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

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प्रश्न

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

संख्यात्मक
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उत्तर

Let N be the normal reaction on the block.

From the free body diagram of the block, it is clear that forces N and mgcosθ pass through the same line. So, there will be no torque due to N and mg cosθ. The only torque will be produced by mg sinθ.

\[\therefore    \overrightarrow{\tau}  =  \overrightarrow{F}  \times  \overrightarrow{r}                                                 \] 

a is the edge of the cube.Therefore, we  have

\[r = \frac{a}{2}\] 

\[ \therefore \tau = mg\sin\theta \times \frac{a}{2}\] 

\[= \frac{1}{2}mga\sin\theta\]

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अध्याय 10: Rotational Mechanics - Exercise [पृष्ठ १९६]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 10 Rotational Mechanics
Exercise | Q 21 | पृष्ठ १९६

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