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Q. A sphere of mass $M$ and radius $R$ is released from the top of an inclined plane of inclination $\theta .$ The minimum coefficient of friction between the plane and the sphere so that it rolls down the plane without sliding is given by

NTA AbhyasNTA Abhyas 2022

Solution:

$\mu \geq \frac{tan \theta }{1 + \frac{R^{2}}{K^{2}}}$
$\Rightarrow \, \mu _{min}=\frac{tan \theta }{1 + \frac{R^{2}}{K^{2}}}$