Two spheres are rolling with same velocity (f | vedclass

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Question

$K=K_{t r}+K_{r a t}=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}=$$\frac{1}{2} m v^{2}+\frac{1}{2} 	imes\left(\frac{2}{5} m r^{2}ight) 	imes\left

Vedclass · Accepted Answer

$2 : 1$

Vedclass · Answer

$K=K_{t r}+K_{r a t}=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}=$$\frac{1}{2} m v^{2}+\frac{1}{2} 	imes\left(\frac{2}{5} m r^{2}ight) 	imes\left(\frac{v}{r}ight)^{2}=\frac{7}{10} m v^{2}$

$\frac{K_{1}}{K_{2}}=\frac{m_{1}}{m_{2}} 	imes\left(\frac{v_{1}}{v_{2}}ight)^{2}$

Given, $v_{1}=v_{2}$

$\Rightarrow \frac{K_{1}}{K_{2}}=\frac{m_{1}}{m_{2}}$

$\Rightarrow \frac{2}{1}=\frac{m_{1}}{m_{2}}$

Thus if the ratio of kinetic energy is $2:1$ then their ratio of masses will also be $2:1$

Two spheres are rolling with same velocity (for their $C. M.$) their ratio of kinetic energy is $2 : 1$ & radius ratio is $2 : 1$, their mass ratio will be :

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