Question

A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of $2.0 \, rad \, s^{-2}$. Its net acceleration in $ms^{-2}$ at the end of 2.0 s is approximately :

• A. 7
• B. 6
• C. 3
• D. 8

Answer 1

The Correct Option is D

Particle Acceleration Calculation 

A particle at the periphery of a rotating object will have both radial and tangential acceleration. Let's calculate both accelerations step by step.

1. Tangential Acceleration:

The tangential acceleration \( a_t \) is given by:

\(a_t = R \alpha = 0.5 \times 2 = 1 \, \text{m/s}^2\)

2. Angular Velocity:

The angular velocity \( \omega \) is calculated using the formula:

\(\omega = \omega_0 + \alpha t\)

Substitute the values:

\(\omega = 0 + 2 \times 2 = 4 \, \text{rad/sec}\)

3. Centripetal (Radial) Acceleration:

The centripetal acceleration \( a_c \) is given by:

\(a_c = \omega^2 R = (4)^2 \times 0.5 = 16 \times 0.5 = 8 \, \text{m/s}^2\)

4. Total Acceleration:

The total acceleration \( a_{\text{total}} \) is the vector sum of tangential and centripetal accelerations, calculated as:

\(a_{\text{total}} = \sqrt{a_t^2 + a_c^2} = \sqrt{1^2 + 8^2} \approx 8 \, \text{m/s}^2\)

Conclusion:

The total acceleration of the particle is approximately \( 8 \, \text{m/s}^2 \).