## In This post you will get answer of Q.

A thin metal disc of radius of $ 0.25text{ }m $ and mass $ 2text{ }kg $ starts from rest and rolls down on an inclined plane. If its rotational kinetic energy is $ 4text{ }J $ at the foot of inclined plane, then the linear velocity at the same point, is in m/s

Solution:

## Rotational kinetic energy $ =frac{1}{2}I,,{{omega }^{2}} $

$ therefore $ Rotational $ KE=frac{1}{2}left[ frac{1}{2}m{{r}^{2}} right]frac{{{v}^{2}}}{{{r}^{2}}} $

$ left( where,,,,I=frac{1}{2},m{{r}^{2}} right) $

$ 4=frac{1}{2}left[ frac{1}{2}(2){{r}^{2}} right]frac{{{v}^{2}}}{{{r}^{2}}} $

$ Rightarrow $ $ {{v}^{2}}=8 $

$ v=2sqrt{2}m/s $