Question 1

A uniform, solid, cylindrical flywheel of radius 1.4 m and mass 15 kg rotates at $2.7 \mathrm { rad } / \mathrm { s }$ What is the magnitude of the flywheel's angular momentum?

Accepted Answer

The angular momentum $ L $ of a solid cylinder is given by $ L = I \omega $, where $ I = \frac{1}{2} m r^2 $ is the moment of inertia. Substituting $ m = 15 \, \text{kg} $, $ r = 1.4 \, \text{m} $, and $ \omega = 2.7 \, \text{rad/s} $, we find $ L = \frac{1}{2} \times 15 \times (1.4)^2 \times 2.7 = 39.69 \approx 40 \, \text{kg} \cdot \text{m}^2/\text{s} $.

Question 2

A solid wood door, 90.0 cm wide by 2.00 m tall has a mass of 35.0 kg. It is open and at rest. A small 500-g ball is thrown perpendicular to the door with a speed of 20.0 m/s and hits the door 60.0 cm from the hinged side. The ball rebounds with a speed of 16.0 m/s along the same line. What is the angular speed of the door just after the collision with the ball?

Accepted Answer

First, we need to calculate the moment of inertia of the door. Assuming the door is a rectangular slab, we have:
$I = \frac{1}{12}m(2h)^2 + \frac{1}{3}m(\frac{1}{2}w)^2$
$I = \frac{1}{12}(35.0	ext{ kg})(2.00	ext{ m})^2 + \frac{1}{3}(35.0	ext{ kg})(0.5	imes0.9	ext{ m})^2$
$I = 9.13	ext{ kg}\cdot	ext{m}^2$

Next, we need to calculate the torque exerted by the ball on the door. The torque is given by:
$	au = rF\sin	heta$
where $r$ is the distance from the hinge to the point of contact between the ball and the door, $F$ is the force exerted by the ball on the door, and $	heta$ is the angle between the vector from the hinge to the point of contact and the force vector.

In this case, $r = 0.6	ext{ m}$, $F = 2	imes 0.5	ext{ kg}	imes 20.0	ext{ m/s} = 20.0	ext{ N}$ (we double the mass of the ball to account for the fact that the ball will rebound with the same speed), and $	heta = 90^\circ$, so $\sin	heta = 1$. Therefore:
$	au = (0.6	ext{ m})(20.0	ext{ N})(1) = 12.0	ext{ N}\cdot	ext{m}$

The impulse delivered to the door by the ball is equal to the change in angular momentum of the door. We use the conservation of momentum to find the initial angular momentum of the ball and the door:
$L_i = r_{ball}m_{ball}v_{ball} + I\omega_i$
$r_{ball}$ is the distance from the hinge to the point where the ball hits the door, which is 0.6 m. $m_{ball}$ is the mass of the ball, which is 0.5 kg. $v_{ball}$ is the speed of the ball, which is 20.0 m/s. $\omega_i$ is the initial angular velocity of the door, which is 0. Since the ball hits the door perpendicularly, we can assume that the ball's initial velocity is perpendicular to the door's initial velocity, so there is no linear momentum transferred to the door. Therefore:
$L_i = (0.6	ext{ m})(0.5	ext{ kg})(20.0	ext{ m/s}) + 0 = 6.0	ext{ kg}\cdot	ext{m}^2/	ext{s}$

After the collision, the ball rebounds with a speed of 16.0 m/s, so we can use the conservation of energy to find the final angular velocity of the door:
$\frac{1}{2}m_{ball}v_{final}^2 = \frac{1}{2}m_{ball}v_{initial}^2$
$v_{final} = \sqrt{\frac{v_{initial}^2}{m_{ball}}} = \sqrt{\frac{(16.0	ext{ m/s})^2}{0.5	ext{ kg}}} = 35.8	ext{ m/s}$

The final angular momentum of the system is given by:
$L_f = r_{ball}m_{ball}v_{final} + I\omega_f$
Since the collision is elastic, angular momentum is conserved, so $L_f = L_i$. We can solve for $\omega_f$:
$\omega_f = \frac{L_i - r_{ball}m_{ball}v_{final}}{I} = \frac{(6.0	ext{ kg}\cdot	ext{m}^2/	ext{s}) - (0.6	ext{ m})(0.5	ext{ kg})(35.8	ext{ m/s})}{9.13	ext{ kg}\cdot	ext{m}^2} = 1.14	ext{ rad/s}$

Therefore, the final angular speed of the door is 1.14 rad/s. The answer is (C).

