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Ch 10: Dynamics of Rotational Motion
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 15th EditionCh 10: Dynamics of Rotational MotionProblem 33b
Chapter 10, Problem 33b

A playground merry-go-round has radius 2.40 m2.40\(\text{ m}\) and moment of inertia 2100 kg m22100\(\text{ kg m}\)^2 about a vertical axle through its center, and it turns with negligible friction. A child applies an 18.0 N18.0\(\text{ N}\) force tangentially to the edge of the merry-go-round for 15.0 s15.0\(\text{ s}\). If the merry-go-round is initially at rest, how much work did the child do on the merry-go-round?

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First, understand that work done on an object is calculated using the formula: \( W = F \cdot d \cdot \cos(\theta) \), where \( F \) is the force applied, \( d \) is the distance over which the force is applied, and \( \theta \) is the angle between the force and the direction of motion. In this case, the force is tangential, so \( \theta = 0 \) degrees, and \( \cos(0) = 1 \).
Next, calculate the distance \( d \) that the edge of the merry-go-round travels. Since the force is applied tangentially, the distance can be found using the arc length formula: \( d = r \cdot \Delta\theta \), where \( r \) is the radius of the merry-go-round and \( \Delta\theta \) is the angular displacement in radians.
To find \( \Delta\theta \), use the relationship between angular acceleration \( \alpha \), time \( t \), and angular displacement: \( \Delta\theta = \frac{1}{2} \cdot \alpha \cdot t^2 \). First, calculate \( \alpha \) using the formula \( \alpha = \frac{\tau}{I} \), where \( \tau \) is the torque and \( I \) is the moment of inertia. Torque \( \tau \) is given by \( \tau = F \cdot r \).
Substitute the values into the torque formula: \( \tau = 18.0 \text{ N} \cdot 2.40 \text{ m} \). Then, use \( \alpha = \frac{\tau}{2100 \text{ kg} \cdot \text{m}^2} \) to find the angular acceleration.
Finally, substitute \( \alpha \) and \( t = 15.0 \text{ s} \) into \( \Delta\theta = \frac{1}{2} \cdot \alpha \cdot t^2 \) to find the angular displacement. Use this \( \Delta\theta \) to calculate \( d = r \cdot \Delta\theta \), and then substitute \( F \), \( d \), and \( \cos(0) = 1 \) into the work formula \( W = F \cdot d \cdot \cos(\theta) \) to find the work done by the child.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Moment of Inertia

Moment of inertia is a measure of an object's resistance to changes in its rotation. It depends on the mass distribution relative to the axis of rotation. For a merry-go-round, it quantifies how difficult it is to change its rotational speed, given its mass and shape.
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Intro to Moment of Inertia

Torque

Torque is the rotational equivalent of force, causing an object to rotate around an axis. It is calculated as the product of force and the radius at which the force is applied. In this scenario, the child's tangential force creates torque, initiating the merry-go-round's rotation.
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Work-Energy Principle

The work-energy principle states that work done on an object results in a change in its energy. For rotational motion, work is the product of torque and angular displacement. The child's work on the merry-go-round increases its rotational kinetic energy, starting from rest.
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Textbook Question

A playground merry-go-round has radius 2.40 m2.40\(\text{ m}\) and moment of inertia 2100 kg m22100\(\text{ kg m}\)^2 about a vertical axle through its center, and it turns with negligible friction. A child applies an 18.0 N18.0\(\text{ N}\) force tangentially to the edge of the merry-go-round for 15.0 s15.0\(\text{ s}\). If the merry-go-round is initially at rest, what is its angular speed after this 15.0 s15.0\(\text{ s}\) interval?

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