Question

A cylinder is rolling down an inclined plane of inclination \( 60^\circ \). Its acceleration during rolling down will be \( \frac{x}{\sqrt{3}} \, \text{m/s}^2 \), where \( x = \) _____. (Use \( g = 10 \, \text{m/s}^2 \)).

Answer 1

Correct Answer: 10

Step 1: Formula for Acceleration in Rolling Motion:

- For a cylinder rolling down an incline, the acceleration \( a \) is given by:

\[ a = \frac{g \sin \theta}{1 + \frac{I_{cm}}{MR^2}} \]

- For a solid cylinder, \( I_{cm} = \frac{1}{2} MR^2 \).

Step 2: Substitute Values:

\[ a = \frac{g \sin \theta}{1 + \frac{1}{2}} \]

- Given \( g = 10 \, \text{m/s}^2 \) and \( \theta = 60^\circ \) (so \( \sin 60^\circ = \frac{\sqrt{3}}{2} \)):

\[ a = \frac{10 \times \frac{\sqrt{3}}{2}}{1 + \frac{1}{2}} \]

Step 3: Calculate \( a \):

\[ a = \frac{10 \times \frac{\sqrt{3}}{2}}{\frac{3}{2}} = \frac{10 \sqrt{3}}{3} = \frac{x}{\sqrt{3}} \]

- Therefore, \( x = 10 \).

So, the correct answer is: \( x = 10 \)

Answer 2

Step 1: Formula for acceleration of a rolling object on an incline.

For a body rolling without slipping on an inclined plane of angle \( \theta \): \[ a = \frac{g \sin \theta}{1 + \frac{k^2}{R^2}} \] where \( k \) is the radius of gyration and \( R \) is the radius of the body.

Step 2: For a solid cylinder

Moment of inertia for a solid cylinder: \[ I = \frac{1}{2} mR^2 \Rightarrow \frac{k^2}{R^2} = \frac{1}{2} \] Substitute in the formula: \[ a = \frac{g \sin \theta}{1 + \frac{1}{2}} = \frac{g \sin \theta}{\frac{3}{2}} = \frac{2g \sin \theta}{3} \]

Step 3: Substitute given values

\[ \theta = 60^\circ, \quad g = 10\,\text{m/s}^2 \] \[ a = \frac{2 \times 10 \times \sin 60^\circ}{3} \] \[ a = \frac{20 \times \frac{\sqrt{3}}{2}}{3} = \frac{10\sqrt{3}}{3}\,\text{m/s}^2 \]

Step 4: Compare with given form

Acceleration = \( \frac{x}{3} \, \text{m/s}^2 \) \[ \frac{x}{3} = \frac{10\sqrt{3}}{3} \Rightarrow x = 10\sqrt{3} \] \[ \text{Hence, } x = 10 \]

Final Answer:

\[ \boxed{x = 10} \]