Question 3

In a certain cyclotron, a proton of mass 1.67 × 10^-27 kg moves in a circle of diameter 1.6 m with an angular speed of 2.0 × 10⁶ rad/s. What is the angular momentum of the proton?

Accepted Answer

The angular momentum of a particle rotating in a circular orbit is given by:

L = mvr

where m is the mass of the particle, v is its speed, and r is the radius of the orbit.

In this case, the diameter of the orbit is given as 1.6 m, so the radius is r = 0.8 m. The speed of the proton can be found from its angular speed:

v = ωr = (2.0 × 10^16 rad/s)(0.8 m) = 1.6 × 10^16 m/s

Substituting the mass of the proton m = 1.67 × 10^-27 kg, we get:

L = (1.67 × 10^-27 kg)(1.6 × 10^16 m/s)(0.8 m) = 2.13 × 10^-21 kg·m^2/s

Therefore, the angular momentum of the proton is 2.1 × 10^-21 kg·m^2/s, which corresponds to choice C.

Question 4

A 24.5-kg child is standing on the outer edge of a horizontal merry-go-round that has a moment of inertia of $989 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$ about a vertical axis through its center and a radius of 2.40 m. The entire system (including the child)is initially rotating at 0.180 rev/s. Find the angular velocity if the child moves to a new position 1.10 m from the center of the merry-go-round.

Accepted Answer

The answer of A 24.5-kg child is standing on the...

Question 5

A ballerina spins initially at 1.5 rev/s when her arms are extended. She then draws in her arms to her body and her moment of inertia becomes 0.88 kg ? m², and her angular speed increases to 4.0 rev/s. What was her initial moment of inertia?

Accepted Answer

The answer of A ballerina spins initially at 1.5 rev/s...

Question 6

A 900-kg car traveling 30.0° south of east at 12.0 m/s suddenly collides with a 750-kg car traveling north at 17.0 m/s. The cars stick together after colliding. What is the speed of the wreckage just after the collision?

Accepted Answer

The answer of A 900-kg car traveling 30.0° south of...

Question 7

A 1500-kg car traveling at 90 km/h toward the east suddenly collides with a 3000-kg car traveling at 60 km/h toward the south. The two cars stick together after the collision. What is the speed of the cars after collision?

Accepted Answer

The answer of A 1500-kg car traveling at 90 km/h...

Question 8

A 1500-kg car traveling at 90 km/h toward the east suddenly collides with a 3000-kg car traveling at 60 km/h toward the south. The two cars stick together after the collision. What is the direction of motion of the cars after collision?

Accepted Answer

To solve this problem, we need to use conservation of momentum. Before the collision, the total momentum of the system is:

P = m1v1 + m2v2

where m1 and v1 are the mass and velocity of the first car, and m2 and v2 are the mass and velocity of the second car. Since the first car is traveling in the x-direction (east) and the second car is traveling in the y-direction (south), we need to use vector components to calculate their momenta. Let's define the positive x-axis to the right and the positive y-axis to the south. Then:

m1v1,x = (1500 kg)(90 km/h) = 37500 kg km/h (to the right)
m1v1,y = 0 kg km/h (no motion in the y-direction)
m2v2,x = 0 kg km/h (no motion in the x-direction)
m2v2,y = (3000 kg)(60 km/h) = 180000 kg km/h (to the south)

The total momentum before the collision is then:

Pbefore = m1v1,x + m1v1,y + m2v2,x + m2v2,y
        = 37500 kg km/h - 180000 kg km/h
        = -142500 kg km/h (to the left)

After the collision, the two cars stick together and move off as one object with a new velocity v. We can use conservation of momentum again to find this velocity. The total momentum after the collision is:

Pafter = (m1 + m2)v

where m1 + m2 is the total mass of the system after the collision. Since the two cars stick together, their masses add up:

m1 + m2 = 1500 kg + 3000 kg = 4500 kg

So we have:

Pafter = (m1 + m2)v

Using the same vector components as before, the x-component of the velocity is the velocity of the two cars after the collision, and the y-component is again zero:

m1v,x = (1500 kg + 3000 kg)v,x = 4500 kg v,x
m1v,y = 0 kg v,y

The total momentum after the collision is then:

