A solid cylinder is placed gently over an incline plane of inclination $60°$. The acceleration of cylinder when it start rolling without slipping is $\frac{g}{\sqrt x}$ where $μ$ is the coefficient of friction. [take $g$ = $10\;m / (s ^ 2)$ ]

Question

A cubic block of mass $ m $ is sliding down on an inclined plane at $ 60^\circ $ with an acceleration of $ \frac{g}{2} $, the value of coefficient of kinetic friction is:

Answer 1

Acceleration of cylinder:

$a=\frac{g\frac{\sqrt3}{2}}{\frac{3}{2}}$

=$\frac{g}{\sqrt3}$

The Correct Answer is : $\frac{g}{\sqrt3}$

A solid cylinder is placed gently over an incline plane of inclination \(60°\). The acceleration of cylinder when it start rolling without slipping is \(\frac{g}{\sqrt x}\) where \(μ\) is the coefficient of friction. [take \(g\) = \(10\;m / (s ^ 2)\) ]

Solution and Explanation

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