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6.System of Particles and Rotational Motion
hard
$A$ ring of mass $m$ is rolling without slipping with linear velocity $v$ as shown is figure. $A$ rod of identical mass is fixed along one of its diameter. The total kinetic energy of the system is :-
A
$\frac{7}{5}m{v^2}$
B
$\frac{2}{5}m{v^2}$
C
$\frac{5}{3}m{v^2}$
D
$\frac{5}{4}m{v^2}$
Solution
$\mathrm{KE}=\frac{1}{2} \mathrm{mV}_{\mathrm{C}}^{2}+\frac{1}{2} \mathrm{I}_{\mathrm{C}} \omega^{2}$
$=\frac{1}{2} \mathrm{mV}^{2}+\frac{1}{2} \mathrm{mR}^{2}\left(\frac{\mathrm{V}^{2}}{\mathrm{R}^{2}}\right)+\frac{1}{2} \mathrm{mV}^{2}+$
$\frac{1}{2} \frac{\mathrm{m}\left(2 \mathrm{R}^{2}\right)}{12} \frac{\mathrm{V}^{2}}{\mathrm{R}^{2}}$
Standard 11
Physics
