Problem 31 - University Physics with Modern Physics

Question

A playground merry-go-round has radius 2.40 \(\mathrm{m}\) and moment of inertia 2100 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) about a vertical axle through its center, and it turns with negligible friction. (a) A child applies an 18.0 -N force tangentially to the edge of the merry-go-round for 15.0 s. If the merry- go-round is initially at rest, what is its angular speed after this 15.0 -s interval? (b) How much work did the child do on the merry-go-round? (c) What is the average power supplied by the child?

Step-by-Step Solution

Answer

(a) 0.3086 rad/s, (b) 99.98 J, (c) 6.67 W.

Step 1: Identify Given Values

We are given several values:- Radius of the merry-go-round, \( r = 2.40 \, \text{m} \)- Moment of inertia, \( I = 2100 \, \text{kg} \cdot \text{m}^2 \)- Force applied, \( F = 18.0 \, \text{N} \)- Time for which force is applied, \( t = 15.0 \, \text{s} \)- Initial angular velocity, \( \omega_0 = 0 \, \text{rad/s} \) (since the merry-go-round starts from rest).

Step 2: Calculate Torque Applied

The torque \( \tau \) applied by a force tangent to a circle is given by the formula:\[ \tau = r \times F \]Substituting the known values:\[ \tau = 2.40 \, \text{m} \times 18.0 \, \text{N} = 43.2 \, \text{N} \cdot \text{m} \]

Step 3: Determine Angular Acceleration

Because torque \( \tau \) is also equal to the moment of inertia \( I \) times the angular acceleration \( \alpha \), we use the formula:\[ \tau = I \alpha \]Thus, the angular acceleration \( \alpha \) is:\[ \alpha = \frac{\tau}{I} = \frac{43.2 \, \text{N} \cdot \text{m}}{2100 \, \text{kg} \cdot \text{m}^2} = 0.02057 \, \text{rad/s}^2 \]

Step 4: Calculate Angular Speed After 15 Seconds

The angular speed \( \omega \) can be found using the formula for rotational motion with constant acceleration:\[ \omega = \omega_0 + \alpha t \]Substituting the initial angular velocity \( \omega_0 \), angular acceleration \( \alpha \), and time \( t \):\[ \omega = 0 + (0.02057 \, \text{rad/s}^2)(15.0 \, \text{s}) = 0.3086 \, \text{rad/s} \]

Step 5: Calculate Work Done by the Child

The work done by a torque is given by:\[ W = \tau \theta \]First, find the angular displacement \( \theta \) using:\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]Substituting the known values:\[ \theta = 0 + \frac{1}{2} \times 0.02057 \, \text{rad/s}^2 \times (15.0 \, \text{s})^2 = 2.314 \, \text{rad} \]Now, calculate the work:\[ W = 43.2 \, \text{N} \cdot \text{m} \times 2.314 \, \text{rad} = 99.98 \, \text{J} \]

Step 6: Determine Average Power Supplied

The average power \( P \) is given by work done over time:\[ P = \frac{W}{t} \]Substituting the values:\[ P = \frac{99.98 \, \text{J}}{15.0 \, \text{s}} = 6.67 \, \text{W} \]

Key Concepts

Torque Calculation

Torque is a fundamental concept in rotational dynamics, which deals with how forces cause rotational motion. Imagine opening a door; the force applied further from the hinges results in a greater turning effect. This is due to torque. In simple terms, torque is the rotational equivalent of linear force.
The formula for calculating torque is:

• \( \tau = r \times F \)

where \(\tau\) is the torque, \(r\) is the radius or distance from the pivot point to where the force is applied, and \(F\) is the force applied.
In our merry-go-round example, the child applies an 18.0 N force at the edge. With a radius of 2.40 m, the torque is calculated as:

• \[ \tau = 2.40 \, \text{m} \times 18.0 \, \text{N} = 43.2 \, \text{N} \cdot \text{m} \]

This demonstrates how a force applied at a distance can cause rotational movement, known as angular acceleration.

Moment of Inertia

The moment of inertia is a property of any object that rotates, determining how much torque is needed for a desired angular acceleration. It’s akin to mass in linear motion but for rotation. More mass spread out from the axis means a higher moment of inertia, making it harder to spin the object.
The formula to relate torque and angular acceleration is:

• \( \tau = I \alpha \)

where \(I\) is the moment of inertia and \(\alpha\) is the angular acceleration.
Given the moment of inertia of the merry-go-round is \(2100 \, \text{kg} \cdot \text{m}^2\), the angular acceleration can be found as:

• \[ \alpha = \frac{\tau}{I} = \frac{43.2 \, \text{N} \cdot \text{m}}{2100 \, \text{kg} \cdot \text{m}^2} = 0.02057 \, \text{rad/s}^2 \]

A higher moment of inertia means any force applied will result in a smaller angular acceleration compared to a smaller moment of inertia.

Work and Power in Physics

In physics, work is done when a force causes displacement. For rotational systems, work involves torque and angular displacement. When the child spins the merry-go-round, they do work to overcome inertia and friction, increasing rotational kinetic energy.
The formula for rotational work is:

• \( W = \tau \theta \)

where \(\theta\) is angular displacement.
Using computed values for torque and \(\theta = 2.314 \, \text{rad}\), the work done is:

• \[ W = 43.2 \, \text{N} \cdot \text{m} \times 2.314 \, \text{rad} = 99.98 \, \text{J} \]

Power, the rate at which work is done, calculated as:

• \( P = \frac{W}{t} = \frac{99.98 \, \text{J}}{15.0 \, \text{s}} = 6.67 \, \text{W} \)

Power informs us about how quickly energy is being used or transferred. Here, it is the average energy per second supplied by the child to spin the merry-go-round.