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Systems of Particles and Rotational Motion — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsSystems of Particles and Rotational Motion2 Marks
Question
State whether the statement given below is true or false giving reason in brief. A ring of mass 0.3kg and radius 0.1m and a solid cylinder of mass 0.4kg and of the same radius are given the same kinetic energy and released simultaneously on a flat horizontal surface such that they begin to roll as soon as released towards a wall which is at the same distance from the ring and cylinder. The rolling friction in both the cases is negligible. The cylinder will reach the wall first.
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Answer

In case of rolling, total kinetic energy $\text{K}=\frac{1}{2}\text{mv}^2+\frac{1}{2}\text{I}\omega^2$ with $\omega=\frac{\text{v}}{\text{r}}$ $\text{K}=\frac{1}{2}\text{mv}^2+\frac{1}{2}\text{I}\frac{\text{v}^2}{\text{r}^2}=\frac{1}{2}$ $=\text{mv}^2\big[1+\frac{\text{I}}{\text{mr}^2}\big]$For a ring, $\text{I}=\text{mr}^2$
$\therefore\text{K}_{\text{r}}=\frac{1}{2}\text{mv}^2(1+1)=\text{mv}^2$ $\text{v}=\sqrt{\frac{\text{Kr}}{\text{m}}}=\sqrt{\frac{\text{K}_{r}}{0.3}}$ For a cylinder, $\text{I}=\frac{1}{2}\text{mr}^2$ $\therefore\text{IK}_{\text{c}}=\frac{3}{4}\text{mv}^2$ $\text{v}=\sqrt{\frac{4\text{K}_{\text{c}}}{3\text{m}}}=\sqrt{\frac{\text{K}_{\text{c}}}{\frac{3}{4}\times0.4}}$ $=\sqrt{\frac{\text{K}_{\text{c}}}{0.3}}$ As $\mathrm{K}_{\mathrm{r}}=\mathrm{K}_{\mathrm{c}}$, therefore, velocity (v) of both the ring and cylinder is same. As the motion is uniform, both the ring and cylinder will reach the wall at the same time. The given statement is false.

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