#78 ### Physics Problem Set**Problem 78: Angular Speed of a Merry-Go-Round**A 3.0-meter-diameter merry-go-round with a mass of 250 kg is spinning at 20 rpm. John runs around the merry-go-round at 5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. John's mass is 30 kg. What is the merry-go-round's angular speed, in rpm, after John jumps on?**Problem 79: Rotational Dynamics of Disks**- Disk A, with a mass of 2.0 kg and a radius of 40 cm, rotates clockwise about a frictionless vertical axle at 30 rev/s. - Disk B, also 2.0 kg but with a radius of 20 cm, rotates counterclockwise about that same axle, but at a greater height than disk A, at 30 rev/s. Disk B slides down the axle until it lands on top of disk A, after which they rotate together. - After the collision, what is their common angular speed (in rev/s) and in which direction do they rotate?**Explanation of Concepts:**To solve these problems, the concepts of conservation of angular momentum and rotational inertia are crucial. Here’s a brief overview:- **Angular Momentum (\(L\))**: This is given by \(L = I \cdot \omega\), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity. For a system with no external torques, angular momentum is conserved.- **Moment of Inertia (\(I\))**: This depends on the mass distribution relative to the axis of rotation. For a solid disk, \(I = \frac{1}{2} m r^2 \).- **Angular Velocity (\(\omega\))**: This is the rotational equivalent of linear velocity and indicates how fast an object spins around a fixed point.In Problem 78, you need to account for the change in the merry-go-round’s angular speed after John jumps on, factoring in both John and the merry-go-round's initial and final moments of inertia. In Problem 79, the total angular momentum before and after the collision is used to find the final common angular speed when the two disks rotate together.These foundational principles assist in analyzing and solving the rotational dynamics problems presented.

The merry-go-round's angular speed after John jumps on is 22.7 rpm.

78. I A 3.0-m-diameter merry-go-round with a mass of 250 kg is spinning at 20 rpm. John runs around the merry-go-round at 5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. John's mass is 30 kg. What is the merry-go- round's angular speed, in rpm, after John jumps on?

78. I A 3.0-m-diameter merry-go-round with a mass of 250 kg is spinning at 20 rpm. John runs around the merry-go-round at 5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. John's mass is 30 kg. What is the merry-go- round's angular speed, in rpm, after John jumps on?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Torque

Torque is the rotational equivalent of linear force in physics and mechanics. Based on the course of study, it is also known as the moment of force, rotational force, or turning effect.

Question

#78

### Physics Problem Set

**Problem 78: Angular Speed of a Merry-Go-Round**

A 3.0-meter-diameter merry-go-round with a mass of 250 kg is spinning at 20 rpm. John runs around the merry-go-round at 5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. John's mass is 30 kg. What is the merry-go-round's angular speed, in rpm, after John jumps on?

**Problem 79: Rotational Dynamics of Disks**

- Disk A, with a mass of 2.0 kg and a radius of 40 cm, rotates clockwise about a frictionless vertical axle at 30 rev/s.
- Disk B, also 2.0 kg but with a radius of 20 cm, rotates counterclockwise about that same axle, but at a greater height than disk A, at 30 rev/s. Disk B slides down the axle until it lands on top of disk A, after which they rotate together.
- After the collision, what is their common angular speed (in rev/s) and in which direction do they rotate?

**Explanation of Concepts:**

To solve these problems, the concepts of conservation of angular momentum and rotational inertia are crucial. Here’s a brief overview:

- **Angular Momentum (\(L\))**: This is given by \(L = I \cdot \omega\), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity. For a system with no external torques, angular momentum is conserved.
- **Moment of Inertia (\(I\))**: This depends on the mass distribution relative to the axis of rotation. For a solid disk, \(I = \frac{1}{2} m r^2 \).
- **Angular Velocity (\(\omega\))**: This is the rotational equivalent of linear velocity and indicates how fast an object spins around a fixed point.

In Problem 78, you need to account for the change in the merry-go-round’s angular speed after John jumps on, factoring in both John and the merry-go-round's initial and final moments of inertia.

In Problem 79, the total angular momentum before and after the collision is used to find the final common angular speed when the two disks rotate together.

These foundational principles assist in analyzing and solving the rotational dynamics problems presented.

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