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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

An ice skater spins with her arms extended and then pulls her arms in and spins faster. Which statement is correct? a) Her kinetic energy of rotation does not change because, by conservation of angular momentum, the fraction by which her angular velocity increases is the same as the fraction by which her rotational inertia decreases. b) Her kinetic energy of rotation increases because of the work she does to pull her arms in. c) Her kinetic energy of rotation decreases because of the decrease in her rotational inertia; she loses energy because she gradually gets tired.

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