A ring of mass 100 kg and diameter 2 m is rotating at the rate of `(300/(pi))` rpm. Then

Options

A. moment of inertia is `100kg-m^(2)`
B. kinetic energy is 5kJ
C. if a retarding torque of 200N-m starts acting then it will come at rest after 5 sec.
D. all of these

Explanation

To solve the problem step by step, we will follow the instructions provided in the video transcript:

### Step 1: Identify the given values
- Mass of the ring, \( m = 100 \, \text{kg} \)
- Diameter of the ring, \( d = 2 \, \text{m} \)
- Radius of the ring, \( r = \frac{d}{2} = 1 \, \text{m} \)
- Rotational speed, \( \text{rpm} = \frac{300}{\pi} \)

### Step 2: Convert rpm to radians per second
To convert the rotational speed from revolutions per minute (rpm) to radians per second, we use the formula:
\[ \omega = \text{rpm} \times \frac{2\pi \, \text{rad}}{1 \, \text{revolution}} \times \frac{1 \, \text{minute}}{60 \, \text{seconds}} \]
Substituting the values:
\[ \omega = \frac{300}{\pi} \times \frac{2\pi}{60} = \frac{300 \times 2}{60} = 10 \, \text{rad/s} \]

### Step 3: Calculate the moment of inertia (I) of the ring
The moment of inertia for a ring is given by the formula:
\[ I = m r^2 \]
Substituting the values:
\[ I = 100 \times (1)^2 = 100 \, \text{kg m}^2 \]

### Step 4: Calculate the rotational kinetic energy (KE)
The formula for rotational kinetic energy is:
\[ KE = \frac{1}{2} I \omega^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \times 100 \times (10)^2 = \frac{1}{2} \times 100 \times 100 = 5000 \, \text{J} = 5 \, \text{kJ} \]

### Step 5: Determine the angular deceleration (α)
If a retarding torque of \( \tau = 200 \, \text{N m} \) is applied, we can find the angular deceleration using the relation:
\[ \tau = I \alpha \]
Rearranging gives:
\[ \alpha = \frac{\tau}{I} \]
Substituting the values:
\[ \alpha = \frac{-200}{100} = -2 \, \text{rad/s}^2 \]

### Step 6: Calculate the time to come to rest
Using the kinematic equation for angular motion:
\[ \omega = \omega_0 + \alpha t \]
Where \( \omega_0 = 10 \, \text{rad/s} \) and \( \omega = 0 \) (when it comes to rest). Rearranging gives:
\[ 0 = 10 - 2t \implies 2t = 10 \implies t = 5 \, \text{s} \]

### Final Answer
All options provided in the question are correct:
1. Moment of inertia: \( 100 \, \text{kg m}^2 \)
2. Kinetic energy: \( 5 \, \text{kJ} \)
3. Time to come to rest: \( 5 \, \text{s} \)