insufficient information
Incorrect: simulation 2
Incorrect: same
Correct: simulation 1
Two identical black masses, m, are hung via massless strings over two pulleys of identical mass M and radius R, but different mass distributions. The bearings in the pulleys are frictionless and the strings do not slip as they unwind from their pulleys. The masses fall with different accelerations as shown in the animation (position is in meters and time is in seconds). The mass in the second simulation hits the floor first.

2) In which simulation is the tension in the rope larger?

Correct: simulation 1
Incorrect: same torque
Incorrect: simulation 2
Incorrect: insufficient information
Two identical black masses, m, are hung via massless strings over two pulleys of identical mass M and radius R, but different mass distributions. The bearings in the pulleys are frictionless and the strings do not slip as they unwind from their pulleys. The masses fall with different accelerations as shown in the animation (position is in meters and time is in seconds). The mass in the second simulation hits the floor first.

3) In which simulation is the torque applied to the pulley (by the rope tension) larger?

Incorrect: simulation 2
Incorrect: same moment of inertia
Correct: simulation 1
Incorrect: insufficient information
Two identical black masses, m, are hung via massless strings over two pulleys of identical mass M and radius R, but different mass distributions. The bearings in the pulleys are frictionless and the strings do not slip as they unwind from their pulleys. The masses fall with different accelerations as shown in the animation (position is in meters and time is in seconds). The mass in the second simulation hits the floor first.

4) In which simulation is the moment of inertia of the pulley larger?

Correct: pulley 2
Incorrect: pulley 1
Incorrect: same net forc
Two identical masses are hung over two pulleys of identical mass M and radius R, but different mass distributions. The mass hung from pulley 1 falls slower than the mass hung from pulley 2.

For which pulley is the net force on the falling mass larger?

Incorrect: same net force
Correct: pulley 1
Incorrect: pulley 2
Two identical masses are hung over two pulleys of identical mass M and radius R, but different mass distributions. The mass hung from pulley 1 falls slower than the mass hung from pulley 2.

For which pulley is the tension in the rope (and therefore torque on the pulley) larger?

Incorrect: pulley 2
Incorrect: same net force
Correct: pulley 1
Two identical masses are hung over two pulleys of identical mass M and radius R, but different mass distributions. The mass hung from pulley 1 falls slower than the mass hung from pulley 2.

For which pulley is the moment of inertia larger?

Incorrect: Krot = 1.4 Ktrans
Incorrect: Krot = 0.2 Ktrans
Incorrect: Krot = 0.8 Ktrans
Correct: Krot = 0.4 Ktrans
Incorrect: none of the above
A solid sphere of radius 3 m and mass 4 kg rolls without slipping on a horizontal surface. What is the relationship between its rotational kinetic energy and its translational kinetic energy?

Remember that the rotational kinetic energy Krot = β mv²/2 and the translational kinetic energy Ktrans = mv²/2. Therefore, Krot = β Ktrans.

Correct: Ktrans = 0.71 Ktot
Incorrect: Ktrans = 1.40 Ktot
Incorrect: Ktrans = 1.67 Ktot
Incorrect: Ktrans = 1.20 Ktot
Incorrect: none of the above
A solid sphere of radius 4 m and mass 4 kg rolls without slipping on a horizontal surface. What is the relationship between its translational kinetic energy and its total kinetic energy?

Remember that the total kinetic energy of the object is (1 + β) mv²/2.

Correct: they reach the bottom at the same time
Incorrect: ring 1 reaches the bottom first
Incorrect: insufficient information to determine
Incorrect: ring 2 reaches the bottom first
A ring 1 (2 kg, r=3 m) and ring 2 (6 kg, r=5 m) are placed beside each other at the top of an inclined ramp. They are both released at the same time and roll without slipping down the ramp. Which object reaches the bottom first?
Correct: cylinder reaches the bottom first
Incorrect: ring reaches the bottom first
Incorrect: insufficient information to determine
Incorrect: they reach the bottom at the same time
A ring (5 kg) and a cylinder (2 kg) are placed beside each other at the top of an inclined ramp. They are both released at the same time and roll without slipping down the ramp. Which object reaches the bottom first?
Incorrect: they reach the bottom at the same time
Correct: sphere reaches the bottom first
Incorrect: insufficient information to determine
Incorrect: ring reaches the bottom first
A ring and a solid sphere are placed beside each other at the top of an inclined ramp. They are both released at the same time and roll without slipping down the ramp. Which object reaches the bottom first?
Incorrect: they reach the bottom at the same time
Incorrect: hollow sphere reaches the bottom first
Incorrect: insufficient information to determine
Correct: solid sphere reaches the bottom first
A solid sphere (5 kg, r=5 m) and a hollow sphere (7 kg, r=7 m) are placed beside each other at the top of an inclined ramp. They are both released at the same time and roll without slipping down the ramp. Which object reaches the bottom first?
Correct: they both have the same change in height
Incorrect: insufficient information to determine
Incorrect: ring 1 has the greater change in height
Incorrect: ring 2 has the greater change in height
A ring 1 (3 kg) and a ring 2 (7 kg) are rolling at the same speed when they encounter an inclined plane. Which object has the greater change in height?
A cylinder and a ring are rolling at the same speed when they encounter an inclined plane. Which object has the greater change in height?
Correct: ring has the greater change in height
Incorrect: cylinder has the greater change in height
Incorrect: insufficient information
Incorrect: they both have the same change in height
A cylinder and a ring are rolling at the same speed when they encounter an inclined plane. Which object has the greater change in height?
Incorrect: kg m2/s
Incorrect: kg m/s
Correct: kg m2/s2 or N*m
Incorrect: kg m/s2
What are the units of torque?
Correct: kg m2/s or N*m*s
Incorrect: kg m/s
Incorrect: m/s
Incorrect: kg m2/s2
What are the units of angular momentum?
Correct: I decreases
Incorrect: I increases
Incorrect: I remains constant
A girl sitting on a bar stool with two equal masses in her outstretched arms is rotating counterclockwise at a speed of ω. She slowly brings her arms directly towards her body and as a result her angular speed increases.

Based on this information which statement is true about the moment of inertia I for the system (girl + masses)?

Incorrect: upwards
Incorrect: downwards
A girl sitting on a bar stool with two equal masses in her outstretched arms is rotating counterclockwise at a speed of ω. She slowly brings her arms directly towards her body and as a result her angular speed increases.

What is the direction of the force applied by her arms?

Incorrect: 0 degrees
Correct: 180 degrees
Incorrect: 270 degrees
Incorrect: 90 degrees
Incorrect: 45 degrees
A girl sitting on a bar stool with two equal masses in her outstretched arms is rotating counterclockwise at a speed of ω. She slowly brings her arms directly towards her body and as a result her angular speed increases.

In the formula for the torque exerted on the masses by the girl (τ = r x F) what is the angle between the force vector F and the position r of the masses?

Incorrect: r F tan(θ)
Correct: r F sin(θ)
Incorrect: r F
Incorrect: r F cos(θ)
The magnitude of the torque on the masses due to the force applied by the girl depends on the angle (θ) between the force and position of the masses in the following way
Incorrect: The angular acceleration of the system results from the torque due to her arms.
Incorrect: The kinetic energy of the system decreases because the moment of inertia decreases.
Correct: There is no net external torque on the system.
Incorrect: The angular momentum of the system increases.
girl sitting on a bar stool with two equal masses in her outstretched arms is rotating counterclockwise at a speed of ω. She slowly brings her arms directly towards her body and as a result her angular speed increases.

Which statement below is true?

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