Pafter = m1v,x + m1v,y
       = 4500 kg v,x

We know that momentum is conserved, so:

Pbefore = Pafter

-142500 kg km/h = 4500 kg v,x

Solving for v,x, we get:

v,x = -31.67 km/h (to the left)

This tells us that the cars are moving to the left after the collision. To find the direction of motion, we need to find the angle between the direction of motion and the positive x-axis (which is to the right). We have:

tan θ = (m2v2,y) / (m1v1,x)

tan θ = (180000 kg km/h) / (37500 kg km/h)

θ = tan⁻¹(4.8) = 76.9°

So the angle between the direction of motion and the positive x-axis is 76.9°. However, we want the angle between the direction of motion and the positive y-axis (which is to the south). We can find this angle by subtracting 90° from the angle we just found:

θ' = 76.9° - 90° = -13.1°

Since this angle is negative, that means the direction of motion is in the fourth quadrant (to the left and downward). However, we want the angle as a positive value, so we take the absolute value:

|θ'| = 13.1°

Finally, we want the direction as an angle measured clockwise from the positive x-axis, which means we need to subtract this angle from 90°:

90° - |θ'| = 76.9°

Therefore, the direction of motion of the cars after the collision is 76.9° south of east. Rounded to one decimal place, this is 53.1° south of east. So the answer is choice C.

Question 9

A puck with a mass m₁ = 50 g moving at 1.0 m/s approaches a stationary puck with a mass m₂ = 100 g on a frictionless air table, and they undergo a sudden two-dimensional elastic collision. After the collision, both pucks have identical speeds, but travel in different directions. What is the angle between the original and final paths of puck with mass m₁?

Accepted Answer

In an elastic collision, momentum and kinetic energy are conserved. Let's denote the velocity of the second puck (mass of 100 g) as v2f and the angle between the original and final paths of the first puck (mass of 50 g) as θ.

Using conservation of momentum:
m1v1i = m1v1f*cos(θ) + m2v2f*cos(180-θ)
0.05 kg * 1.0 m/s = 0.05 kg * v1f * cos(θ) + 0.1 kg * v2f * cos(180-θ)
0.05 m/s = 0.05 v1f * cos(θ) - 0.1 v2f * cos(θ)
v1f - 2v2f = 0          (Equation 1)

Using conservation of kinetic energy:
(1/2)*m1*v1i^2 = (1/2)*m1*v1f^2 + (1/2)*m2*v2f^2
0.025 J = 0.025/2 * v1f^2 + 0.05/2 * v2f^2
v1f^2 + 2v2f^2 = 1           (Equation 2)

From Equation 1, we get v1f = 2v2f. Substituting in Equation 2:
(2v2f)^2 + 2v2f^2 = 1
6v2f^2 = 1
v2f = sqrt(1/6) m/s

Now, from Equation 1:
cos(θ) = (0.05 v1f + 0.1 v2f) / 0.05 m/s
cos(θ) = (0.05 * 2v2f + 0.1 v2f) / 0.05 m/s
cos(θ) = 0.4

Taking the inverse cosine:
θ = 66.4 degrees

However, we are interested in the angle between the original and final paths of the first puck, which is 90 - θ. Therefore:
θ = 66.4 degrees
90 - θ = 23.6 degrees

The angle between the original and final paths of the first puck is 23.6 degrees, which is closest to answer choice C (90 degrees).

Question 10

A car heading north suddenly collides at an intersection with a truck of the same mass heading east. If they lock together and travel at $28 \mathrm {~m} / \mathrm { s }$ at 15° north of east just after the collision, how fast was the car initially traveling?

Accepted Answer

The momentum of the car before the collision is equal to the momentum of the car and truck after the collision. The momentum of the car before the collision is $mv_c$, where $m$ is the mass of the car and $v_c$ is the velocity of the car. The momentum of the car and truck after the collision is $(m+m)v_{ct}$, where $v_{ct}$ is the velocity of the car and truck after the collision. Since the car and truck lock together after the collision, their velocities are the same. Therefore, we have:$$mv_c=(m+m)v_{ct}$$Solving for $v_c$, we get:$$v_c=\frac{m+m}{m}v_{ct}$$$$v_c=2v_{ct}$$Since the car and truck are traveling at $28 \mathrm {~m} / \mathrm { s }$ at 15° north of east just after the collision, we have:$$v_{ct}=28 \mathrm {~m} / \mathrm { s }\cos15°$$$$v_{ct}=26.6 \mathrm {~m} / \mathrm { s }$$Therefore, the car was initially traveling at:$$v_c=2v_{ct}=2(26.6 \mathrm {~m} / \mathrm { s })=53.2 \mathrm {~m} / \mathrm { s }$$

Question 11

A uniform 135-g meter stick rotates about an axis perpendicular to the stick passing through its center with an angular speed of 3.50 rad/s. What is the magnitude of the angular momentum of the stick?

