Question:

A uniform disc is acted by two equal forces of magnitude $F$. One of them, acts tangentially to the disc, while other one is acting at the central point of the disc. The friction between disc surface and ground surface is $n F$. If $r$ be the radius of the disc, then the value of $n$ would be (in $N$)

Updated On: Jul 2, 2022
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The Correct Option is A

Solution and Explanation

Let $f_{r}$ be the friction exerting between disc surface and ground surface, then for the motion of the disc, we can write $2 F-f_{r} =m a $ ... (i) and $\left(F+f_{r}\right) r =I \omega $ $\Rightarrow \left(F+f_{r}\right) r =\frac{1}{2} m r^{2} \frac{a}{r}$...(ii) Here, $a=$ linear acceleration of the disc. Solving Eqs. (i) and (ii) we get, $f_{r}=0$
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Rotational Motion Examples:

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