Question

A uniform solid circular cylinder of radius r is placed on a rough horizontal surface and given a linear velocity $v=2{\omega }_{o}r$ and angular velocity ${\omega }_o$. The speed of the cylinder when it starts rolling :

• A. $5/2{\omega }_{o}R$
  
• B. $3/2{\omega }_{o}R$
• C. $5/3{\omega }_{o}R$
• D. $2/3{\omega }_{o}R$

Answer 1

The correct option is C $5/3{\omega }_{o}R$

Time taken by circular clyinder to start pure rotation is

$t=\frac{{\omega }_{0}R}{\mu g}\left[\frac{\frac{I}{m{R}^2}}{1+\frac{I}{m{R}^2}}\right]$

Here,

${\omega }_0=$ Initial angular velocity

R$=$ Radius of clyinder

$\mu =$ Coefficient of friction between surface and clyinder

I$=$ moment of inertia of the clyinder about an axis passing through its centre to its length $=\frac{1}{2}m{R}^2$

m$=$ Mass of the clyinder

Putting its value we get,

$t=\frac{{\omega }_{0}R}{\mu g}\left[\frac{\frac{\frac{1}{2}m{R}^2}{m{R}^2}}{1+\frac{\frac{1}{2}m{R}^2}{m{R}^2}}\right]$

$=\frac{{\omega }_{0}R}{\mu g}\left[\frac{\frac{1}{2}}{1+\frac{1}{2}}\right]$

$=\frac{{\omega }_{0}R}{3\mu g}$

From figure it is seen that the point of contact has a linear velocity equals to

$⟹(v-{\omega }_{0}R)=2{\omega }_{0}R-{\omega }_{0}R$

$={\omega }_{0}R$ in forward direction. So friction will act in backward direction at this point.So, frictional force will tend to decrease the velocity of clyinder.

The frictional force is given by

$F=\mu mg$

So, decceleration of body $=\frac{\mu mg}{m}=\mu g$

Then the time t velocity of clyinder is

$=v-\mu gt$

$=2{\omega }_{0}R-\mu g\frac{{\omega }_{0}R}{\mu g}$

$=\frac{5{\omega }_{0}R}{3}$