### Physics Problem: Rotational Motion of a Merry-Go-Round#### Problem Statement:A certain merry-go-round is accelerated uniformly from rest and attains an angular speed of 2 radians per second in the first 5 seconds. If the net applied torque is 1200 N·m, what is the moment of inertia of the merry-go-round?#### Explanation:This problem involves the concepts of rotational motion and requires the application of the equations of motion to solve for the moment of inertia.**Given Data:**- Angular acceleration (\(\alpha\)): To be calculated- Final angular speed (\(\omega\)): 2 rad/s- Time interval (\(t\)): 5 seconds- Net applied torque (\(\tau\)): 1200 N·m**Steps to solve:**1. **Calculate Angular Acceleration:** Using the rotational kinematic equation: \[ \omega = \alpha t \] Rearrange to solve for angular acceleration (\(\alpha\)): \[ \alpha = \frac{\omega}{t} = \frac{2 \, \text{rad/s}}{5 \, \text{s}} = 0.4 \, \text{rad/s}^2 \]2. **Compute the Moment of Inertia:** Using Newton's second law for rotation: \[ \tau = I \alpha \] Rearrange to solve for the moment of inertia (\(I\)): \[ I = \frac{\tau}{\alpha} = \frac{1200 \, \text{N·m}}{0.4 \, \text{rad/s}^2} = 3000 \, \text{kg·m}^2 \]**Final Answer:**The moment of inertia of the merry-go-round is \(3000 \, \text{kg·m}^2\).---This explanation is designed to help students understand the approach and solution to problems involving rotational motion and torque in physics.

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A certain merry- go-round is accele rated uniformiy from rest and atains an angular speed 1.2 rad ls inthe first 18 seconds. If the net applied torq ue Is 1200 N. M, what is the moment of inertia of the merry-go-

A certain merry- go-round is accele rated uniformiy from rest and atains an angular speed 1.2 rad ls inthe first 18 seconds. If the net applied torq ue Is 1200 N. M, what is the moment of inertia of the merry-go-

College Physics

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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

Angular speed, acceleration and displacement

Angular acceleration is defined as the rate of change in angular velocity with respect to time. It has both magnitude and direction. So, it is a vector quantity.

Angular Position

Before diving into angular position, one should understand the basics of position and its importance along with usage in day-to-day life. When one talks of position, it’s always relative with respect to some other object. For example, position of earth with respect to sun, position of school with respect to house, etc. Angular position is the rotational analogue of linear position.

Question

### Physics Problem: Rotational Motion of a Merry-Go-Round

#### Problem Statement:
A certain merry-go-round is accelerated uniformly from rest and attains an angular speed of 2 radians per second in the first 5 seconds. If the net applied torque is 1200 N·m, what is the moment of inertia of the merry-go-round?

#### Explanation:
This problem involves the concepts of rotational motion and requires the application of the equations of motion to solve for the moment of inertia.

**Given Data:**
- Angular acceleration (\(\alpha\)): To be calculated
- Final angular speed (\(\omega\)): 2 rad/s
- Time interval (\(t\)): 5 seconds
- Net applied torque (\(\tau\)): 1200 N·m

**Steps to solve:**

1. **Calculate Angular Acceleration:**
Using the rotational kinematic equation:
\[
\omega = \alpha t
\]
Rearrange to solve for angular acceleration (\(\alpha\)):
\[
\alpha = \frac{\omega}{t} = \frac{2 \, \text{rad/s}}{5 \, \text{s}} = 0.4 \, \text{rad/s}^2
\]

2. **Compute the Moment of Inertia:**
Using Newton's second law for rotation:
\[
\tau = I \alpha
\]
Rearrange to solve for the moment of inertia (\(I\)):
\[
I = \frac{\tau}{\alpha} = \frac{1200 \, \text{N·m}}{0.4 \, \text{rad/s}^2} = 3000 \, \text{kg·m}^2
\]

**Final Answer:**
The moment of inertia of the merry-go-round is \(3000 \, \text{kg·m}^2\).

---
This explanation is designed to help students understand the approach and solution to problems involving rotational motion and torque in physics.

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