Accepted Answer

The answer of A uniform 135-g meter stick rotates about...

Question 12

An ice skater has a moment of inertia of 5.0 kg ∙ m² when her arms are outstretched, and at this time she is spinning at 3.0 rev/s. If she pulls in her arms and decreases her moment of inertia to 2.0 kg ∙ m², how fast will she be spinning?

Accepted Answer

We can use the conservation of angular momentum to solve this problem: L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity. Since the angular momentum is conserved, we have: I1ω1 = I2ω2, where the subscripts 1 and 2 refer to the initial and final conditions, respectively. Plugging in the given values, we get: (5.0 kg ∙ m²)(3.0 rev/s) = (2.0 kg ∙ m²)ω2. Solving for ω2, we get: ω2 = (5.0 kg ∙ m²)(3.0 rev/s) / (2.0 kg ∙ m²) = 7.5 rev/s. Therefore, the answer is (C) 7.5 rev/s.

Question 13

Two dogs, Rover and Fido, run on a level frictionless surface. Rover runs eastward with a momentum of 24 kg . m/s, and Fido runs northward with momentum 10 kg . m/s. They make a sudden perfectly inelastic collision. What is the magnitude of their combined momentum after the collision?

Accepted Answer

In a perfectly inelastic collision, the total momentum before the collision is equal to the total momentum after the collision. Since momentum is a vector quantity, we calculate the resultant momentum using the Pythagorean theorem, given the eastward and northward components are perpendicular: √(24^2 + 10^2) = √(576 + 100) = √676 = 26 kg.m/s.

Question 14

A 900-kg car traveling east at 15.0 m/s suddenly collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together after the collision. In what direction does the wreckage move just after the collision?

Accepted Answer

The answer of A 900-kg car traveling east at 15.0...

Question 15

Three solid, uniform, cylindrical flywheels, each of mass 65.0 kg and radius 1.47 m, rotate independently around a common axis through their centers. Two of the flywheels rotate in one direction at 8.94 rad/s, but the other one rotates in the opposite direction at 3.42 rad/s. Calculate the magnitude of the net angular momentum of the system.

Accepted Answer

The net angular momentum is calculated by finding the angular momentum of each flywheel and then summing them, taking into account their directions. The angular momentum $L$ of a solid cylindrical flywheel is given by $L = I\omega$, where $I = \frac{1}{2}mr^2$ is the moment of inertia of the cylinder, $m$ is the mass, $r$ is the radius, and $\omega$ is the angular velocity. For the two flywheels rotating in the same direction, $L_{1+2} = 2 \times \frac{1}{2} \times 65.0 \times (1.47)^2 \times 8.94$. For the one rotating in the opposite direction, $L_{3} = \frac{1}{2} \times 65.0 \times (1.47)^2 \times (-3.42)$. The net angular momentum is the sum of these, which gives approximately 1020 kg·m^2/s.

Question 16

A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg ∙ m². If he pulls in his arms so his moment of inertia decreases to 1.80 kg ∙ m², what will be his new angular speed?

Accepted Answer

This problem involves conservation of angular momentum. The initial angular momentum of the skater, L1, is equal to the final angular momentum of the skater, L2 (assuming no external torques act on the system). 
L1 = L2 
I1ω1 = I2ω2 
where I1 and I2 are the initial and final moments of inertia, and ω1 and ω2 are the initial and final angular speeds. 
Solving for ω2, we get:
ω2 = (I1/I2)ω1
Substituting the given values, we get:
ω2 = (2.25 kg∙m²)/(1.80 kg∙m²) × 5.00 rad/s ≈ 6.25 rad/s
Therefore, the new angular speed is 6.25 rad/s, which corresponds to choice C.

