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 $N$)

• A. 0
• B. 1.2
• C. 2.0
• D. 3.2

Answer 1

Answer: (A)

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,
$f_{r}=0$