A solid sphere of mass M and radius R is atta | vedclass

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(C) The total mechanical energy $E$ of the system is the sum of translational kinetic energy,rotational kinetic energy,and potential energy of the spr

Vedclass · Accepted Answer

$2 \pi \sqrt{\frac{7 M}{5 K}}$

Vedclass · Answer

(C) The total mechanical energy $E$ of the system is the sum of translational kinetic energy,rotational kinetic energy,and potential energy of the spring: $E = \frac{1}{2} M V^2 + \frac{1}{2} I \omega^2 + \frac{1}{2} K x^2$ Since the sphere rolls without slipping,$V = R \omega$,so $\omega = V/R$. The moment of inertia of a solid sphere is $I = \frac{2}{5} M R^2$. Substituting these: $E = \frac{1}{2} M V^2 + \frac{1}{2} (\frac{2}{5} M R^2) (\frac{V^2}{R^2}) + \frac{1}{2} K x^2 = \frac{1}{2} M V^2 + \frac{1}{5} M V^2 + \frac{1}{2} K x^2 = \frac{7}{10} M V^2 + \frac{1}{2} K x^2$ Since total energy is conserved,$\frac{dE}{dt} = 0$: $\frac{d}{dt} (\frac{7}{10} M V^2 + \frac{1}{2} K x^2) = 0$ $\frac{7}{10} M (2 V \frac{dV}{dt}) + \frac{1}{2} K (2 x \frac{dx}{dt}) = 0$ Since $V = \frac{dx}{dt}$ and $a = \frac{dV}{dt}$: $\frac{7}{5} M V a + K V x = 0$ $\frac{7}{5} M a + K x = 0 \implies a = -(\frac{5 K}{7 M}) x$ Comparing with the $SHM$ equation $a = -\omega^2 x$,we get $\omega^2 = \frac{5 K}{7 M}$,so $\omega = \sqrt{\frac{5 K}{7 M}}$. The time period $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{7 M}{5 K}}$.

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