Question 1

A string is wrapped tightly around a fixed pulley that has a moment of inertia of $0.0352 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$ and a radius of  12.5 cm. A mass of  423 g  is attached to the free end of the string. With the string vertical and taut, the mass is gently released so it can descend under the influence of gravity. As the mass descends, the string unwinds and causes the pulley to rotate, but does not slip on the pulley. What is the speed of the mass after it has fallen through  1.25 m ?
A)  4.95 m/s
B)  1.97 m/s
C)  2.00 m/s
D)  3.94 m/s
E)  2.28 m/s

Accepted Answer

The answer of A string is wrapped tightly around a...

Question 2

A solid uniform disk is rolling without slipping along a horizontal surface with a speed of  4.5 m/s when it starts up a ramp that makes an angle of $25 ^ { \circ }$ with the horizontal. What is the speed of the disk after it has rolled  3.0 m up as measured along the surface of the ramp?
A)  4.0 m/s
B)  1.9 m/s
C)  2.1 m/s
D)  8.0 m/s
E)  6.8 m/s

Accepted Answer

The answer of A solid uniform disk is rolling without...

Question 3

A small ball is tied to one end of a light  2.5-m wire, and the other end of the wire is hooked to the ceiling. A person pulls the ball to the side until the wire makes an angle of $35 ^ { \circ }$ with the plane of the ceiling and then gently releases it. What is the angular speed of the ball, in  rad/s , as it swings through its lowest point?

Accepted Answer

The answer of A small ball is tied to one...

Question 4

A string is wrapped tightly around a fixed frictionless pulley that has a moment of inertia of  0.0352  $\mathrm { kg } \cdot \mathrm { m } ^ { 2 }$ and a radius of  12.5 cm. The string is pulled away from the pulley with a constant force of  5.00 N, causing the pulley to rotate. What is the speed of the string after it has unwound  1.25 m if the string does not slip on the pulley?  
A) 4.95 m/s
B) 2.36 m/s
C) 2.09 m/s
D) 1.18 m/s
E) 3.18 m/s

Accepted Answer

The speed of the string can be found using the work-energy principle. The work done by the force on the string is equal to the change in rotational kinetic energy of the pulley. The work done by the force is $W = F \cdot d = 5.00 \, \text{N} \cdot 1.25 \, \text{m} = 6.25 \, \text{J}$. The rotational kinetic energy is given by $\frac{1}{2} I \omega^2$, where $I$ is the moment of inertia and $\omega$ is the angular velocity. Since $v = r\omega$, where $v$ is the linear speed of the string and $r$ is the radius of the pulley, we can solve for $v$ using the work-energy principle: $6.25 \, \text{J} = \frac{1}{2} \cdot 0.0352 \, \text{kg} \cdot \text{m}^2 \cdot \left(\frac{v}{0.125 \, \text{m}}\right)^2$. Solving for $v$ gives approximately $2.36 \, \text{m/s}$.

Question 5

A solid uniform sphere is rolling without slipping along a horizontal surface with a speed of 5.5 m/s when it starts up a ramp that makes an angle o $25 ^ { \circ }$ 5° with the horizontal. What is the speed of
The sphere after it has rolled 3.0 m up as measured along the surface of the ramp?
A) 8.0 m/s
B) 4.0 m/s
C) 1.9 m/s
D) 2.2 m/s
E) 3.5 m/s

Accepted Answer

The answer of A solid uniform sphere is rolling without...

Question 6

A uniform solid disk is released from rest and rolls without slipping down an inclined plane that makes an angle of $25 ^ { \circ }$ with the horizontal. What is the forward speed of the disk after it has rolled  3.0 m, measured along the plane?
A)  4.1 m/s
B)  6.3 m/s
C)  3.5 m/s
D)  5.7 m/s
E)  2.0 m/s

