Question

A metal ring of radius \(10\) cm and mass \(0.5\) kg is rolling down an inclined plane from rest without slipping. The inclined plane makes an angle of \(30^\circ\) with the horizontal. What is the linear acceleration of the ring? \[ g=10\ \text{m s}^{-2} \]

• A. \(2.5\ \text{m s}^{-2}\)
• B. \(5\ \text{m s}^{-2}\)
• C. \(10\ \text{m s}^{-2}\)
• D. \(3.33\ \text{m s}^{-2}\)

Hint:

For rolling bodies: \[ a=\frac{g\sin\theta} {1+\dfrac{I}{mR^2}} \] For a ring, \[ I=mR^2 \] which gives \[ a=\frac{g\sin\theta}{2}. \]

Answer 1

The Correct Option is A

Concept: For rolling without slipping, \[ a=\frac{g\sin\theta} {1+\dfrac{I}{mR^2}} \] For a ring, \[ I=mR^2. \]

Step 1: Substitute the moment of inertia of a ring. \[ a = \frac{g\sin\theta} {1+\dfrac{mR^2}{mR^2}} \] \[ = \frac{g\sin\theta}{2} \]

Step 2: Use the given values. \[ g=10\ \text{m s}^{-2} \] \[ \theta=30^\circ \] \[ \sin30^\circ=\frac12 \] Hence, \[ a = \frac{10\times\frac12}{2} \] \[ = \frac52 \] \[ = 2.5\ \text{m s}^{-2} \] \[\begin{aligned} \boxed{2.5\ \text{m s}^{-2}} \end{aligned}\] Hence, option \(\mathbf{(A)}\) is correct.