Problem 19 - College Physics

Question

A playground merry-go-round has a radius of 4.40 \(\mathrm{m}\) and a moment of inertia of 245 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) and turns with negligible friction about a vertical axle through its center. (a) A child applies a 25.0 \(\mathrm{N}\) force tangentially to the edge of the merry-go-round for 20.0 s. If the merry-go-round is initially at rest, what is its angular velocity after this 20.0 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) 8.98 rad/s, (b) 9,878 J, (c) 493.9 W.

Step 1: Understanding the Problem

In this problem, we need to find the angular velocity of a merry-go-round given a force applied by a child, the work done by the child, and the average power supplied.

Step 2: Calculate Torque

The torque (τ) applied about the center of the merry-go-round can be calculated using the formula: \( \tau = F \times r \)where \( F = 25.0 \ \mathrm{N} \) and \( r = 4.40 \ \mathrm{m} \). Plugging in these values gives:\( \tau = 25.0 \ \mathrm{N} \times 4.40 \ \mathrm{m} = 110.0 \ \mathrm{N} \cdot \mathrm{m} \).

Step 3: Calculate Angular Acceleration

Using the moment of inertia \( I = 245 \ \mathrm{kg} \cdot \mathrm{m}^2 \), the angular acceleration \( \alpha \) can be found using the relation: \( \tau = I \times \alpha \) which rearranges to: \( \alpha = \frac{\tau}{I} = \frac{110.0}{245} \approx 0.449 \ \mathrm{rad/s}^2 \).

Step 4: Calculate Angular Velocity

Using the angular acceleration \( \alpha \) and the time \( t = 20.0 \ \mathrm{s} \), the angular velocity \( \omega \) after 20 seconds can be found using: \( \omega = \omega_0 + \alpha \times t \)where \( \omega_0 = 0 \ \mathrm{rad/s} \) (initially at rest):\( \omega = 0 + 0.449 \times 20.0 = 8.98 \ \mathrm{rad/s} \).

Step 5: Calculate Work Done

The work done by the child, \( W \), can be found using the formula:\( W = \tau \times \theta \)The angle \( \theta \) in radians is given by:\( \theta = \frac{1}{2} \times \alpha \times t^2 \)So, \( \theta = \frac{1}{2} \times 0.449 \times (20.0)^2 = 89.8 \ \mathrm{rad} \).Thus, \( W = 110.0 \times 89.8 = 9,878 \ \mathrm{J} \).

Step 6: Calculate Average Power

The average power \( P \) is the work done divided by time:\( P = \frac{W}{t} = \frac{9,878}{20.0} = 493.9 \ \mathrm{W} \).

Key Concepts

Torque

Torque is a measure of how much a force acting on an object causes that object to rotate. It's essential in understanding rotational motion. Torque (\( \tau \)) depends on two main factors:

• Magnitude of the force (\( F \)).
• Distance from the pivot point, which is called the lever arm (\( r \)).

The formula to calculate torque is:\[ \tau = F \times r \]where, in our merry-go-round example, a 25 \( \mathrm{N} \) force is applied at a 4.40 \( \mathrm{m} \) lever arm. Therefore, the torque generated is 110 \( \mathrm{N} \cdot \mathrm{m} \). This calculated torque helps determine how the merry-go-round spins in response to the force.

Moment of Inertia

Moment of inertia (\( I \)) is a property of a rotating body, which determines how much torque is needed for a desired angular acceleration. Think of it as the "rotational mass," representing resistance to change in rotational motion. The formula relating torque and moment of inertia is:\[ \tau = I \times \alpha \]where:

• \( \tau \) is the torque.
• \( I \) is the moment of inertia.
• \( \alpha \) is the angular acceleration.

In our example, the merry-go-round has a moment of inertia of 245 \( \mathrm{kg} \cdot \mathrm{m}^2 \). This value is used with the torque to find the angular acceleration. As a result, we calculated an angular acceleration of approximately 0.449 \( \mathrm{rad/s}^2 \). This explains how quickly the merry-go-round picks up speed when the force is applied.

Work Done

Work done on a rotating object is similar to the linear work concept but involves torque and angular displacement. It represents the energy transferred when a force acts over a distance.The work done (\( W \)) in the context of rotation is given by:\[ W = \tau \times \theta \]where:

• \( \tau \) is the torque.
• \( \theta \) is the angular displacement, in radians.

In this exercise, the angular displacement is found using the formula:\[ \theta = \frac{1}{2} \times \alpha \times t^2 \]Using the known values, the work done by the child on the merry-go-round is approximately 9,878 \( \mathrm{J} \). This amount of energy has been transferred to the merry-go-round to get it spinning from rest.

Average Power

Average power (\( P \)) measures how quickly work is done or energy is transferred over time. It is an essential concept in understanding how efficient the energy transfer is.The formula to calculate average power is:\[ P = \frac{W}{t} \]where:

• \( W \) is the work done.
• \( t \) is the time over which the work is done.

With the work done by the child being 9,878 \( \mathrm{J} \) and the time being 20 seconds, the average power supplied by the child is 493.9 \( \mathrm{W} \). This value helps in understanding how much energy per second the child transfers to the merry-go-round, giving insight into the child's contribution to its motion.