A solid cylinder of mass `50 kg` and radius `0.5 m` is free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to it and other end hanging freely. Tension in the string required to produce an angular acceleration of `2` revolution `s^(-2)` is

Options

A. `25 N`
B. `50 N`
C. `78.5 N`
D. `157 N`

Explanation

To solve the problem, we need to find the tension in the string that produces an angular acceleration of 2 revolutions per second squared for a solid cylinder. Here’s a step-by-step breakdown of the solution:

### Step 1: Convert Angular Acceleration to Radians
The angular acceleration \(\alpha\) is given in revolutions per second squared. We need to convert this to radians per second squared.

\[ \alpha = 2 \text{ revolutions/s}^2 \times \frac{2\pi \text{ radians}}{1 \text{ revolution}} = 4\pi \text{ radians/s}^2 \]

### Step 2: Calculate the Moment of Inertia of the Cylinder
The moment of inertia \(I\) for a solid cylinder rotating about its central axis is given by the formula:

\[ I = \frac{1}{2} m r^2 \]

Where:
- \(m = 50 \text{ kg}\) (mass of the cylinder)
- \(r = 0.5 \text{ m}\) (radius of the cylinder)

Substituting the values:

\[ I = \frac{1}{2} \times 50 \text{ kg} \times (0.5 \text{ m})^2 = \frac{1}{2} \times 50 \times 0.25 = 6.25 \text{ kg m}^2 \]

### Step 3: Relate Torque to Angular Acceleration
The torque \(\tau\) acting on the cylinder due to the tension \(T\) in the string is given by:

\[ \tau = T \cdot r \]

According to Newton's second law for rotation, torque is also related to angular acceleration:

\[ \tau = I \cdot \alpha \]

### Step 4: Set the Two Expressions for Torque Equal
Equating the two expressions for torque:

\[ T \cdot r = I \cdot \alpha \]

### Step 5: Solve for Tension \(T\)
Rearranging the equation to solve for \(T\):

\[ T = \frac{I \cdot \alpha}{r} \]

Substituting the values we have:

- \(I = 6.25 \text{ kg m}^2\)
- \(\alpha = 4\pi \text{ radians/s}^2\)
- \(r = 0.5 \text{ m}\)

\[ T = \frac{6.25 \text{ kg m}^2 \cdot 4\pi \text{ radians/s}^2}{0.5 \text{ m}} = \frac{25\pi \text{ kg m}^2/s^2}{0.5} = 50\pi \text{ N} \]

### Step 6: Calculate the Numerical Value of Tension
Using \(\pi \approx 3.14\):

\[ T \approx 50 \times 3.14 \approx 157 \text{ N} \]

### Final Answer
The tension in the string required to produce an angular acceleration of \(2\) revolutions per second squared is approximately \(157 \text{ N}\).

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