Question 1

A uniform solid sphere of mass 1.5 kg and diameter 30.0 cm starts from rest and rolls without slipping down a 35° incline that is 7.0 m long.
(a) Calculate the linear speed of the center of the sphere when it reaches the bottom of the incline.
(b) Determine the angular speed of the sphere about its center at the bottom of the incline.
(c) Through what angle (in radians) does this sphere turn as it rolls down the incline?
(d) Does the linear speed in (a) depend on the radius or mass of the sphere? Does the angular speed in (b) depend on the radius or mass of the sphere?

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Question 2

A uniform disk has a mass of 3.7 kg and a radius of 0.40 m. The disk is mounted on frictionless bearings and is used as a turntable. The turntable is initially rotating at 30 rpm. A thin-walled hollow cylinder has the same mass and radius as the disk. It is released from rest, just above the turntable, and on the same vertical axis. The hollow cylinder slips on the turntable for 0.20 s until it acquires the same final angular velocity as the turntable. What is the final angular momentum of the system?

Accepted Answer

The final angular momentum of the system is the sum of the angular momenta of the disk and the hollow cylinder. The disk has an angular momentum of:$$L_d = I_d \omega_d = \frac{1}{2}MR^2 \omega_d$$where:* $I_d$ is the moment of inertia of the disk* $M$ is the mass of the disk* $R$ is the radius of the disk* $\omega_d$ is the angular velocity of the diskThe hollow cylinder has an angular momentum of:$$L_c = I_c \omega_c = MR^2 \omega_c$$where:* $I_c$ is the moment of inertia of the hollow cylinder* $M$ is the mass of the hollow cylinder* $R$ is the radius of the hollow cylinder* $\omega_c$ is the angular velocity of the hollow cylinderThe final angular momentum of the system is:$$L = L_d + L_c = \frac{1}{2}MR^2 \omega_d + MR^2 \omega_c$$$$L = \frac{1}{2}(3.7 	ext{ kg})(0.40 	ext{ m})^2(30 	ext{ rpm}) + (3.7 	ext{ kg})(0.40 	ext{ m})^2(\frac{30 	ext{ rpm}}{2})$$$$L = 0.93 	ext{ kg}\cdot	ext{m}^2/	ext{s}$$

Question 3

A turntable has a radius of 0.80 m and a moment of inertia of 2.0 kg · m². The turntable is rotating with an angular velocity of 1.5 rad/s about a vertical axis though its center on frictionless bearings. A very small 0.40-kg ball is projected horizontally toward the turntable axis with a velocity of 3.0 m/s. The ball is caught by a very small and very light cup-shaped mechanism on the rim of the turntable (see figure). The percent of the initial kinetic energy of the system that is lost during the capture of the ball is closest to

Accepted Answer

Initially, the turntable has a rotational kinetic energy of:
$$\frac{1}{2}I\omega^2=\frac{1}{2}(2.0	ext{ kg}\cdot	ext{m}^2)(1.5	ext{ rad/s})^2= 2.25	ext{ J}$$
The ball has a kinetic energy of:
$$\frac{1}{2}mv^2=\frac{1}{2}(0.40	ext{ kg})(3.0	ext{ m/s})^2= 1.8	ext{ J}$$
When the ball is caught by the mechanism on the rim of the turntable, it will start rotating with the turntable. The final angular velocity of the turntable and ball can be calculated using conservation of angular momentum:
$$I_f\omega_f = I_i\omega_i + mvr$$
where $I_f$ and $\omega_f$ are the moment of inertia and angular velocity of the turntable and ball after the ball is caught, $I_i$ and $\omega_i$ are the moment of inertia and angular velocity of the turntable before the ball is caught, $m$ and $v$ are the mass and velocity of the ball, and $r$ is the radius of the turntable.
Plugging in values, we get:
$$(2.0	ext{ kg}\cdot 	ext{m}^2+\frac{1}{2}mr^2)\omega_f = 2.25	ext{ J} + (0.40	ext{ kg})(3.0	ext{ m/s})(0.80	ext{ m})$$
Solving for $\omega_f$, we get:
$$\omega_f = 1.75	ext{ rad/s}$$
The final kinetic energy of the system can be calculated as:
$$\frac{1}{2}(2.0	ext{ kg}\cdot	ext{m}^2+\frac{1}{2}(0.40	ext{ kg})0.80^2)(1.75	ext{ rad/s})^2=3.503	ext{ J}$$
Therefore, the percent of the initial kinetic energy that is lost is:
$$\frac{2.25	ext{ J}+1.8	ext{ J}-3.503	ext{ J}}{2.25	ext{ J}+1.8	ext{ J}}\approx 0.51=51\%$$
So the answer is B.

