The uniform speed of a body is the same as seen from any point in the body. A light cord is wrapped around the rim of the disc and mass of `1 kg` is tied to the free end. If it is released from rest,

Options

A. the tension in the cord is `5 N`
B. in the first `4 s` the angular displacement of the disc is `40 rad`
C. the work done by the torque on the disc in the first `4 s` is `200J`
D. the increase in kinetic energy of the disc in the first `4 s` is `200 J`

Explanation

To solve the problem step by step, we will analyze the forces acting on the system and apply the relevant equations of motion for both linear and rotational dynamics.

### Step 1: Analyze the Forces
When the mass is released, it experiences two forces:
1. The gravitational force (weight) acting downward, \( F_g = mg \).
2. The tension in the cord acting upward, \( T \).

For the mass \( m = 1 \, \text{kg} \):
- Weight, \( F_g = mg = 1 \times 9.81 = 9.81 \, \text{N} \).

### Step 2: Apply Newton's Second Law
Using Newton's second law for the mass, we can write:
\[ mg - T = ma \]
Where \( a \) is the linear acceleration of the mass.

### Step 3: Relate Linear Acceleration to Angular Acceleration
The linear acceleration \( a \) of the mass is related to the angular acceleration \( \alpha \) of the disc by the equation:
\[ a = r \alpha \]
Where \( r \) is the radius of the disc.

### Step 4: Calculate Torque on the Disc
The torque \( \tau \) acting on the disc due to the tension in the cord is given by:
\[ \tau = T \cdot r \]
The moment of inertia \( I \) for a disc about its center is:
\[ I = \frac{1}{2} m_{\text{disc}} r^2 \]
Assuming the mass of the disc is \( 2 \, \text{kg} \).

### Step 5: Apply Newton's Second Law for Rotation
Using Newton's second law for rotation:
\[ \tau = I \alpha \]
Substituting the expressions for torque and moment of inertia:
\[ T \cdot r = \frac{1}{2} m_{\text{disc}} r^2 \alpha \]

### Step 6: Substitute \( \alpha \) in Terms of \( a \)
Substituting \( \alpha = \frac{a}{r} \) into the torque equation:
\[ T \cdot r = \frac{1}{2} m_{\text{disc}} r^2 \left(\frac{a}{r}\right) \]
This simplifies to:
\[ T = \frac{1}{2} m_{\text{disc}} a \]
Substituting \( m_{\text{disc}} = 2 \, \text{kg} \):
\[ T = \frac{1}{2} \cdot 2 \cdot a = a \]

### Step 7: Substitute \( T \) Back into the Linear Equation
Now substituting \( T = a \) back into the linear motion equation:
\[ mg - a = ma \]
This gives:
\[ mg = ma + a \]
\[ mg = a(m + 1) \]
Solving for \( a \):
\[ a = \frac{mg}{m + 1} = \frac{9.81}{2} = 4.905 \, \text{m/s}^2 \]

### Step 8: Calculate Tension
Since \( T = a \):
\[ T = 4.905 \, \text{N} \]

### Step 9: Calculate Angular Displacement
Using the formula for angular displacement:
\[ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 \]
Since the initial angular velocity \( \omega_0 = 0 \):
\[ \theta = \frac{1}{2} \alpha t^2 \]
Where \( \alpha = \frac{a}{r} = \frac{4.905}{1} = 4.905 \, \text{rad/s}^2 \) and \( t = 4 \, \text{s} \):
\[ \theta = \frac{1}{2} \cdot 4.905 \cdot (4^2) = \frac{1}{2} \cdot 4.905 \cdot 16 = 39.24 \, \text{radians} \]

### Step 10: Calculate Work Done by Torque
The work done by torque is given by:
\[ W = \tau \cdot \theta \]
Where \( \tau = T \cdot r = 4.905 \cdot 1 = 4.905 \, \text{N m} \):
\[ W = 4.905 \cdot 39.24 = 192.3 \, \text{J} \]

### Conclusion
1. The tension in the cord is approximately \( 4.905 \, \text{N} \).
2. The angular displacement of the disc in the first 4 seconds is approximately \( 39.24 \, \text{radians} \).
3. The work done by torque on the disc is approximately \( 192.3 \, \text{J} \).