/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 A disk of clay is rotating with ... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 13

A disk of clay is rotating with angular velocity \(\omega .\) A blob of clay is stuck to the outer rim of the disk, and it has a mass \(\frac{1}{10}\) of that of the disk. If the blob detaches and flies off tangent to the outer rim of the disk, what is the angular velocity of the disk after the blob separates? a) \(\frac{5}{6} \omega\) b) \(\frac{10}{11} \omega\) c) \(\omega\) d) \(\frac{11}{10} \omega\) e) \(\frac{6}{5} \omega\)

Short Answer

Expert verified
Answer: a) \(\frac{5}{6} \omega\)

Step by step solution

01

Define the initial and final angular momentum of the system

Let's call the initial mass of the disk \(M_d\), the mass of the blob of clay \(M_b\), and the radius of the disk \(R\). The initial angular momentum of the system (\(L_i\)) is the sum of the angular momentum of the disk and the angular momentum of the blob. The final angular momentum of the system (\(L_f\)) is just the angular momentum of the disk after the blob has separated since the blob is flying off tangent to the outer rim of the disk. We know the mass of the blob is \(\frac{1}{10}\) of that of the disk, so \(M_b = \frac{M_d}{10}\).
02

Calculate the initial and final moments of inertia

To find the initial and final angular momentum of the system, we need to calculate the initial and final moments of inertia. Let's denote the moment of inertia of the disk as \(I_d\) and the moment of inertia of the blob as \(I_b\). Initially, both the disk and the blob rotate around the center of the disk, so the initial moment of inertia of the blob (\(I_{b_i}\)) includes the effect of its rotation around the disk center: \(I_d = \frac{1}{2} M_d R^2\) (Moment of inertia for a disk) \(I_{b_i} = M_b R^2\) (Moment of inertia for a point mass at a distance R) The final moment of inertia for the disk is the same as the initial moment of inertia since its shape and mass haven't changed: \(I_{d_f} = I_d = \frac{1}{2} M_d R^2\)
03

Calculate the initial and final angular momentum of the system

We can now calculate the initial and final angular momentum of the system: Initial angular momentum: \(L_i = (\frac{1}{2} M_d R^2 + M_b R^2)\omega = (\frac{1}{2} M_d R^2 + \frac{M_d}{10} R^2)\omega\) Final angular momentum: \(L_f = I_{d_f} \omega_f\)
04

Apply conservation of angular momentum

Since no external torques act on the system, the total angular momentum must be conserved. Therefore, \(L_i = L_f\). We can plug in the expressions for the initial and final angular momenta and solve for \(\omega_f\): \((\frac{1}{2} M_d R^2 + \frac{M_d}{10} R^2)\omega = \frac{1}{2} M_d R^2 \omega_f\) Simplifying the equation and solving for \(\omega_f\), we get: \(\omega_f = \frac{5}{6}\omega\)
05

Answer the question

The angular velocity of the disk after the blob separates is \(\frac{5}{6} \omega\). Therefore, the correct answer is: a) \(\frac{5}{6} \omega\)

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Most popular questions from this chapter

Consider a cylinder and a hollow cylinder, rotating about an axis going through their centers of mass. If both objects have the same mass and the same radius, which object will have the larger moment of inertia? a) The moment of inertia will be the same for both objects. b) The solid cylinder will have the larger moment of inertia because its mass is uniformly distributed. c) The hollow cylinder will have the larger moment of inertia because its mass is located away from the axis of rotation.

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 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 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 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?
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