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 $nF$. 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
$2F-f_r=ma$ ... (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,
$f_r=0$