A 0.45 kg tetherball is attached to a pole and rotating in a horizontal circle of radius r₁ = 1.4 m and is circling at angular speed = 1.42 rad/s. As the rope wraps around the pole the radius of the circle shortens and became r2 = 0.9 m, that decreases the moment of inertia of the rotating tetherball. Neglecting air resistance what will be the angular speed of the ball after the rope is wrapped around the pole measured in kgm2/s (answer with 3 decimal places)?

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**Problem Statement: Conservation of Angular Momentum in Tetherball**

A 0.45 kg tetherball is attached to a pole and rotating in a horizontal circle of radius \( r_1 = 1.4 \) m and is circling at angular speed \( \omega_1 = 1.42 \) rad/s. As the rope wraps around the pole, the radius of the circle shortens and becomes \( r_2 = 0.9 \) m, which decreases the moment of inertia of the rotating tetherball. Neglecting air resistance, what will be the angular speed of the ball after the rope is wrapped around the pole, measured in kgm²/s (answer with 3 decimal places)?

**Analysis & Solution:**

To solve this problem, we use the principle of conservation of angular momentum. The angular momentum \(L\) of the tetherball must remain constant because there are no external torques acting on it.

The formula for angular momentum \(L\) is given by:
\[ L = I \omega \]

Where:
- \( I \) is the moment of inertia
- \( \omega \) is the angular speed

For a point mass \( m \) rotating at a radius \( r \):
\[ I = m r^2 \]

Initially:
\[ L_1 = I_1 \omega_1 = m r_1^2 \omega_1 \]

After the radius changes:
\[ L_2 = I_2 \omega_2 = m r_2^2 \omega_2 \]

Since angular momentum is conserved:
\[ L_1 = L_2 \]
\[ m r_1^2 \omega_1 = m r_2^2 \omega_2 \]

Solving for \( \omega_2 \):
\[ \omega_2 = \frac{r_1^2 \omega_1}{r_2^2} \]

Substituting the given values:
\[ m = 0.45 \, \text{kg} \]
\[ r_1 = 1.4 \, \text{m} \]
\[ \omega_1 = 1.42 \, \text{rad/s} \]
\[ r_2 = 0.9 \, \text{m} \]

\[ \omega_2 = \frac{(1.4 \, \text{m})^2 \
Transcribed Image Text:**Problem Statement: Conservation of Angular Momentum in Tetherball** A 0.45 kg tetherball is attached to a pole and rotating in a horizontal circle of radius \( r_1 = 1.4 \) m and is circling at angular speed \( \omega_1 = 1.42 \) rad/s. As the rope wraps around the pole, the radius of the circle shortens and becomes \( r_2 = 0.9 \) m, which decreases the moment of inertia of the rotating tetherball. Neglecting air resistance, what will be the angular speed of the ball after the rope is wrapped around the pole, measured in kgm²/s (answer with 3 decimal places)? **Analysis & Solution:** To solve this problem, we use the principle of conservation of angular momentum. The angular momentum \(L\) of the tetherball must remain constant because there are no external torques acting on it. The formula for angular momentum \(L\) is given by: \[ L = I \omega \] Where: - \( I \) is the moment of inertia - \( \omega \) is the angular speed For a point mass \( m \) rotating at a radius \( r \): \[ I = m r^2 \] Initially: \[ L_1 = I_1 \omega_1 = m r_1^2 \omega_1 \] After the radius changes: \[ L_2 = I_2 \omega_2 = m r_2^2 \omega_2 \] Since angular momentum is conserved: \[ L_1 = L_2 \] \[ m r_1^2 \omega_1 = m r_2^2 \omega_2 \] Solving for \( \omega_2 \): \[ \omega_2 = \frac{r_1^2 \omega_1}{r_2^2} \] Substituting the given values: \[ m = 0.45 \, \text{kg} \] \[ r_1 = 1.4 \, \text{m} \] \[ \omega_1 = 1.42 \, \text{rad/s} \] \[ r_2 = 0.9 \, \text{m} \] \[ \omega_2 = \frac{(1.4 \, \text{m})^2 \
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