A billiard ball of mass m and radius r, when hit in a horizontal direction by a cue at a height h above its centre, acquired a linear velocity `v_(0).` The angular velocity `omega_(0)` acquired by the ball is

Options

A. `(2v_(0)h)/(5r^(2))`
B. `(5v_(0)h)/(2r^(2))`
C. `(5v_(0)r^(2))/(5h)`
D. `(5v_(0)r^(2))/(2h)`

Answer

To find the angular velocity \( \omega_0 \) acquired by the billiard ball when it is hit, we can follow these steps:

### Step 1: Understand the problem
The billiard ball is hit at a height \( h \) above its center, resulting in a linear velocity \( v_0 \). We need to determine the angular velocity \( \omega_0 \) that the ball acquires.

### Step 2: Identify the forces and torques
When the ball is struck, a force \( F \) is applied at a distance \( h \) from the center of mass. This creates a torque \( \tau \) about the center of mass.

### Step 3: Calculate the torque
The torque \( \tau \) can be calculated using the formula:
\[ \tau = F \cdot h \]

### Step 4: Relate torque to angular acceleration
The torque is also related to the moment of inertia \( I \) and angular acceleration \( \alpha \) by the equation:
\[ \tau = I \cdot \alpha \]
For a solid sphere, the moment of inertia \( I \) is given by:
\[ I = \frac{2}{5} m r^2 \]
Thus, we can write:
\[ F \cdot h = \frac{2}{5} m r^2 \cdot \alpha \]

### Step 5: Relate linear and angular acceleration
The angular acceleration \( \alpha \) can be expressed in terms of angular velocity \( \omega_0 \) and time \( t \):
\[ \alpha = \frac{\omega_0}{t} \]
Substituting this into the torque equation gives:
\[ F \cdot h = \frac{2}{5} m r^2 \cdot \frac{\omega_0}{t} \]

### Step 6: Express force in terms of linear acceleration
The force \( F \) can be expressed as:
\[ F = m \cdot a \]
where \( a \) is the linear acceleration. Since the ball acquires a linear velocity \( v_0 \) over time \( t \), we can write:
\[ a = \frac{v_0}{t} \]
Thus, substituting for \( F \):
\[ F = m \cdot \frac{v_0}{t} \]

### Step 7: Substitute back into the torque equation
Now substituting \( F \) back into the torque equation:
\[ m \cdot \frac{v_0}{t} \cdot h = \frac{2}{5} m r^2 \cdot \frac{\omega_0}{t} \]

### Step 8: Cancel common terms
We can cancel \( m \) and \( t \) from both sides:
\[ v_0 \cdot h = \frac{2}{5} r^2 \cdot \omega_0 \]

### Step 9: Solve for angular velocity \( \omega_0 \)
Rearranging the equation to solve for \( \omega_0 \):
\[ \omega_0 = \frac{5 v_0 h}{2 r^2} \]

### Final Result
The angular velocity \( \omega_0 \) acquired by the billiard ball is:
\[ \omega_0 = \frac{5 v_0 h}{2 r^2} \]

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