Question

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 hanging freely. Tension in the string required to produce an angular acceleration of 2 revolutions $ s^ {-2} $ is

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

Answer 1

The Correct Option is D

To solve this problem, we need to determine the tension in the string that produces an angular acceleration of 2 revolutions per second squared in the solid cylinder.

1. First, let's convert the angular acceleration from revolutions per second squared to radians per second squared:

\text{Angular acceleration, } \alpha = 2 \times 2\pi = 4\pi \, \text{rad/s}^2

2. Next, compute the moment of inertia (I) for a solid cylinder rotating about its central axis (perpendicular to its length). The moment of inertia is given by:

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, we get:

I = \frac{1}{2} \times 50 \times (0.5)^2 = \frac{1}{2} \times 50 \times 0.25 = \frac{1}{2} \times 12.5 = 6.25 \, \text{kg m}^2

3. The torque (\tau) required to produce an angular acceleration is given by the equation:

\tau = I \alpha

Substitute the values:

\tau = 6.25 \times 4\pi = 25\pi \, \text{Nm}

4. The relation between torque and tension in the string is given by:

\tau = T \cdot r

Solving for tension (T), we have:

T = \frac{\tau}{r} = \frac{25\pi}{0.5} = 50\pi \, \text{N}

Approximating \pi \approx 3.14, the tension becomes:

T \approx 50 \times 3.14 = 157 \, \text{N}

5. Thus, the tension in the string required to produce an angular acceleration of 2 revolutions per second squared is 157 N.

The correct answer is therefore 157 N.