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In figure $E$ and $v_{\mathrm{cm}}$ represent the total energy and speed of centre of mass of an object of mass $1 \mathrm{~kg}$ in pure rolling. The object is
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The correct answer is:
disc
As we know, kinetic energy of an object in pure rolling motion is given as
$E=\frac{1}{2} m v_{\mathrm{cm}}^{2}\left(1+\frac{k^{2}}{R^{2}}\right)$
where, $k$ is radius of gyration and $m$ is the mass of the object.
$\Rightarrow \quad \frac{E}{v_{\mathrm{cm}}^{2}}=\frac{1}{2}\left[1+\frac{k^{2}}{R^{2}}\right] ...(i)$ $[\because$ Given, $m=1 \mathrm{~kg}$ ]
From the given graph, substituting the value of $\frac{E}{v_{\mathrm{cm}}^{2}}$ in Eq. (i), we get
$\begin{aligned}
\frac{3}{4} &=\frac{1}{2}\left[1+\frac{k^{2}}{R^{2}}\right] \\
\Rightarrow \quad \frac{3}{2}-1 &=\frac{k^{2}}{R^{2}} \\
\Rightarrow \quad \frac{k^{2}}{R^{2}} &=\frac{1}{2}
\end{aligned}$
Since, for a
(a) Sphere, $\frac{k^{2}}{R^{2}}=\frac{2}{5}$
(b) Ring, $\frac{k^{2}}{R^{2}}=1$
(c) Hollow cylinder, $\frac{k^{2}}{R^{2}}=1$
(d) Disc, $\frac{k^{2}}{R^{2}}=\frac{1}{2}$
So, the given object is disc.
$E=\frac{1}{2} m v_{\mathrm{cm}}^{2}\left(1+\frac{k^{2}}{R^{2}}\right)$
where, $k$ is radius of gyration and $m$ is the mass of the object.
$\Rightarrow \quad \frac{E}{v_{\mathrm{cm}}^{2}}=\frac{1}{2}\left[1+\frac{k^{2}}{R^{2}}\right] ...(i)$ $[\because$ Given, $m=1 \mathrm{~kg}$ ]
From the given graph, substituting the value of $\frac{E}{v_{\mathrm{cm}}^{2}}$ in Eq. (i), we get
$\begin{aligned}
\frac{3}{4} &=\frac{1}{2}\left[1+\frac{k^{2}}{R^{2}}\right] \\
\Rightarrow \quad \frac{3}{2}-1 &=\frac{k^{2}}{R^{2}} \\
\Rightarrow \quad \frac{k^{2}}{R^{2}} &=\frac{1}{2}
\end{aligned}$
Since, for a
(a) Sphere, $\frac{k^{2}}{R^{2}}=\frac{2}{5}$
(b) Ring, $\frac{k^{2}}{R^{2}}=1$
(c) Hollow cylinder, $\frac{k^{2}}{R^{2}}=1$
(d) Disc, $\frac{k^{2}}{R^{2}}=\frac{1}{2}$
So, the given object is disc.
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