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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 professor doing a lecture demonstration stands at the center of a frictionless turntable, holding 5.00 -kg masses in each hand with arms extended so that each mass is \(1.20 \mathrm{~m}\) from his centerline. A (carefully selected!) student spins the professor up to a rotational speed of \(1.00 \mathrm{rpm} .\) If he then pulls his arms in by his sides so that each mass is \(0.300 \mathrm{~m}\) from his centerline, what is his new rotation rate? Assume that his rotational inertia without the masses is \(2.80 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\), and neglect the effect on the rotational inertia of the position of his arms, since their mass is small compared to the mass of the body.

Two solid steel balls, one small and one large, are on an inclined plane. The large ball has a diameter twice as large as that of the small ball. Starting from rest, the two balls roll without slipping down the incline until their centers of mass are \(1 \mathrm{~m}\) below their starting positions. What is the speed of the large ball \(\left(v_{\mathrm{L}}\right)\) relative to that of the small ball \(\left(v_{\mathrm{S}}\right)\) after rolling \(1 \mathrm{~m} ?\) a) \(v_{\mathrm{L}}=4 v_{\mathrm{S}}\) d) \(v_{\mathrm{L}}=0.5 v_{\mathrm{S}}\) b) \(v_{\mathrm{L}}=2 v_{\mathrm{S}}\) e) \(v_{\mathrm{L}}=0.25 v_{\mathrm{S}}\) c) \(v_{\mathrm{L}}=v_{\mathrm{S}}\)

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.

A circular platform of radius \(R_{p}=4.00 \mathrm{~m}\) and mass \(M_{\mathrm{p}}=400 .\) kg rotates on frictionless air bearings about its vertical axis at 6.00 rpm. An 80.0 -kg man standing at the very center of the platform starts walking \((\) at \(t=0)\) radially outward at a speed of \(0.500 \mathrm{~m} / \mathrm{s}\) with respect to the platform. Approximating the man by a vertical cylinder of radius \(R_{\mathrm{m}}=0.200 \mathrm{~m}\) determine an equation (specific expression) for the angular velocity of the platform as a function of time. What is the angular velocity when the man reaches the edge of the platform?

A uniform solid sphere of radius \(R, \operatorname{mass} M,\) and moment of inertia \(I=\frac{2}{5} M R^{2}\) is rolling without slipping along a horizontal surface. Its total kinetic energy is the sum of the energies associated with translation of the center of mass and rotation about the center of mass. Find the fraction of the sphere's total kinetic energy that is attributable to rotation.
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