A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its center of mass is K. If radius of the ball be R, then the fraction of total energy associated with its rotational energy will be

Question

A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its center of mass is K. If radius of the ball be R, then the fraction of total energy associated with its rotational energy will be (A) (K2/K2 +R2) (B) (R2/K2 +R2) (C) (K2 +R2/R2) (D) (K2/R2)

Tardigrade Admin · Accepted Answer

Total energy =(1/2) I ω2+(1/2) m v2 =(1/2) m v2(1+K2 / R2) Required fraction =(K2 / R2/1+K2 / R2)=(K2/R2+K2).

Q. A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its center of mass is $K$ . If radius of the ball be $R,$ then the fraction of total energy associated with its rotational energy will be

$\frac{K ^{2}}{K ^{2} + R ^{2}}$

$\frac{R ^{2}}{K ^{2} + R ^{2}}$

$\frac{K ^{2} + R ^{2}}{R ^{2}}$

$\frac{K ^{2}}{R ^{2}}$

Total energy
$= \frac{1}{2} I ω^{2} + \frac{1}{2} m v^{2}$
$= \frac{1}{2} m v^{2} (1 + K^{2} / R^{2})$
Required fraction $= \frac{K ^{2} / R ^{2}}{1 + K ^{2} / R ^{2}} = \frac{K ^{2}}{R ^{2} + K ^{2}}$ .

Q. A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its center of mass is K. If radius of the ball be R, then the fraction of total energy associated with its rotational energy will be

K2+R2K2​

K2+R2R2​

R2K2+R2​

R2K2​