An angular impulse of 20 Nms is applied to a hollow cylinder of mass 2 kg and radius 20 cm. The change in its angular speed is :

Options

A. 25 rad/s
B. 2.5 rad/s
C. 250 rad/s
D. 2500 rad/s

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between angular impulse and angular momentum
Angular impulse (J) is equal to the change in angular momentum (ΔL). Mathematically, we can express this as:
\[ J = \Delta L \]
Where:
- \( J \) is the angular impulse.
- \( \Delta L \) is the change in angular momentum.

### Step 2: Express angular momentum in terms of moment of inertia and angular speed
The angular momentum (L) of a rotating object can be expressed as:
\[ L = I \cdot \omega \]
Where:
- \( I \) is the moment of inertia.
- \( \omega \) is the angular speed.

The change in angular momentum can be expressed as:
\[ \Delta L = I \cdot \Delta \omega \]
Where:
- \( \Delta \omega \) is the change in angular speed.

### Step 3: Relate angular impulse to change in angular speed
From the above relationships, we can write:
\[ J = I \cdot \Delta \omega \]
Thus, we can rearrange this to find the change in angular speed:
\[ \Delta \omega = \frac{J}{I} \]

### Step 4: Calculate the moment of inertia for a hollow cylinder
The moment of inertia (I) for a hollow cylinder (ring) about its central axis is given by:
\[ I = m \cdot r^2 \]
Where:
- \( m \) is the mass of the cylinder.
- \( r \) is the radius of the cylinder.

Given:
- Mass \( m = 2 \, \text{kg} \)
- Radius \( r = 20 \, \text{cm} = 0.2 \, \text{m} \)

Substituting the values:
\[ I = 2 \cdot (0.2)^2 = 2 \cdot 0.04 = 0.08 \, \text{kg m}^2 \]

### Step 5: Substitute values to find the change in angular speed
Now we can substitute the values of angular impulse and moment of inertia into the equation for change in angular speed:
Given \( J = 20 \, \text{Nms} \):
\[ \Delta \omega = \frac{20}{0.08} = 250 \, \text{rad/s} \]

### Final Answer
The change in angular speed is \( 250 \, \text{rad/s} \).

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