A merry-go-round (radius \(R\), rotational inertia \(I_{\mathrm{i}}\) ) spins with negligible friction. Its initial angular velocity is \(\omega_{i}\) A child (mass \(m\) ) on the merry-go-round moves from the center out to the rim. (a) Calculate the angular velocity after the child moves out to the rim. (b) Calculate the rotational kinetic energy and angular momentum of the system (merry-go-round + child) before and after.

Short Answer

Expert verified
Question: A merry-go-round with moment of inertia \(I_i\) is spinning with an initial angular velocity \(\omega_i\). A child initially standing at the center moves out to the rim of the merry-go-round, a distance \(R\) away from the center. Calculate the new angular velocity, rotational kinetic energy, and angular momentum before and after the child moves to the rim, given the values for \(R\), \(I_{\mathrm{i}}\), \(m\), and \(\omega_{i}\). Answer: First, calculate the final angular velocity using the conservation of angular momentum equation: \(\omega_{f} = \frac{I_{\mathrm{i}}\omega_{i}}{I_{\mathrm{i}} + mR^2}\) Next, calculate the rotational kinetic energy before and after the child moves: Before: \(K_{\text{initial}} = \frac{1}{2} I_{\mathrm{i}}\omega_{i}^2\) After: \(K_{\text{final}} = \frac{1}{2} I_{\mathrm{i}}\omega_{f}^2 + \frac{1}{2} mR^2\omega_{f}^2\) Finally, calculate the angular momentum before and after: Initial angular momentum: \(L_{\text{initial}} = I_{\mathrm{i}}\omega_{i}\) Final angular momentum: \(L_{\text{final}} = I_{\mathrm{i}}\omega_{\text{f}} + mR^2\omega_{\text{f}}\) Plug in the given values for \(R\), \(I_{\mathrm{i}}\), \(m\), and \(\omega_{i}\) to find the final angular velocity, rotational kinetic energy, and angular momentum before and after the child moves to the rim.

Step by step solution

01

Identify the initial angular momentum

Initially, the system consists of the merry-go-round and the child at the center. The total initial angular momentum (L_initial) is given by the sum of the merry-go-round's angular momentum and the child's angular momentum. Since the child is initially at the center, its angular momentum is 0. So the total initial angular momentum is just the merry-go-round's angular momentum: \(L_{\text{initial}} = I_{\mathrm{i}}\omega_{i}\)
02

Identify the final angular momentum

After the child moves to the rim, two components contribute to the total angular momentum: the merry-go-round and the child. The child is now a distance \(R\) away from the axis of rotation, and the merry-go-round's moment of inertia is still \(I_{\mathrm{i}}\). Let's use \(\omega_f\) to denote the final angular velocity. The total final angular momentum (L_final) is given by: \(L_{\text{final}} = I_{\mathrm{i}}\omega_{\text{f}} + mR^2\omega_{\text{f}}\)
03

Applying the conservation of angular momentum

We know that the total angular momentum is conserved, which means that \(L_{\text{initial}} = L_{\text{final}}\). We can use this to find the final angular velocity: \(I_{\mathrm{i}}\omega_{i} = I_{\mathrm{i}}\omega_{\text{f}} + mR^2\omega_{\text{f}}\) Now, we can solve for \(\omega_{\text{f}}\): \(\omega_{f} = \frac{I_{\mathrm{i}}\omega_{i}}{I_{\mathrm{i}} + mR^2}\)
04

Calculate the rotational kinetic energy before and after

Before the child moves, the rotational kinetic energy (K_initial) is just the energy of the spinning merry-go-round: \(K_{\text{initial}} = \frac{1}{2} I_{\mathrm{i}}\omega_{i}^2\) After the child moves, the total rotational kinetic energy (K_final) is the sum of the kinetic energy of the merry-go-round and the child: \(K_{\text{final}} = \frac{1}{2} I_{\mathrm{i}}\omega_{f}^2 + \frac{1}{2} mR^2\omega_{f}^2\)
05

Calculate the angular momentum before and after

We have already calculated the initial and final angular momentum in steps 1 and 2. To summarize: Initial angular momentum: \(L_{\text{initial}} = I_{\mathrm{i}}\omega_{i}\) Final angular momentum: \(L_{\text{final}} = I_{\mathrm{i}}\omega_{\text{f}} + mR^2\omega_{\text{f}}\) Now you can plug in the given values for \(R\), \(I_{\mathrm{i}}\), \(m\), and \(\omega_{i}\) to calculate the final angular velocity, rotational kinetic energy, and angular momentum before and after the child moves to the rim.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A hoop of \(2.00-\mathrm{m}\) circumference is rolling down an inclined plane of length \(10.0 \mathrm{m}\) in a time of \(10.0 \mathrm{s} .\) It started out from rest. (a) What is its angular velocity when it arrives at the bottom? (b) If the mass of the hoop, concentrated at the rim, is \(1.50 \mathrm{kg},\) what is the angular momentum of the hoop when it reaches the bottom of the incline? (c) What force(s) supplied the net torque to change the hoop's angular momentum? Explain. [Hint: Use a rotation axis through the hoop's center. \(]\) (d) What is the magnitude of this force?
A uniform diving board, of length \(5.0 \mathrm{m}\) and mass \(55 \mathrm{kg}\) is supported at two points; one support is located \(3.4 \mathrm{m}\) from the end of the board and the second is at \(4.6 \mathrm{m}\) from the end (see Fig. 8.19 ). What are the forces acting on the board due to the two supports when a diver of mass \(65 \mathrm{kg}\) stands at the end of the board over the water? Assume that these forces are vertical. (W) tutorial: plank) [Hint: In this problem, consider using two different torque equations about different rotation axes. This may help you determine the directions of the two forces.]
How long would a braking torque of \(4.00 \mathrm{N}\).m have to act to just stop a spinning wheel that has an initial angular momentum of $6.40 \mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s} ?$
A lawn sprinkler has three spouts that spray water, each \(15.0 \mathrm{cm}\) long. As the water is sprayed, the sprinkler turns around in a circle. The sprinkler has a total rotational inertia of $9.20 \times 10^{-2} \mathrm{kg} \cdot \mathrm{m}^{2} .\( If the sprinkler starts from rest and takes \)3.20 \mathrm{s}$ to reach its final speed of 2.2 rev/s, what force does each spout exert on the sprinkler?
A uniform rod of length \(L\) is free to pivot around an axis through its upper end. If it is released from rest when horizontal, at what speed is the lower end moving at its lowest point? [Hint: The gravitational potential energy change is determined by the change in height of the center of gravity.]
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free