Question 17

A 5.0-m radius playground merry-go-round with a moment of inertia of 2000 kg ∙ m² is rotating freely with an angular speed of 1.0 rad/s. Two people, each of mass 60 kg, are standing right outside the edge of the merry-go-round and suddenly step onto the edge with negligible speed relative to the ground. What is the angular speed of the merry-go-round right after the two people have stepped on?

Accepted Answer

The answer of A 5.0-m radius playground merry-go-round with a...

Question 18

A 40.0-kg child running at 3.00 m/s suddenly jumps onto a stationary playground merry-go-round at a distance 1.50 m from the axis of rotation of the merry-go-round. The child is traveling tangential to the edge of the merry-go-round just before jumping on. The moment of inertia about its axis of rotation is 600 kg ∙ m² and very little friction at its rotation axis. What is the angular speed of the merry-go-round just after the child has jumped onto it?

Accepted Answer

First, we need to calculate the initial tangential velocity of the child:
$v_{tangential} = 3.00 m/s$
Next, we can use the principle of conservation of angular momentum. The initial angular momentum of the system (child + merry-go-round) is:
$L_{initial} = L_{child} + L_{merry-go-round} = mv_{tangential}r + I\omega = (40.0 kg)(3.00 m/s)(1.50 m) + (600 kg*m^2)\omega$
Since there is no external torque acting on the system, angular momentum is conserved. Therefore, the final angular momentum of the system is equal to the initial angular momentum:
$L_{final} = L_{initial}$
When the child jumps onto the merry-go-round, the system becomes one solid object rotating about the same axis. The final moment of inertia about this axis is the sum of the moment of inertia of the original merry-go-round and the child:
$I_{final} = I_{merry-go-round} + I_{child} = (600 kg*m^2) + (40.0 kg)(1.50 m)^2 = 660 kg*m^2$
Plugging this into the expression for conservation of angular momentum and solving for $\omega$, we get:
$(40.0 kg)(3.00 m/s)(1.50 m) + (600 kg*m^2)\omega = (660 kg*m^2)\omega$
Simplifying and solving for $\omega$, we get:
$\omega = 0.261 rad/s$

Question 19

A 900-kg car traveling east at 15.0 m/s suddenly collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together after the collision. What is the speed of the wreckage just after the collision?

Accepted Answer

The answer of A 900-kg car traveling east at 15.0...

Question 20

A light-weight potter's wheel, having a moment of inertia of $24 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$ is spinning freely at 40.0 rpm. The potter drops a small but dense lump of clay onto the wheel, where it sticks a distance 1.2 m from the rotational axis. If the subsequent angular speed of the wheel and clay is 32 rpm, what is the mass of the clay?

Accepted Answer

Explanation:The angular momentum of the system is conserved, so:$$I_i \omega_i = I_f \omega_f$$where $I_i$ and $\omega_i$ are the initial moment of inertia and angular speed, and $I_f$ and $\omega_f$ are the final moment of inertia and angular speed.The initial moment of inertia is:$$I_i = 24 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$$The final moment of inertia is:$$I_f = 24 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 } + m r^2$$where $m$ is the mass of the clay and $r$ is the distance from the rotational axis.The initial angular speed is:$$\omega_i = 40.0 \mathrm {~rpm} = 4.19 \mathrm {~rad/s}$$The final angular speed is:$$\omega_f = 32 \mathrm {~rpm} = 3.35 \mathrm {~rad/s}$$Substituting these values into the conservation of angular momentum equation, we get:$$24 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 } \times 4.19 \mathrm {~rad/s} = (24 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 } + m (1.2 \mathrm {~m})^2) \times 3.35 \mathrm {~rad/s}$$Solving for $m$, we get:$$m = 4.2 \mathrm {~kg}$$

Quiz 9: Momentum

A uniform, solid, cylindrical flywheel of radius 1.4 m and mass 15 kg rotates at 2.7rad/s2.7 \mathrm { rad } / \mathrm { s }2.7rad/s What is the magnitude of the flywheel's angular momentum?

In a certain cyclotron, a proton of mass 1.67 × 10-27 kg moves in a circle of diameter 1.6 m with an angular speed of 2.0 × 106 rad/s. What is the angular momentum of the proton?

A ballerina spins initially at 1.5 rev/s when her arms are extended. She then draws in her arms to her body and her moment of inertia becomes 0.88 kg ? m2, and her angular speed increases to 4.0 rev/s. What was her initial moment of inertia?