A uniform sphere of radius `R/16` starts rolling down without slipping from the top of another sphere of radius `R=1m`. The angular velocity of the sphere of in rad`s^(-1)`, after it leaves the surface of the larger sphere is `8 xn`. Where `n`=--`

Answer

5

Explanation

The equation of motion for the centre of the sphere at the moment of breaking off, `N=0` is `(mv^(2))/(R+r)=mg cos theta`
Where `v` is the speed of the centre of the sphere at that moment and `theta` is the corresponding angle. The speed `v` can be found by using the law of conservation energy,

According to which `mgh=(mv^(2))/2+(Iomega^(2))/2`
where `I=2/5 mr^(2),v=romega` and `h=(R+r)(1-cos theta)`
From the equations we get
`omega=sqrt((10g(R+r))/(17r^(2)))`
`rArromega=40 rads^(-1)`