Question

The body B can be a disc, cylinder or sphere of mass M & radius R. It lies on a floor having the coefficient of friction as $\mu$, & a force ${}^{\prime }{F}^{\prime }$ acts at a height ${}^{\prime }{h}^{\prime }$ from the centre. .If 'B' is a solid cylinder & h $=$R. Find the frictional force $f$.

Answer 1

The body B can be a disc, cylinder or sphere of mass $=M$

Radius $=R$

Co-efficient of friction $=\mu$

Fore$=F$ acts at a height h from the centre.

If ${}^{\prime }{B}^{\prime }$ is a solid cylinder $\&h=R$

Find $=$ frictional force$=f$

Solution

Velocity of centre of mass is $V$ and angular velocity of sphere is $0$ initially.

Friction will act in backward direction... now

$f=Kmg(\because f=$ friction force)

$ma=Kmg$

$a=Kg(a$ is retardation of centre of mass)

Now, torque $=fR=I$(angularacceleration)

Angular acceleration $fR/I$

$=\frac{5Kg}{2R}(I_{sphere}=\frac{2M{R}^2}{5})$

Now, after time $t$ let, velocity of centre ofmass is $V$, then

$v=u+at$(linear velocity is $v$)

$v=u-Kgt\to \left(1\right)$

Let, at the time angular velocity$0$ is $v$ then,

$w=w_0+\left(angularvelocity\right)t$

(initial angular velocity is $0$)

$w=\frac{5Kg}{2R}\to \left(2\right)$

Now, if pure rolling has started then

$v=wRSo,v-Kgt=5Kgt/2t=2v/7Kg$