Question

A solid disc of radius 'a' and mass 'm' rolls down without slipping on an inclined plane making an angle \(\theta\) with the horizontal. The acceleration of the disc will be \(\frac{2}{b}\) g sin\(\theta\) where b is ____________. (Round off to the Nearest Integer) (g = acceleration due to gravity, \(\theta\) = angle as shown in figure)

Hint:

For rolling without slipping, \(a = \frac{g\sin\theta}{1 + I/(ma^2)}\) with \(I = \frac{1}{2}ma^2\) for a solid disc.

Answer 1

Correct Answer: 3

Let f be the friction force up the incline and A the acceleration of the centre.
\(mg\sin\theta - f = mA\) and torque: \(f a = I\alpha = \frac{1}{2}ma^2 \cdot \frac{A}{a}\), so \(f = \frac{mA}{2}\).
Substituting: \(mg\sin\theta = mA + \frac{mA}{2} = \frac{3}{2}mA\).
\(A = \frac{2}{3}g\sin\theta\).
Comparing with \(\frac{2}{b}g\sin\theta\) gives b = 3.
Answer: 3