Question

A disc rolls down a smooth inclined plane without slipping. The inclined plane makes an angle of \(60^\circ\) with the vertical. The linear acceleration of the disc along the inclined plane is (Given: \(g\) is acceleration due to gravity, \(\sin 30^\circ = \cos 60^\circ = \frac{1}{2}\), \(\sin 60^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2}\))

• A. \( \dfrac{g}{18} \)
• B. \( \dfrac{g}{3} \)
• C. \( \dfrac{g}{6} \)
• D. \( \dfrac{g}{9} \)

Hint:

Always convert angle with vertical into angle with horizontal before applying formulas.

Answer 1

The Correct Option is B

Step 1: Angle of inclination with horizontal.
Given plane makes \(60^\circ\) with vertical, hence angle with horizontal is
\[ \theta = 30^\circ \]
Step 2: Acceleration of a rolling disc.
For a solid disc rolling without slipping:
\[ a = \frac{g \sin\theta}{1 + \frac{I}{mR^2}} \]
Step 3: Moment of inertia of disc.
\[ I = \frac{1}{2}mR^2 \]
Step 4: Substituting values.
\[ a = \frac{g \sin 30^\circ}{1 + \frac{1}{2}} = \frac{g \times \frac{1}{2}}{\frac{3}{2}} \] \[ a = \frac{g}{3} \]
Step 5: Conclusion.
The linear acceleration of the disc is \( \dfrac{g}{3} \).