Solve the following :
Question:

A cubical block of mass $m$ and edge a slides down a rough inclined plane of inclination $\theta$ with a uniform speed. Find the torque of the normal force acting on the block about its centre.

Solution:

Since,

acceleration $=0$

ff $=m g^{\sin \theta}$

The normal contact will shift as block is not toppling.

So, $\tau_{\text {Normal }}=\tau_{f f}$

$=\mathrm{mg}^{\sin \theta\left(\frac{a}{2}\right)}$

HC VERMA Solutions for Class 11 Physics Chapter 10 Rotational Mechanics

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