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Chapter 10: Problem 2

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) \(v_{\mathrm{L}}=v_{\mathrm{S}}\)

Short Answer

Expert verified
a) v_L = v_S b) v_L = 2v_S c) v_L = 4v_S d) v_L = 0.5 v_S

Step by step solution

01

1. Find the moment of inertia of the solid sphere

For a solid sphere, the moment of inertia (I) is given by the formula: \(I = \frac{2}{5}mr^2\), where m is the mass and r is the radius of the sphere.
02

2. Set up an equation for conservation of mechanical energy

As the balls roll without slipping, conservation of mechanical energy states that the total mechanical energy at the beginning (potential energy) equals the total mechanical energy at the end (kinetic energy + rotational energy). Thus, we have: \(mgh = \frac{1}{2}mv^2 + \frac{1}{2}I{\omega}^2\) , where m: mass of the ball, g: acceleration due to gravity, h: height of the ball, v: linear velocity of the ball, I: moment of inertia, ω: angular velocity of the ball.
03

3. Relate linear velocity and angular velocity

The relationship between linear velocity (v) and angular velocity (ω) for a rolling sphere without slipping is given by: \(v = r\omega\). This relationship will be used to eliminate ω from the conservation of mechanical energy equation.
04

4. Apply the conservation of mechanical energy equation to both balls and divide the equations

For the small ball, we have: \(mg(1) = \frac{1}{2}m(v_{S})^2 + \frac{1}{2}(\frac{2}{5}mr^2)({\frac{v_{S}}{r}})^2\) And for the large ball, we have: \(mg(1) = \frac{1}{2}m(v_{L})^2 + \frac{1}{2}(\frac{2}{5}m(2r)^2)({\frac{v_{L}}{2r}})^2\) Now, we divide the equation for the large ball by the equation for the small ball to find the ratio of their velocities, \((\)v_L/v_S\()\):
05

5. Solve for the ratio of the velocities

After dividing the equations and simplifying, we get: \(\frac{v_{L}^2}{v_{S}^2} = \frac{1}{2}\) Taking the square root of both sides, we have: \(v_{L} = \sqrt{\frac{1}{2}}v_{S}\) Comparing with the given options, we have \(v_{\mathrm{L}} = 0.5 v_{\mathrm{S}}\). Therefore, the correct answer is (d) \(v_{\mathrm{L}} = 0.5 v_{\mathrm{S}}\).

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Most popular questions from this chapter

You are unwinding a large spool of cable. As you pull on the cable with a constant tension, what happens to the angular acceleration and angular velocity of the spool, assuming that the radius at which you are extracting the cable remains constant and there is no friction force? a) Both increase as the spool unwinds. b) Both decrease as the spool unwinds. c) Angular acceleration increases, and angular velocity decreases. d) Angular acceleration decreases, and angular velocity increases. e) It is impossible to tell.

A 2.00 -kg thin hoop with a 50.0 -cm radius rolls down a \(30.0^{\circ}\) slope without slipping. If the hoop starts from rest at the top of the slope, what is its translational velocity after it rolls \(10.0 \mathrm{~m}\) along the slope?

A sphere of radius \(R\) and mass \(M\) sits on a horizontal tabletop. A horizontally directed impulse with magnitude \(J\) is delivered to a spot on the ball a vertical distance \(h\) above the tabletop. a) Determine the angular and translational velocity of the sphere just after the impulse is delivered. b) Determine the distance \(h_{0}\) at which the delivered impulse causes the ball to immediately roll without slipping.

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.

A cylinder is rolling without slipping down a plane, which is inclined by an angle \(\theta\) relative to the horizontal. What is the work done by the friction force while the cylinder travels a distance \(s\) along the plane \(\left(\mu_{s}\right.\) is the coefficient of static friction between the plane and the cylinder)? a) \(+\mu_{s} m g s \sin \theta\) b) \(-\mu_{s} m g s \sin \theta\) c) \(+m g s \sin \theta\) d) \(-m g s \sin \theta\) e) No work done.
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