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 concepts of rotational motion and kinematics.

### Step 1: Understand the Problem
We have a disc that is initially rotating with an angular velocity \( \omega_0 \). A constant retarding torque is applied to stop the disc. After \( n \) rotations, the angular velocity of the disc reduces to \( \frac{\omega_0}{2} \). We need to find out how many more rotations the disc will make before coming to rest.

### Step 2: Use the Kinematic Equation for Rotational Motion
We can use the kinematic equation for rotational motion:
\[ \omega^2 = \omega_0^2 + 2\alpha\theta \]
where:
- \( \omega \) = final angular velocity
- \( \omega_0 \) = initial angular velocity
- \( \alpha \) = angular acceleration (retardation in this case)
- \( \theta \) = angular displacement (in rotations)

### Step 3: Apply the Equation for the First Case
For the first case, where the disc slows down from \( \omega_0 \) to \( \frac{\omega_0}{2} \) after \( n \) rotations:
- \( \omega = \frac{\omega_0}{2} \)
- \( \theta_1 = n \) (in rotations)

Substituting these values into the kinematic equation:
\[ \left(\frac{\omega_0}{2}\right)^2 = \omega_0^2 + 2\alpha n \]
This simplifies to:
\[ \frac{\omega_0^2}{4} = \omega_0^2 - 2\alpha n \]
Rearranging gives:
\[ 2\alpha n = \omega_0^2 - \frac{\omega_0^2}{4} \]
\[ 2\alpha n = \frac{3\omega_0^2}{4} \]
Thus:
\[ \alpha = \frac{3\omega_0^2}{8n} \]

### Step 4: Apply the Equation for the Second Case
Now we consider the second case, where the disc goes from \( \frac{\omega_0}{2} \) to \( 0 \):
- \( \omega = 0 \)
- \( \theta_2 \) = additional rotations before coming to rest

Using the kinematic equation again:
\[ 0 = \left(\frac{\omega_0}{2}\right)^2 + 2\alpha \theta_2 \]
Substituting \( \alpha \):
\[ 0 = \frac{\omega_0^2}{4} - 2\left(\frac{3\omega_0^2}{8n}\right) \theta_2 \]
This simplifies to:
\[ \frac{\omega_0^2}{4} = \frac{3\omega_0^2}{4n} \theta_2 \]
Dividing both sides by \( \frac{\omega_0^2}{4} \):
\[ 1 = \frac{3\theta_2}{4n} \]
Solving for \( \theta_2 \):
\[ \theta_2 = \frac{4n}{3} \]

### Step 5: Total Rotations
The total number of rotations made by the disc before coming to rest is:
\[ \text{Total Rotations} = n + \theta_2 = n + \frac{4n}{3} = n + \frac{4n}{3} = \frac{3n + 4n}{3} = \frac{7n}{3} \]

### Final Answer
The disc will make \( \frac{4n}{3} \) more rotations before coming to rest.