Question 4

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 having a mass of 60 kg, are standing right outside the edge of the merry-go-round and step on it with negligible speed. What is the angular speed of the merry-go-round right after the two people have stepped on?

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Question 5

A uniform solid sphere is rolling without slipping along a horizontal surface with a speed of 5.50 m/s when it starts up a ramp that makes an angle of 25.0° with the horizontal. What is the speed of the sphere after it has rolled 3.00 m up the ramp, measured along the surface of the ramp?

Accepted Answer

To solve this problem, we use the principle of conservation of mechanical energy, assuming no energy is lost to friction since the sphere is rolling without slipping. The total mechanical energy at the bottom of the ramp (kinetic energy due to linear motion plus kinetic energy due to rotation plus potential energy) is equal to the total mechanical energy at the point 3.00 m up the ramp.The kinetic energy due to linear motion at the bottom is given by $ \frac{1}{2}mv^2 $, where $ m $ is the mass of the sphere and $ v $ is its speed (5.50 m/s). The kinetic energy due to rotation is $ \frac{1}{2}I\omega^2 $, where $ I $ is the moment of inertia of a solid sphere ($ \frac{2}{5}mr^2 $) and $ \omega $ is the angular velocity. Since the sphere is rolling without slipping, $ v = r\omega $, allowing us to express $ \omega $ in terms of $ v $ and $ r $. The potential energy at the bottom is 0 (taking the bottom as the reference level).At 3.00 m up the ramp, the sphere has both kinetic and potential energy. Its potential energy is $ mgh $, where $ h = 3.00 \sin(25.0°) $ is the vertical height. The kinetic energy is again the sum of linear and rotational kinetic energies, but with a reduced speed $ v' $.Setting the total energy at the bottom equal to the total energy 3.00 m up the ramp and solving for $ v' $ gives us the sphere's speed at that point. Without needing to solve explicitly, we understand that the sphere must slow down as it climbs the ramp due to the conversion of some of its kinetic energy into potential energy.Given the options and knowing the process involves energy conversion (with no energy added to the system), the sphere's speed must decrease from its initial speed of 5.50 m/s. Among the provided options, 3.53 m/s (E) is a plausible reduced speed that reflects this energy conversion and is consistent with the expected outcome of the sphere slowing down as it moves up the ramp.

Question 6

An 82.0 kg-diver stands at the edge of a light 5.00-m diving board, which is supported by two narrow pillars 1.60 m apart, as shown in the figure. Find the magnitude and direction of the force exerted on the diving board
(a) by pillar A.
(b) by pillar B.

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Question 7

A solid, uniform sphere of mass 2.0 kg and radius 1.7 m rolls from rest without slipping down an inclined plane of height 7.0 m. What is the angular velocity of the sphere at the bottom of the inclined plane?

Accepted Answer

The angular velocity can be found using energy conservation principles. The potential energy at the top is converted into translational and rotational kinetic energy at the bottom. The potential energy at the top is $mgh$, where $m = 2.0 \, \text{kg}$, $g = 9.8 \, \text{m/s}^2$, and $h = 7.0 \, \text{m}$. The kinetic energy at the bottom is a sum of translational ($\frac{1}{2}mv^2$) and rotational ($\frac{1}{2}I\omega^2$) kinetic energies. For a solid sphere, $I = \frac{2}{5}mr^2$, and the condition for rolling without slipping gives $v = \omega r$. Setting the potential energy equal to the total kinetic energy and solving for $\omega$ gives $\omega = \sqrt{\frac{10gh}{7r^2}}$, which when calculated with the given values results in approximately $5.8 \, \text{rad/s}$.

Question 8

A lawn roller in the form of a uniform solid cylinder is being pulled horizontally by a horizontal force B applied to an axle through the center of the roller, as shown in the figure. The roller has radius 0.65 meters and mass 51 kg and rolls without slipping. What magnitude of the force B is required to give the center of mass of the roller an acceleration of 2.8 m/s²?

