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

To solve the problem, we need to find the net linear acceleration of a point on the rim of a uniform circular disc subjected to an angular acceleration. Here's how we can approach this:

1. Given the radius of the disc, r = 50 \, \text{cm} = 0.5 \, \text{m}.
2. The constant angular acceleration provided is \alpha = 2.0 \, \text{rad/s}^2.
3. The formula to calculate the angular velocity \omega after a time t with constant angular acceleration is:
   \omega = \omega_0 + \alpha t
   • Where \omega_0 = 0 \, \text{rad/s} (since the disc is initially at rest).
   • Substituting the given values: \omega = 0 + 2.0 \times 2 = 4.0 \, \text{rad/s}.
4. The linear (tangential) acceleration a_t is related to the angular acceleration by:
   a_t = r \cdot \alpha
   • Substituting the values: a_t = 0.5 \times 2.0 = 1.0 \, \text{m/s}^2.
5. The centripetal (radial) acceleration a_c is given by:
   a_c = r \cdot \omega^2
   • Substituting the values: a_c = 0.5 \times (4.0)^2 = 8.0 \, \text{m/s}^2.
6. The net acceleration a_{\text{net}} is the vector sum of the tangential and centripetal accelerations, which are perpendicular to each other:
   a_{\text{net}} = \sqrt{a_t^2 + a_c^2}
   • Substituting the values: a_{\text{net}} = \sqrt{(1.0)^2 + (8.0)^2} = \sqrt{1 + 64} = \sqrt{65} \approx 8 \, \text{m/s}^2.

Thus, the net acceleration of a point on the rim of the disc at the end of 2.0 seconds is approximately 8 \text{ms}^{-2}.