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Chapter 10: Problem 23

A solid sphere of radius \(R\) and mass \(M\) is placed at a height \(h_{0}\) on an inclined plane of slope \(\theta\). When released, it rolls without slipping to the bottom of the incline. Next, a cylinder of same mass and radius is released on the same incline. From what height \(h\) should it be released in order to have the same speed as the sphere at the bottom?

Short Answer

Expert verified
Answer: The cylinder should be released from a height of \(\frac{15}{14}h_0\), where \(h_0\) is the initial height of the solid sphere.

Step by step solution

01

Write the Conservation of Mechanical Energy for the Sphere

The total mechanical energy is the sum of gravitational potential energy and kinetic energy. For the sphere, the initial mechanical energy at height \(h_0\) is all gravitational potential energy. At the bottom of the incline, this potential energy has been converted into kinetic energy (both translational and rotational). The equation for the conservation of mechanical energy is: Initial Energy = Final Energy $$Mgh_0 = \frac{1}{2}Mv_{s}^2 + \frac{1}{2}I_{sphere}\omega^2$$
02

Find the Moment of Inertia and Angular Velocity for the Sphere and the Cylinder

Moment of inertia for a solid sphere is \(I_{sphere} = \frac{2}{5}MR^2\), and for a cylinder it is \(I_{cylinder} = \frac{1}{2}MR^2\). Since both objects are rolling without slipping, we can relate their linear and angular velocities by: \(v = R\omega\). We can substitute the moment of inertia and convert angular velocity to linear velocity: $$Mgh_0 = \frac{1}{2}Mv_{s}^2 + \frac{1}{2}\left(\frac{2}{5}MR^2\right)\left(\frac{v_{s}}{R}\right)^2$$
03

Simplify the Energy Conservation Equation for the Sphere

Now, we can simplify the equation for the sphere by cancelling out terms and collecting like terms: $$gh_0 = \frac{1}{2}v_{s}^2 + \frac{1}{5}v_{s}^2$$ Combine the fractions: $$gh_0 = \frac{7}{10}v_{s}^2$$ Solving for the final velocity of the sphere: $$v_{s} = \sqrt{\frac{10}{7}gh_0}$$
04

Write the Conservation of Mechanical Energy for the Cylinder

Now we need to repeat the process for the cylinder. The initial mechanical energy at height \(h\) is all gravitational potential energy. At the bottom of the incline, this potential energy has been converted into kinetic energy (both translational and rotational). The equation for the conservation of mechanical energy is: Initial Energy = Final Energy $$Mgh = \frac{1}{2}Mv_{c}^2 + \frac{1}{2}I_{cylinder}\omega^2$$ Substitute the moment of inertia and convert angular velocity to linear velocity: $$Mgh = \frac{1}{2}Mv_{c}^2 + \frac{1}{2}\left(\frac{1}{2}MR^2\right)\left(\frac{v_{c}}{R}\right)^2$$
05

Simplify the Energy Conservation Equation for the Cylinder

Now, we can simplify the equation for the cylinder by cancelling out terms and collecting like terms: $$gh = \frac{1}{2}v_{c}^2 + \frac{1}{4}v_{c}^2$$ Combine the fractions: $$gh = \frac{3}{4}v_{c}^2$$
06

Set the Sphere's and Cylinder's Final Velocities Equal

Since we want the cylinder to have the same speed as the sphere at the bottom, we can set their final velocities equal: $$v_{s} = v_{c}$$ $$\sqrt{\frac{10}{7}gh_0} = \sqrt{\frac{4}{3}gh}$$
07

Solve for the Cylinder's Release Height

Now, we can simplify the equation and solve for the cylinder's release height: $$\frac{10}{7}gh_0 = \frac{4}{3}gh$$ Dividing both sides by \(g\) and multiplying by \(\frac{3}{4}\), we find the value for \(h\): $$h = \frac{3}{4}\left(\frac{10}{7}h_0\right)$$ $$h = \frac{15}{14}h_0$$ So the cylinder should be released from a height of \(\frac{15}{14}h_0\) to have the same speed as the sphere at the bottom of the incline.

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Most popular questions from this chapter

A ballistic pendulum consists of an arm of mass \(M\) and length \(L=0.48 \mathrm{~m} .\) One end of the arm is pivoted so that the arm rotates freely in a vertical plane. Initially, the arm is motionless and hangs vertically from the pivot point. A projectile of the same mass \(M\) hits the lower end of the arm with a horizontal velocity of \(V=3.6 \mathrm{~m} / \mathrm{s}\). The projectile remains stuck to the free end of the arm during their subsequent motion. Find the maximum angle to which the arm and attached mass will swing in each case: a) The arm is treated as an ideal pendulum, with all of its mass concentrated as a point mass at the free end. b) The arm is treated as a thin rigid rod, with its mass evenly distributed along its length.

In another race, a solid sphere and a thin ring roll without slipping from rest down a ramp that makes angle \(\theta\) with the horizontal. Find the ratio of their accelerations, \(a_{\text {ring }} / a_{\text {sphere }}\)

A uniform solid sphere of mass \(m\) and radius \(r\) is placed on a ramp inclined at an angle \(\theta\) to the horizontal. The coefficient of static friction between sphere and ramp is \(\mu_{s} .\) Find the maximum value of \(\theta\) for which the sphere will roll without slipping, starting from rest, in terms of the other quantities.

Does a particle traveling in a straight line have an angular momentum? Explain.

A sheet of plywood \(1.3 \mathrm{~cm}\) thick is used to make a cabinet door \(55 \mathrm{~cm}\) wide by \(79 \mathrm{~cm}\) tall, with hinges mounted on the vertical edge. A small 150 - \(\mathrm{g}\) handle is mounted \(45 \mathrm{~cm}\) from the lower hinge at the same height as that hinge. If the density of the plywood is \(550 \mathrm{~kg} / \mathrm{m}^{3},\) what is the moment of inertia of the door about the hinges? Neglect the contribution of hinge components to the moment of inertia.
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