Question

A circular disc rolls down an inclined plane. The ratio of the total kinetic energy to the rotational kinetic energy is (a) 1 : 3 (b) 3 : 1 (c) 2 : 3 (d) 3 : 2

Answer 1

Answer:

2:3 is right answer if you want full solution then follow to me

Answer 2

• Rotational Kinetic Energy, $\boxed{k_R = \frac{1}{2}l\omega^2}$

$k_R = \frac{1}{2}\frac{MR^2}{2}\omega^2 = \frac{1}{2}Mv^2$         (∵ $v=R\omega$)

• Translational Kinetic Energy, $\boxed{k_R = \frac{1}{2}mv^2}$

Total Kinetic Energy = $k_T + k_R$

$KE_(total) = \frac{1}{2}Mv^2 + \frac{1}{4}Mv^2$

$\bold{KE_(total) = \frac{3}{4}Mv^2}$

Therefore,

Ratio = $\bold{\frac{\frac{3}{4}Mv^2}{\frac{1}{4}Mv^2}}$ = $\frac{3}{1}$ = ${\bold{3:1}}$

Therefore the ratio is  $\boxed{\bold{3:1}}$