Question

A uniform disc of mass $10kg$ radius $1m$ is placed on a rough horizontal surface. The co-efficient of friction between the disc and the surface is $\mathrm{0.2.}A$ horizontal time varying force is applied on the centre of the disc whose variation with time is shown in graph.

Answer 1

$a=R\alpha =R\left(\frac{\tau }{I}\right)$
$\therefore \frac{F-20}{10}=\left(1\right)\left(\frac{20\times 1}{\frac{1}{2}\times 10\times 1^2}\right)$
Solving this equation, we get $F=60N$
So, when $F=60N$, friction reaches its maximum value or slipping will start. $F$ becomes $60N$ at 6 s.
d) $a=R\alpha =R\left(\frac{\tau }{I}\right)$
or $\frac{F-10}{10}=\left(1\right)\left[\frac{10\times 1}{\frac{1}{2}\times 10\times 1^2}\right]$
$F=30N$ and $F$ becomes $30N$ at 3s.