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A solid uniform cylinder is gently released on an inclined surface AB whose angle of inclination with horizontal can be varied. If coefficient of static friction between the cylinder and the inclined surface AB is $\frac{1}{3 \sqrt{3}}$ , The maximum possible linear acceleration of the centre of mass of the cylinder for pure rolling, is (Take g = 10 m/s²)

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3.33 m/s²

5 m/s²

2.67 m/s²

4.33 m/s²

answer is A.

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Detailed Solution

For pure rolling down the inclined plane, maximum possible angle of inclination is given by, $\tan θ_{\max} = μ (1 + \frac{{mR}^{2}}{I_{C}})$

For solid cylinder,

$\tan θ_{\max} = \frac{1}{3 \sqrt{3}} (1 + \frac{{mR}^{2}}{\frac{1}{2} {mR}^{2}}) = \frac{1}{\sqrt{3}}$

$\Rightarrow θ_{\max} = 30^{0}$

$∴ a_{\max} = \frac{g \sin θ_{\max}}{(1 + \frac{I_{C}}{{mR}^{2}})} = \frac{10 \times \sin 30^{0}}{(1 + \frac{\frac{1}{2} {mR}^{2}}{{mR}^{2}})} m / s^{2}$

$\Rightarrow a_{\max} = \frac{10}{3} m / s^{2} = 3 .33 m / s^{2}$

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