In the figure,E and v_{cm} represent the tota | vedclass

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(C) The total kinetic energy of an object in pure rolling motion is given by:$E = \frac{1}{2} m v_{cm}^{2} \left(1 + \frac{k^{2}}{R^{2}}\right)$where

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disc

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(C) The total kinetic energy of an object in pure rolling motion is given by:$E = \frac{1}{2} m v_{cm}^{2} \left(1 + \frac{k^{2}}{R^{2}}\right)$where $k$ is the radius of gyration,$R$ is the radius,and $m$ is the mass of the object.Given $m = 1 \ kg$,the expression becomes:$\frac{E}{v_{cm}^{2}} = \frac{1}{2} \left(1 + \frac{k^{2}}{R^{2}}\right) \quad ...(i)$From the given graph,the slope of the line is $\frac{E}{v_{cm}^{2}} = \frac{3}{4}$.Substituting this value into Eq. $(i)$:$\frac{3}{4} = \frac{1}{2} \left(1 + \frac{k^{2}}{R^{2}}\right)$$\frac{3}{2} = 1 + \frac{k^{2}}{R^{2}}$$\frac{k^{2}}{R^{2}} = \frac{3}{2} - 1 = \frac{1}{2}$We know that for a disc,the moment of inertia $I = \frac{1}{2} m R^{2}$,so $k^{2} = \frac{1}{2} R^{2}$,which means $\frac{k^{2}}{R^{2}} = \frac{1}{2}$.Thus,the object is a disc.

In the figure,$E$ and $v_{cm}$ represent the total energy and speed of the centre of mass of an object of mass $1 \ kg$ in pure rolling. The object is

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