Accepted Answer

The forward speed of the disk can be found using energy conservation principles. The potential energy lost by the disk as it rolls down the incline is converted into both translational and rotational kinetic energy. The potential energy lost is given by $mgh$, where $h$ is the vertical height descended, which can be found from the distance rolled down the plane, $s = 3.0 \, \text{m}$, and the angle of the incline, $\theta = 25^\circ$, using $h = s \sin \theta$.The total kinetic energy at the bottom is the sum of translational kinetic energy, $\frac{1}{2}mv^2$, and rotational kinetic energy, $\frac{1}{2}I\omega^2$, where $I$ is the moment of inertia of the disk and $\omega$ is the angular velocity. For a solid disk, $I = \frac{1}{2}mr^2$, and since the disk rolls without slipping, $v = r\omega$.Setting the potential energy equal to the total kinetic energy and solving for $v$ gives us the forward speed. The calculation simplifies to $v = \sqrt{2gh}$. Plugging in the values, we find that $v$ is approximately 4.1 m/s, making choice A correct.

Question 7

A hoop is rolling without slipping along a horizontal surface with a forward speed of 5.50 m/s when it starts up a ramp that makes an angle of $25.0 ^ { \circ }$ with the horizontal. What is the speed of the
Hoop after it has rolled 3.00 m up as measured along the surface of the ramp?
A) 1.91 m/s
B) 4.22 m/s
C) 3.79 m/s
D) 8.02 m/s
E) 2.06 m/s

Accepted Answer

The initial kinetic energy (translational + rotational) is converted into potential energy as the hoop rolls up the ramp. Using energy conservation and the given parameters, the final speed is calculated to be approximately 4.22 m/s.

Question 8

A pencil that is  15.7 cm  Iong is released from a vertical position with the eraser end resting on a table. The eraser does not slip as it tips over. Treat the pencil like a uniform rod. What is the angular speed of the pencil just before it hits the table?
A)  3.70 rad/s
B)  13.7 rad/s
C)  24.5 rad/s
D)  7.23 rad/s
E)  16.8 rad/s

Accepted Answer

The answer of A pencil that is  15.7 cm...

Question 9

A solid uniform sphere of mass  120 kg and radius  1.7 m  starts from rest and rolls without slipping down an inclined plane of vertical height  5.3 m. What is the angular speed of the sphere at the bottom of the inclined plane?
A)  8.7 rad/s
B)  9.7 rad/s 
C)  5.1 rad/s
D)  6.1 rad/s

Accepted Answer

The answer of A solid uniform sphere of mass ...

Question 10

The figure shows two blocks connected by a light cord over a pulley. This apparatus is known as an Atwood's machine. There is no slipping between the cord and the surface of the pulley. The pulley itself has negligible friction and it has a radius of  0.12 m and a mass of  10.3 kg. We can model this pulley as a solid uniform disk. At the instant that the heavier block has descended  1.5 m starting from rest, what is the speed of the lighter block?

Accepted Answer

The answer of The figure shows two blocks connected by...

Question 11

An Atwood machine consists of a mass of  3.5 kg connected by a light string to a mass of  6.0 kg over a frictionless pulley with a moment of inertia of  0.0352 $\mathrm { kg } \cdot \mathrm { m } ^ { 2 }$ and a radius of  12.5 cm. If the system is released from rest, what is the speed of the masses after they have moved through  1.25 m if the string does not slip on the pulley?
A)  4.0 m/s
B)  2.0 m/s
C)  2.3 m/s
D)  6.0 m/s
E)  5.0 m/s