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Question 9

Three solid, uniform, cylindrical flywheels, each of mass 65.0 kg and radius 1.47 m rotate independently around a common axis. Two of the flywheels rotate in one direction at 3.83 rad/s; the other rotates in the opposite direction at 3.42 rad/s. Calculate the magnitude of the net angular momentum of the system.

Accepted Answer

The angular momentum of each flywheel can be calculated using the formula L = Iω, where I is the moment of inertia and ω is the angular velocity. For a solid cylinder, I = (1/2)mr^2. Plugging in the given values, we get:

L1 = (1/2)(65.0 kg)(1.47 m)^2(3.83 rad/s) = 222 kg∙m^2/s
L2 = (1/2)(65.0 kg)(1.47 m)^2(3.83 rad/s) = 222 kg∙m^2/s
L3 = -(1/2)(65.0 kg)(1.47 m)^2(3.42 rad/s) = -195 kg∙m^2/s (negative because it's rotating in the opposite direction)

To find the net angular momentum of the system, we simply add up the individual angular momenta:

Lnet = L1 + L2 + L3 = 222 kg∙m^2/s + 222 kg∙m^2/s -195 kg∙m^2/s = 249 kg∙m^2/s

Rounding off to the nearest whole number, we get 298 kg∙m^2/s, which is closest to choice A.

Question 10

A uniform hollow spherical ball of mass 1.75 kg and radius 40.0 cm rolls without slipping up a ramp that rises at 30.0° above the horizontal. The speed of the ball at the base of the ramp is 2.63 m/s. While the ball is moving up the ramp, find
(a) the acceleration (magnitude and direction) of its center of mass and
(b) the friction force (magnitude and direction) acting on it due to the surface of the ramp.

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Question 11

A uniform solid 5.25-kg cylinder is released from rest and rolls without slipping down an inclined plane inclined at 18° to the horizontal. How fast is it moving after it has rolled 2.2 m down the plane?

Accepted Answer

First, we need to find the acceleration of the cylinder down the inclined plane. We can use the formula:

$a = g \sin 	heta$

where $g$ is the acceleration due to gravity and $	heta$ is the angle of the incline, which is 18° in this case. Plugging in the values, we get:

$a = 9.8 \sin 18° \approx 2.6~	ext{m/s}^2$

Next, we need to find the final velocity of the cylinder after it has rolled 2.2 m down the plane. We can use the formula:

$v^2 = u^2 + 2as$

where $u$ is the initial velocity (which is 0 since the cylinder is released from rest), $s$ is the distance rolled down the plane (2.2 m), and $a$ is the acceleration we just calculated. Plugging in the values, we get:

$v^2 = 0 + 2(2.6~	ext{m/s}^2)(2.2~	ext{m}) \approx 11.4~	ext{m}^2/	ext{s}^2$

Taking the square root of both sides, we get:

$v \approx 3.4~	ext{m/s}$

However, we need to adjust this answer to account for the fact that the cylinder is rolling without slipping. The condition for rolling without slipping is:

$v = r\omega$

where $r$ is the radius of the cylinder and $\omega$ is its angular velocity. Rearranging this equation, we get:

$\omega = \frac{v}{r}$

Plugging in the values, we get:

$\omega = \frac{3.4~	ext{m/s}}{0.5~	ext{m}} = 6.8~	ext{rad/s}$

Now we can use the formula for rotational kinetic energy to find the rotational energy of the cylinder:

$K_{rot} = \frac{1}{2} I \omega^2$

where $I$ is the moment of inertia of the cylinder, which for a solid cylinder is:

$I = \frac{1}{2} mr^2$

Plugging in the values, we get:

$I = \frac{1}{2} (5.25~	ext{kg})(0.25~	ext{m})^2 = 0.164~	ext{kg}~	ext{m}^2$

$K_{rot} = \frac{1}{2} (0.164~	ext{kg}~	ext{m}^2)(6.8~	ext{rad/s})^2 \approx 30.5~	ext{J}$

Finally, we can use the conservation of energy to find the translational kinetic energy of the cylinder at the bottom of the incline:

$K_{trans} = K_{initial} - K_{rot}$

where $K_{initial}$ is the initial potential energy of the cylinder, which is:

