Explanation
apply consecration of energy
$${E_i} = mgh$$
$${E_i} = m10(6\sin 30^\circ )$$
$${E_i} = 30\,m$$
$${E_f} = {1 \over 2}m{v^2} + {1 \over 2}l{\omega ^2}$$
$${E_f} = {1 \over 2}m{R^2}{\omega ^2} + {1 \over 2}\left( {{2 \over 3}m{R^2}} \right){\omega ^2}$$
$${E_f} = {5 \over 6}m{R^2}{\omega ^2}$$
$${E_i} = {E_f}$$
$${R^2}{\omega ^2} = 36$$
$${v_{cm}} = 6$$
Kinetic energy is given by
K=V2/(2M)……………(1)
When KE increases by 300%, the value of KE becomes 4 times it's original value.
from equation (1) that v becomes 2V
So, V increases by 100%.
Please disable the adBlock and continue. Thank you.