A uniform ring of mass m and radius R is performing pure rolling motion on a horizontal surface. The velocity of centre of the ring is `V_(0)`. If at the given instant the kinetic energy of the semi circular are AOB is lambda `mv_(0)_(2)`, then find the value of `11lambda ("take" pi=(22)/(7))`

Answer

2

Explanation

`V=2v_(0) sin .(theta)/(2) dk=(1)/(2)dmv^(2)`

`K=intdk`
`K= int (1)/(2)(M)/(2piR)Rd theta (2V_(0)sin. (theta)/(2))^(2)`
`=(M)/(pi)V_(0)^(2)underset(-pi/2)overset(pi//2)int sin^(2),(theta)/(2)d theta`
`=(MV_(0)^(2))/(pi)(pi/(2)-1)=MV_(0)^(2)((1)/(2)-(1)/(pi))`
`E=(2)/(11) MV_(0)^(2)`
`lambda=(2)/(11)`
`11lambda=2`