Question

-2. An impulse J is applied on a ring of mass m along
a line passing through its centre 0. The ring is
placed on a rough horizontal surface. The linear
velocity of centre of ring once it starts rolling
without slipping is
​

Answer 1

Answer:

The linear velocity of center of ring once it starts rolling without slipping is

J/2m

Explanation:

Let v be the velocity of ring just after the impulse is applied and v' its velocity when pure rolling starts. Angular velocity ω of the ring at this instant will be ω=v′r

From impulse =change in linear momentum, we have  

J=mv  

or v=J/m.....i  

Now force of friction on the ring acts backwards. So angular momentum of the ring about bottom most point will remain conserved  

∴Li=Lf  

or mr=mv'r+Iω  

mv'r+(mr2)(v′r)=2mg'r  

v'=v/2=J/2m (from equation i)