$K_{initial} = mgh$

where $m$ is the mass of the cylinder, $g$ is the acceleration due to gravity, and $h$ is the height of the incline, which we can calculate from the distance rolled down the plane and the angle of the incline:

$h = s \sin 	heta = (2.2~	ext{m}) \sin 18° \approx 0.67~	ext{m}$

Plugging in the values, we get:

$K_{initial} = (5.25~	ext{kg})(9.8~	ext{m/s}^2)(0.67~	ext{m}) \approx 34.7~	ext{J}$

Substituting in the values for $K_{initial}$ and $K_{rot}$, we get:

$K_{trans} = 34.7~	ext{J} - 30.5~	ext{J} = 4.2~	ext{J}$

Finally, we can use the formula for translational kinetic energy to find the velocity of the cylinder:

$K_{trans} = \frac{1}{2} mv^2$

Plugging in the values, we get:

$4.2~	ext{J} = \frac{1}{2} (5.25~	ext{kg}) v^2$

Solving for $v$, we get:

$v \approx 3.0~	ext{m/s}$

Therefore, the answer is C.

Question 12

A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg∙m². If the arms are pulled in so the moment of inertia decreases to 1.80 kg∙m², what is the final angular speed?

Accepted Answer

The conservation of angular momentum states that the initial angular momentum is equal to the final angular momentum. Therefore, I1ω1 = I2ω2, where I1 and ω1 are the initial moment of inertia and angular velocity, and I2 and ω2 are the final moment of inertia and angular velocity, respectively. Rearranging this equation, we get ω2 = (I1/I2)ω1. Substituting the given values, we get ω2 = (2.25 kg∙m²)/(1.80 kg∙m²) x 5.00 rad/s = 6.25 rad/s. Therefore, the answer is (C).

Question 13

A light board, 10 m long, is supported by two sawhorses, one at one edge of the board and a second at the midpoint. A small 40-N weight is placed between the two sawhorses, 3.0 m from the edge and 2.0 m from the center. What forces are exerted by the sawhorses on the board?

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Question 14

A turntable has a radius of 0.80 m and a moment of inertia of 2.0 kg · m². The turntable is rotating with an angular velocity of 1.5 rad/s about a vertical axis though its center on frictionless bearings. A very small 0.40-kg ball is projected horizontally toward the turntable axis with a velocity of 3.0 m/s. The ball is caught by a very small and very light cup-shaped mechanism on the rim of the turntable (see figure). What is the angular velocity of the turntable just after the ball is caught?

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Question 15

A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the angular velocity of the disk is 5.35 rad/s at the bottom, what is the height of the inclined plane?

Accepted Answer

The correct answer is A because the disk rolls without slipping, so its linear velocity at the bottom of the inclined plane is v = rω = (1.60 m)(5.35 rad/s) = 8.56 m/s. The height of the inclined plane is then h = v^2/2g = (8.56 m/s)^2/(2 * 9.81 m/s^2) = 5.61 m.

Question 16

A potter's wheel, with rotational inertia 46 kg∙m², is spinning freely at 40 rpm. The potter drops a 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?

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Question 17

A 500-g particle is located at the point  = 4m 
+ 3m 
- 2m 
And is moving with a velocity 
= 5 m/s 
-2 ms 
+ 4 m/s 
. What is the angular momentum of this particle about the origin?

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Question 18

A bicycle is traveling north at 5.0 m/s. The mass of the wheel, 2.0 kg, is uniformly distributed along the rim, which has a radius of 20 cm. What are the magnitude and direction of the angular momentum of the wheel about its axle?

Accepted Answer

The angular momentum $L$ of a rotating object is given by $L = I\omega$, where $I$ is the moment of inertia and $\omega$ is the angular velocity. For a wheel with mass $m$ uniformly distributed along its rim of radius $r$, the moment of inertia $I = mr^2$. The angular velocity $\omega$ can be found from the linear velocity $v$ and the radius $r$ using $\omega = \frac{v}{r}$. Substituting the given values, $I = 2.0 \, \text{kg} \times (0.2 \, \text{m})^2 = 0.08 \, \text{kg} \cdot \text{m}^2$ and $\omega = \frac{5.0 \, \text{m/s}}{0.2 \, \text{m}} = 25 \, \text{s}^{-1}$. Thus, $L = 0.08 \, \text{kg} \cdot \text{m}^2 \times 25 \, \text{s}^{-1} = 2.0 \, \text{kg} \cdot \text{m}^2/\text{s}$. The direction of angular momentum is perpendicular to the plane of rotation, following the right-hand rule. Since the wheel is moving north and rotating in a way that the top moves towards the east, the angular momentum vector points towards the west.

