Question

0.6. A solid cylinder, of mass 1.5 kg and radius 0.1 m, rolls down an inclined plane
from a height of 3 m. Calculate the rotational energy at the bottom of the plane. ​

Answer 1

According to Conservation of mechanical Energy

PE = KE

MgH = 1/2mV^2 { 1+K^2/R^2}

v = 20m/s

Energy at bottom is KE

1/2 mV^2 {1+K^2/R^2}

K^2/R^2 = 1/2

put Value of V

we Get energy =450J

Thanks for Reading