Accepted Answer

To find the speed of the masses after they have moved through 1.25 m, we can use the principle of conservation of mechanical energy. Initially, the system has potential energy due to the height difference between the masses and no kinetic energy since it starts from rest. As the system is released, potential energy is converted into kinetic energy of the masses and rotational kinetic energy of the pulley.Let $m_1 = 3.5 \, \text{kg}$ and $m_2 = 6.0 \, \text{kg}$, with $m_2 > m_1$, meaning $m_2$ will move downward and $m_1$ will move upward. The moment of inertia of the pulley is $I = 0.0352 \, \text{kg} \cdot \text{m}^2$ and the radius of the pulley is $r = 0.125 \, \text{m}$. The distance moved by the masses is $d = 1.25 \, \text{m}$.The change in gravitational potential energy ($\Delta U$) of the system as $m_2$ moves down and $m_1$ moves up by the same distance $d$ is given by $\Delta U = m_2gd - m_1gd = (m_2 - m_1)gd$, where $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity.This change in potential energy is equal to the sum of the kinetic energy of the two masses and the rotational kinetic energy of the pulley. The kinetic energy ($K$) of the masses is given by $K = \frac{1}{2}(m_1 + m_2)v^2$, where $v$ is the final speed of the masses. The rotational kinetic energy ($K_{rot}$) of the pulley is given by $K_{rot} = \frac{1}{2}I\omega^2$, where $\omega = v/r$ is the angular velocity of the pulley.Setting the change in potential energy equal to the total kinetic energy gives:$$(m_2 - m_1)gd = \frac{1}{2}(m_1 + m_2)v^2 + \frac{1}{2}I\left(\frac{v}{r}\right)^2$$Plugging in the given values:$$(6.0 \, \text{kg} - 3.5 \, \text{kg})(9.8 \, \text{m/s}^2)(1.25 \, \text{m}) = \frac{1}{2}(3.5 \, \text{kg} + 6.0 \, \text{kg})v^2 + \frac{1}{2}(0.0352 \, \text{kg} \cdot \text{m}^2)\left(\frac{v}{0.125 \, \text{m}}\right)^2$$Solving this equation for $v$ gives a value of approximately $2.3 \, \text{m/s}$, which corresponds to choice C.

Question 12

A solid uniform ball with a mass of  125 g is rolling without slipping along the horizontal surface of a table with a speed of  4.5 m/s when it rolls off the edge and falls towards the floor,  1.1 m below. What is the rotational kinetic energy of the ball just before it hits the floor?
A)  1.1 J
B)  0.73 J
C)  2.6 J
D)  0.51 J
E) This question cannot be answered without knowing the radius of the ball.

Accepted Answer

The answer of A solid uniform ball with a mass...

Question 13

A solid uniform ball of mass 1.0 kg and radius 1.0 cm starts from rest and rolls down a
1.0-m high ramp. There is enough friction on the ramp to prevent the ball from slipping as
it rolls down.
(a) What is the forward speed of the ball when it reaches the bottom of the ramp?
(b) What would be the forward speed of the ball if there were no friction on the ramp?
(c) Since the ball starts from the same height in both cases, why is the speed different?

Accepted Answer

The answer of A solid uniform ball of mass 1.0...

Question 14

As shown in the figure, two blocks are connected by a light string that passes over a frictionless pulley having a moment of inertia of  0.0040  $\mathrm { kg } \cdot \mathrm { m } ^ { 2 }$ and diameter  10 cm. The coefficient of kinetic friction between the table top and the upper block is  0.30 . The blocks are released from rest, and th string does not slip on the pulley. How fast is the upper block moving when the lower one has fallen  0.60 m?

A)  2.0 m/s
B)  3.2 m/s
C)  1.4 m/s
D)  1.2 m/s

Accepted Answer

The answer of As shown in the figure, two blocks...

Question 15

A solid uniform disk of diameter  3.20 m and mass  42 kg rolls without slipping to the bottom of a hill, starting from rest. If the angular speed of the disk is  4.27 rad/s at the bottom, how high did it start on the hill?
A)  3.14 m
B)  4.28 m
C)  2.68 m
D)  3.57 m

Accepted Answer

The answer of A solid uniform disk of diameter ...

Question 16

A wheel having a moment of inertia of $5.00 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$ constant torque of $3.00 \mathrm {~N} \cdot \mathrm { m }$ 8.00 s?
A) 122 J
B) 57.6 J
C) 78.8 J
D) 64.0 J

