Problem 85 - Fundamentals of Physics

Question

A girl of mass \(M\) stands on the rim of a frictionless merrygo-round of radius \(R\) and rotational inertia \(I\) that is not moving. She throws a rock of mass \(m\) horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is \(v .\) Afterward, what are (a) the angular speed of the merry-go-round and (b) the linear speed of the girl?

Step-by-Step Solution

Answer

(a) \( \omega = - \frac{mRv}{I} \); (b) \( v_{girl} = -\frac{mR^2v}{I} \).

Step 1: Understand the Problem

A girl stands on a merry-go-round, which is initially at rest, and throws a rock tangentially. We need to find the effects of this action on the merry-go-round's angular speed and the girl's linear speed.

Step 2: Apply the Law of Conservation of Angular Momentum

Initially, the angular momentum is zero because the merry-go-round is at rest. The angular momentum will be conserved when the girl throws the rock. Therefore, the initial angular momentum equals the final angular momentum. Let \( L_i = L_f \), where \( L_f = I \omega + mRv \). Here, \( \omega \) is the angular speed of the merry-go-round, and \( mRv \) is the angular momentum of the rock.

Step 3: Solve for the Angular Speed of the Merry-Go-Round

Set up the equation from the conservation of angular momentum: \[ 0 = I \omega + mRv \] Rearrange to solve for \( \omega \): \[ \omega = - \frac{mRv}{I} \]. This negative sign indicates that the merry-go-round rotates in the opposite direction of the rock's throw.

Step 4: Determine the Linear Speed of the Girl

The linear speed of the girl relative to the ground can be determined by considering the relationship between angular speed and linear speed. The formula for this relationship is \( v = \omega R \). Substitute the expression found for \( \omega \): \[ v_{girl} = -\frac{mRv}{I} \times R \]Simplifying, we find: \[ v_{girl} = -\frac{mR^2v}{I} \].

Key Concepts

Angular Speed

When you think about **angular speed**, you might imagine how quickly something spins or rotates. In this exercise, we're focusing on a merry-go-round. Angular speed is how fast the merry-go-round spins after the girl throws the rock. It's important because it tells us the effect of the throw.
First, we need to apply the law of conservation of angular momentum, which means the total angular momentum in a closed system remains constant. Initially, the merry-go-round is not spinning, so its angular momentum is zero.
When the girl throws the rock, it gains angular momentum equal to \( mRv \), where \( m \) is the mass of the rock, \( R \) is the radius of the merry-go-round, and \( v \) is the speed of the rock. This angular momentum is transferred to the merry-go-round, causing it to spin with angular speed \( \omega \).
Using the formula \( I \omega + mRv = 0 \), where \( I \) is the rotational inertia of the merry-go-round, we solve for \( \omega \):

• \( \omega = - \frac{mRv}{I} \)

The negative sign indicates the merry-go-round spins in the opposite direction of the rock's throw.

Linear Speed

Linear speed refers to how fast something moves along a straight path. In our case, we want to know how fast the girl moves after she throws the rock. Although she stands on the revolving edge, her motion can still be described using concepts of linear speed.
To find the girl's linear speed, we use the relationship between angular speed and linear speed. This can be formulated as \( v = \omega R \). By substituting the expression for \( \omega \) derived from conservation laws, we get:

• \( v_{girl} = -\frac{mRv}{I} \times R \)

Which simplifies to:

• \( v_{girl} = -\frac{mR^2v}{I} \)

This formula tells us how fast the girl is moving in a straight line after the rock is thrown
The negative sign tells us she moves opposite to the throw, consistent with the angular momentum conservation principle.

Rotational Inertia

**Rotational inertia**, also known as the moment of inertia, is a measure of an object's resistance to changes in its rotation. Think of it as the rotational equivalent of mass.
In this problem, the rotational inertia \( I \) of the merry-go-round helps determine how easily it starts to spin when the rock is thrown. A large rotational inertia means the merry-go-round won't start spinning quickly, while a smaller rotational inertia means it will spin more easily.
When the rock is thrown, the angular momentum gained by the merry-go-round depends on its rotational inertia. The formula \( I \omega + mRv = 0 \), derived from the law of conservation of angular momentum, shows this relation clearly.
Distributing the values properly helps us to understand the dynamics:

• The magnitude of rotational inertia determines how much angular speed \( \omega \) is produced for a given amount of angular momentum.
• The greater the inertia \( I \), the smaller the angular speed \( \omega \) for the same impulse.

Understanding rotational inertia is key to predicting how objects behave when subjected to rotational forces.