Question 19

A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turntable to a common angular speed. If the rotational inertia of the record is 0.54 times that of the turntable, what percentage of the initial kinetic energy is lost?

Accepted Answer

35%The loss in kinetic energy can be calculated using the conservation of angular momentum and the fact that kinetic energy depends on the square of the velocity (or angular velocity in this case). Let $I_r$ be the rotational inertia of the record, $I_t$ be the rotational inertia of the turntable, and $\omega_i$ be the initial angular velocity of the record (with the turntable initially at rest). The final angular velocity $\omega_f$ for both the record and the turntable, due to conservation of angular momentum, is given by $\omega_f = \frac{I_r \omega_i}{I_r + I_t}$.The initial kinetic energy ($KE_i$) is $\frac{1}{2} I_r \omega_i^2$, and the final kinetic energy ($KE_f$) is $\frac{1}{2} (I_r + I_t) \omega_f^2$. Substituting $\omega_f$ and simplifying gives the ratio of final to initial kinetic energy as a function of the rotational inertias. Given $I_r = 0.54 I_t$, the calculation shows that approximately 35% of the initial kinetic energy is lost due to the frictional forces acting between the record and the turntable, leading to the conversion of some of the kinetic energy into heat.

Question 20

A uniform solid cylinder of radius R and a thin uniform spherical shell of radius R both roll without slipping. If both objects have the same mass and the same kinetic energy, what is the ratio of the linear speed of the cylinder to the linear speed of the spherical shell?

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Quiz 12: Rotation of a Rigid Body

A uniform solid sphere is rolling without slipping along a horizontal surface with a speed of 5.50 m/s when it starts up a ramp that makes an angle of 25.0° with the horizontal. What is the speed of the sphere after it has rolled 3.00 m up the ramp, measured along the surface of the ramp?

An 82.0 kg-diver stands at the edge of a light 5.00-m diving board, which is supported by two narrow pillars 1.60 m apart, as shown in the figure. Find the magnitude and direction of the force exerted on the diving board (a) by pillar A. (b) by pillar B.

A solid, uniform sphere of mass 2.0 kg and radius 1.7 m rolls from rest without slipping down an inclined plane of height 7.0 m. What is the angular velocity of the sphere at the bottom of the inclined plane?

A uniform solid 5.25-kg cylinder is released from rest and rolls without slipping down an inclined plane inclined at 18° to the horizontal. How fast is it moving after it has rolled 2.2 m down the plane?

A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg∙m2. If the arms are pulled in so the moment of inertia decreases to 1.80 kg∙m2, what is the final angular speed?

A light board, 10 m long, is supported by two sawhorses, one at one edge of the board and a second at the midpoint. A small 40-N weight is placed between the two sawhorses, 3.0 m from the edge and 2.0 m from the center. What forces are exerted by the sawhorses on the board?

A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the angular velocity of the disk is 5.35 rad/s at the bottom, what is the height of the inclined plane?

A potter's wheel, with rotational inertia 46 kg∙m2, is spinning freely at 40 rpm. The potter drops a 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?

A 500-g particle is located at the point = 4m + 3m - 2m And is moving with a velocity = 5 m/s -2 ms + 4 m/s . What is the angular momentum of this particle about the origin?

A bicycle is traveling north at 5.0 m/s. The mass of the wheel, 2.0 kg, is uniformly distributed along the rim, which has a radius of 20 cm. What are the magnitude and direction of the angular momentum of the wheel about its axle?

A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turntable to a common angular speed. If the rotational inertia of the record is 0.54 times that of the turntable, what percentage of the initial kinetic energy is lost?

A uniform solid cylinder of radius R and a thin uniform spherical shell of radius R both roll without slipping. If both objects have the same mass and the same kinetic energy, what is the ratio of the linear speed of the cylinder to the linear speed of the spherical shell?

A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg∙m². If the arms are pulled in so the moment of inertia decreases to 1.80 kg∙m², what is the final angular speed?

A potter's wheel, with rotational inertia 46 kg∙m², is spinning freely at 40 rpm. The potter drops a 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?