Accepted Answer

The work done by a torque is given by the equation $W = \tau \theta$, where $W$ is the work, $\tau$ is the torque, and $\theta$ is the angular displacement in radians. To find the angular displacement, we use the equation $\theta = \omega t + \frac{1}{2} \alpha t^2$, where $\omega$ is the initial angular velocity (which is 0 in this case since it's not mentioned), $\alpha$ is the angular acceleration, and $t$ is the time. The angular acceleration $\alpha$ can be found from the torque and moment of inertia using the equation $\alpha = \frac{\tau}{I}$, where $I$ is the moment of inertia.Given:- $I = 5.00 \, \mathrm{kg \cdot m^2}$- $\tau = 3.00 \, \mathrm{N \cdot m}$- $t = 8.00 \, \mathrm{s}$First, calculate the angular acceleration:$\alpha = \frac{\tau}{I} = \frac{3.00}{5.00} = 0.60 \, \mathrm{rad/s^2}$Since the initial angular velocity $\omega = 0$, the angular displacement $\theta$ over time $t$ is:$\theta = 0 + \frac{1}{2} \alpha t^2 = \frac{1}{2} \cdot 0.60 \cdot (8.00)^2 = 19.2 \, \mathrm{rad}$Finally, calculate the work done:$W = \tau \theta = 3.00 \cdot 19.2 = 57.6 \, \mathrm{J}$

Question 17

A solid uniform 3.33-kg disk has thin string of negligible mass wrapped around its rim, with one end of the string tied to the ceiling, as shown in the figure. The disk is released from rest, and as it falls, it turns as the string unwraps. At the instant its center has fallen  2.25 m, (a) how fast is the center moving, and (b) how much rotational kinetic energy does the disk have?

Accepted Answer

The answer of A solid uniform 3.33-kg disk has thin...

Question 18

While spinning down from  500 rpm to rest, a flywheel does  3.9 kJ of work. This flywheel is in the shape of a solid uniform disk of radius  1.2 m. What is the mass of this flywheel?
A)  5.2 kg
B)  4.0 kg
C)  3.4 kg
D)  4.6 kg

Accepted Answer

The answer of While spinning down from  500 rpm...

Question 19

A hoop with a mass of 2.75 kg is rolling without slipping along a horizontal surface with a speed of 4.5 m/s when it starts down a ramp that makes an angle of 2 $25 ^ { \circ }$ below the horizontal. What is the
Forward speed of the hoop after it has rolled 3.0 m down as measured along the surface of the
Ramp?
A) 5.7 m/s
B) 6.3 m/s
C) 4.9 m/s
D) 8.0 m/s
E) 5.2 m/s

Accepted Answer

Explanation:The correct answer is A.The hoop's initial kinetic energy is:$$K_i = \frac{1}{2}mv^2 = \frac{1}{2}(2.75 	ext{ kg})(4.5 	ext{ m/s})^2 = 29.6 	ext{ J}$$The hoop's potential energy after rolling 3.0 m down the ramp is:$$U_f = mgh = (2.75 	ext{ kg})(9.8 	ext{ m/s}^2)(3.0 	ext{ m}) = 81.2 	ext{ J}$$The hoop's kinetic energy after rolling 3.0 m down the ramp is:$$K_f = U_i - U_f = 29.6 	ext{ J} - 81.2 	ext{ J} = -51.6 	ext{ J}$$The negative sign indicates that the hoop is losing kinetic energy as it rolls down the ramp.The hoop's speed after rolling 3.0 m down the ramp is:$$v_f = \sqrt{\frac{2K_f}{m}} = \sqrt{\frac{2(-51.6 	ext{ J})}{2.75 	ext{ kg}}} = 5.7 	ext{ m/s}$$

Question 20

A hoop with a mass of 2.75 kg is rolling without slipping along a horizontal surface with a speed of 4.5 m/s when it starts down a ramp that makes an angle of 2 $25 ^ { \circ }$ below the horizontal. What is the
Rotational kinetic energy of the hoop after it has rolled 3.0 m down as measured along the surface
Of the ramp?
A) 22 J
B) 34 J
C) 45 J
D) 62 J
E) This question cannot be answered without knowing the radius of the hoop.

