A hollow sphere of mass m filled with a non-v | vedclass

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Question

(A) The total mass of the system is $M = m + m = 2m$. Since the liquid is non-viscous and the inner surface is frictionless,the liquid does not rotate

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) The total mass of the system is $M = m + m = 2m$. Since the liquid is non-viscous and the inner surface is frictionless,the liquid does not rotate with the sphere. Thus,the moment of inertia of the system about the center of mass is only due to the hollow sphere,$I = \frac{2}{3}mR^2$.For pure rolling,$v = \omega R$,so $\omega = v/R$.The translational kinetic energy is $K_{trans} = \frac{1}{2} M v^2 = \frac{1}{2} (2m) v^2 = mv^2$.The rotational kinetic energy is $K_{rot} = \frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{2}{3} mR^2) (v/R)^2 = \frac{1}{3} mv^2$.The ratio is $\frac{K_{trans}}{K_{rot}} = \frac{mv^2}{\frac{1}{3} mv^2} = \frac{3}{1}$.Thus,$x = 3$ and $y = 1$. The value of $(x+y) = 3+1 = 4$.

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