Question 1

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 2

For the apparatus shown in the figure,there is no slipping between the cord and the surface of the pulley.The blocks have mass of 3.0 kg and 5.7 kg,and the pulley has a radius of  and a mass of 10.3 kg.At the instant the 5.7 kg mass has fallen 1.5 m starting from rest,find the speed of each block.(Assume the pulley is in the shape of a uniform solid disk and has no friction in its axle.)

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Question 3

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  the other rotates in the opposite direction at 3.42 rad/s.Calculate the magnitude of the net angular momentum of the system.

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Question 4

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 5

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

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Question 6

In the figure,a mass of 31.77 kg is attached to a light string that is wrapped around a cylindrical spool of radius 10.0 cm and moment of inertia 4.00 kg • m².The spool is suspended from the ceiling,and the mass is then released from rest a distance 5.70 m above the floor.How long does it take to reach the floor?

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Question 7

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  down the plane?

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Question 8

A thin cylindrical shell is released from rest and rolls without slipping down an inclined ramp that makes an angle of 30° with the horizontal.How long does it take it to travel the first 3.1 m?

Accepted Answer

The acceleration of a rolling object without slipping is (2/3)g*sin(θ). For a 30° incline, this becomes (2/3)g*(1/2) = g/3. Using the equation s = (1/2)at^2, where s = 3.1 m and a = g/3, we solve for t to find t ≈ 1.6 s.

Question 9

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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Question 10

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 11

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 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 12

A very thin horizontal,2.00-m long,5.00-kg uniform beam that lies along the east-west direction is acted on by two forces.At the east end of the beam,a 200-N force pushes downward.At the west end of the beam,a 200-N force pushes upward.What is the angular acceleration of the beam?

Accepted Answer

The net torque on the beam is 400 N·m, and the moment of inertia of the beam about an axis through its center of mass and perpendicular to its length is (1/12)ML² = (1/12)(5.00 kg)(2.00 m)² = 1.67 kg·m². Therefore, the angular acceleration of the beam is α = τ/I = (400 N·m)/(1.67 kg·m²) = 240 rad/s² north.

Question 13

A uniform solid cylindrical log begins rolling without slipping down a ramp that rises 28.0° above the horizontal.After it has rolled 4.20 m along the ramp,what is the magnitude of the linear acceleration of its center of mass?

Accepted Answer

The correct answer is D.The linear acceleration of the center of mass of a rolling object is given by a = g*sin(theta), where g is the acceleration due to gravity and theta is the angle of the incline. In this case, theta = 28.0°, so a = 9.80 m/s^2 * sin(28.0°) = 4.60 m/s^2.

Question 14

A very thin uniform rod,2.40 m long and of weight 135 N,has a frictionless hinge at its lower end.It starts out vertically from rest and falls,pivoting about the hinge.Just as it has rotated through an angle of 55.0°,what is the downward acceleration of the end farthest from the hinge?

Accepted Answer

The downward acceleration of the end farthest from the hinge can be found using the formula for angular acceleration and the relationship between linear and angular acceleration. The angular acceleration can be calculated using the torque due to gravity and the moment of inertia of the rod. The linear acceleration is then given by the product of the angular acceleration and the distance from the hinge.

Question 15

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 16

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  ?

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Question 17

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  at the bottom,what is the height of the inclined plane?

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Question 18

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 19

A wheel has a radius of 0.40 m and is mounted on frictionless bearings.A block is suspended from a rope that is wound on the wheel and attached to it (see figure).The wheel is released from rest and the block descends 1.5 m in 2.00 s without any slipping of the rope.The tension in the rope during the descent of the block is 20 N.What is the moment of inertia of the wheel?

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Question 20

In the figure,a very light rope is wrapped around a wheel of radius R = 2.0 meters and does not slip.The wheel is mounted with frictionless bearings on an axle through its center.A block of mass 14 kg is suspended from the end of the rope.When the system is released from rest it is observed that the block descends 10 meters in 2.0 seconds.What is the moment of inertia of the wheel?

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Quiz 10: Dynamics of Rotational Motion

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 the other rotates in the opposite direction at 3.42 rad/s.Calculate the magnitude of the net angular momentum of the system.

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?

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

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 down the plane?