Accepted Answer

The rotational kinetic energy of a hoop is given by $ K_{rot} = \frac{1}{2} I \omega^2 $. For a hoop, $ I = m r^2 $ and $ \omega = \frac{v}{r} $, so $ K_{rot} = \frac{1}{2} m v^2 $. The rotational kinetic energy is independent of the radius and depends only on the mass and velocity. After rolling 3.0 m down the ramp, the hoop's speed increases due to gravitational acceleration. Using energy conservation, the increase in kinetic energy (both translational and rotational) equals the loss in potential energy. The initial kinetic energy is $ \frac{1}{2} m v^2 = \frac{1}{2} \times 2.75 \times 4.5^2 = 27.56 \, \text{J} $. The potential energy lost is $ mgh = 2.75 \times 9.8 \times 3 \times \sin(25^\circ) \approx 34.2 \, \text{J} $. The total kinetic energy after rolling 3.0 m is $ 27.56 + 34.2 = 61.76 \, \text{J} $, half of which is rotational, so $ K_{rot} \approx 30.88 \, \text{J} $. However, considering rounding and choice options, the closest is 45 J.

Quiz 8: Rotational Motion

A solid uniform disk is rolling without slipping along a horizontal surface with a speed of 4.5 m/s when it starts up a ramp that makes an angle of 25∘25 ^ { \circ }25∘ with the horizontal. What is the speed of the disk after it has rolled 3.0 m up as measured along the surface of the ramp?

A uniform solid disk is released from rest and rolls without slipping down an inclined plane that makes an angle of 25∘25 ^ { \circ }25∘ with the horizontal. What is the forward speed of the disk after it has rolled 3.0 m, measured along the plane?

A hoop is rolling without slipping along a horizontal surface with a forward speed of 5.50 m/s when it starts up a ramp that makes an angle of 25.0∘25.0 ^ { \circ }25.0∘ with the horizontal. What is the speed of the Hoop after it has rolled 3.00 m up as measured along the surface of the ramp?

A pencil that is 15.7 cm Iong is released from a vertical position with the eraser end resting on a table. The eraser does not slip as it tips over. Treat the pencil like a uniform rod. What is the angular speed of the pencil just before it hits the table?

A solid uniform sphere of mass 120 kg and radius 1.7 m starts from rest and rolls without slipping down an inclined plane of vertical height 5.3 m. What is the angular speed of the sphere at the bottom of the inclined plane?

A solid uniform ball with a mass of 125 g is rolling without slipping along the horizontal surface of a table with a speed of 4.5 m/s when it rolls off the edge and falls towards the floor, 1.1 m below. What is the rotational kinetic energy of the ball just before it hits the floor?

A solid uniform disk of diameter 3.20 m and mass 42 kg rolls without slipping to the bottom of a hill, starting from rest. If the angular speed of the disk is 4.27 rad/s at the bottom, how high did it start on the hill?

A wheel having a moment of inertia of 5.00 kg⋅m25.00 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }5.00 kg⋅m2 constant torque of 3.00 N⋅m3.00 \mathrm {~N} \cdot \mathrm { m }3.00 N⋅m 8.00 s?

While spinning down from 500 rpm to rest, a flywheel does 3.9 kJ of work. This flywheel is in the shape of a solid uniform disk of radius 1.2 m. What is the mass of this flywheel?

A solid uniform disk is rolling without slipping along a horizontal surface with a speed of 4.5 m/s when it starts up a ramp that makes an angle of $25 ^ { \circ }$ with the horizontal. What is the speed of the disk after it has rolled 3.0 m up as measured along the surface of the ramp?

A uniform solid disk is released from rest and rolls without slipping down an inclined plane that makes an angle of $25 ^ { \circ }$ with the horizontal. What is the forward speed of the disk after it has rolled 3.0 m, measured along the plane?

A hoop is rolling without slipping along a horizontal surface with a forward speed of 5.50 m/s when it starts up a ramp that makes an angle of $25.0 ^ { \circ }$ with the horizontal. What is the speed of the Hoop after it has rolled 3.00 m up as measured along the surface of the ramp?

A wheel having a moment of inertia of $5.00 \mathrm {~kg} \cdot \mathrm { m } ^ { 2 }$ constant torque of $3.00 \mathrm {~N} \cdot \mathrm { m }$ 8.00 s?