A disc is rotaing with an angular velocity `omega_(0)`. A constant retarding torque is applied on it to stop the disc. The angular velocity becomes `(omega_(0))/(2)` after n rotations. How many more rotations will it make before coming to rest ?

Options

A. n
B. 2n
C. `(n)/(2)`
D. `(n)/(3)`

Answer

To solve the problem step by step, we will use the equations of rotational motion. The problem states that a disc is initially rotating with an angular velocity \( \omega_0 \) and experiences a constant retarding torque until it comes to rest. We need to find out how many additional rotations it will make after reaching an angular velocity of \( \frac{\omega_0}{2} \).

### Step 1: Understand the given data
- Initial angular velocity \( \omega_0 \)
- Final angular velocity after \( n \) rotations \( \omega = \frac{\omega_0}{2} \)
- We need to find the additional rotations before the disc comes to rest.

### Step 2: Use the equation of motion for rotation
We can use the rotational motion equation analogous to linear motion:
\[ \omega^2 = \omega_0^2 - 2\alpha \theta \]
where:
- \( \omega \) is the final angular velocity,
- \( \omega_0 \) is the initial angular velocity,
- \( \alpha \) is the angular retardation,
- \( \theta \) is the angular displacement in radians.

### Step 3: Apply the equation for the first phase (from \( \omega_0 \) to \( \frac{\omega_0}{2} \))
Substituting the known values:
\[ \left(\frac{\omega_0}{2}\right)^2 = \omega_0^2 - 2\alpha \theta_1 \]
This simplifies to:
\[ \frac{\omega_0^2}{4} = \omega_0^2 - 2\alpha n \]
Rearranging gives:
\[ 2\alpha n = \omega_0^2 - \frac{\omega_0^2}{4} \]
\[ 2\alpha n = \frac{3\omega_0^2}{4} \]
\[ \alpha = \frac{3\omega_0^2}{8n} \]

### Step 4: Apply the equation for the second phase (from \( \frac{\omega_0}{2} \) to rest)
Now, we need to find the angular displacement \( \theta_2 \) when the disc comes to rest from \( \frac{\omega_0}{2} \):
\[ 0 = \left(\frac{\omega_0}{2}\right)^2 - 2\alpha \theta_2 \]
Substituting for \( \alpha \):
\[ 0 = \frac{\omega_0^2}{4} - 2\left(\frac{3\omega_0^2}{8n}\right)\theta_2 \]
This simplifies to:
\[ \frac{\omega_0^2}{4} = \frac{3\omega_0^2}{4n}\theta_2 \]
Cancelling \( \omega_0^2 \) (assuming \( \omega_0 \neq 0 \)):
\[ \frac{1}{4} = \frac{3}{4n}\theta_2 \]
Multiplying both sides by \( 4n \):
\[ n = 3\theta_2 \]
Thus, we find:
\[ \theta_2 = \frac{n}{3} \]

### Step 5: Calculate the total rotations
The total number of rotations made by the disc before coming to rest is:
- The first phase: \( n \) rotations
- The second phase: \( \frac{n}{3} \) rotations

### Final Answer
The total additional rotations before coming to rest is:
\[ \text{Total additional rotations} = \frac{n}{3} \]