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(D) For an object rolling without slipping,the total initial kinetic energy is the sum of translational and rotational kinetic energy: $K_i = \frac{1}

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(D) For an object rolling without slipping,the total initial kinetic energy is the sum of translational and rotational kinetic energy: $K_i = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2(1 + \frac{k^2}{r^2})$.At the maximum height $h$,the kinetic energy is converted into potential energy: $K_i = mgh$.Equating the two: $\frac{1}{2}mv^2(1 + \frac{k^2}{r^2}) = mgh$.Given $h = \frac{3v^2}{4g}$,we substitute this into the equation:$\frac{1}{2}v^2(1 + \frac{k^2}{r^2}) = g(\frac{3v^2}{4g})$$\frac{1}{2}(1 + \frac{k^2}{r^2}) = \frac{3}{4}$$1 + \frac{k^2}{r^2} = \frac{3}{2}$$\frac{k^2}{r^2} = \frac{1}{2}$.Since the radius of gyration $k$ for a disc is given by $k^2 = \frac{1}{2}r^2$,the object is a disc.

$A$ small object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches up to a maximum height of $\frac{3 v^2}{4 g}$ with respect to the initial position. The object is

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