A ring of mass m is rolling without slipping | vedclass

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Question

(C) The total kinetic energy of the system is the sum of the kinetic energy of the ring and the kinetic energy of the rod.For rolling without slipping

Vedclass · Accepted Answer

$\frac{5}{3}mv^2$

Vedclass · Answer

(C) The total kinetic energy of the system is the sum of the kinetic energy of the ring and the kinetic energy of the rod.For rolling without slipping,$\omega = \frac{v}{R}$.Kinetic energy of the ring $(KE_{ring})$: $KE_{ring} = \frac{1}{2}mv^2 + \frac{1}{2}I_{ring}\omega^2 = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.Kinetic energy of the rod $(KE_{rod})$: The rod rotates about the center of the ring. The moment of inertia of a rod of length $2R$ about its center is $I_{rod} = \frac{m(2R)^2}{12} = \frac{4mR^2}{12} = \frac{1}{3}mR^2$.$KE_{rod} = \frac{1}{2}mv^2 + \frac{1}{2}I_{rod}\omega^2 = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{3}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{6}mv^2 = \frac{3+1}{6}mv^2 = \frac{4}{6}mv^2 = \frac{2}{3}mv^2$.Total kinetic energy = $KE_{ring} + KE_{rod} = mv^2 + \frac{2}{3}mv^2 = \frac{5}{3}mv^2$.

$A$ ring of mass $m$ is rolling without slipping with linear velocity $v$ as shown in the figure. $A$ rod of identical mass $m$ is fixed along one of its diameters. The total kinetic energy of the system is:

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