A small object of uniform density rolls up a curved surface with an initial velocity 'v'. It reaches upto a maximum height of 3v²4g with respect to the initial position. The object is (a) hollow sphere (b) disc (c) ring (d) solid sphere

Answer: (b). Energy: 1/2mv²(1 + (k²)/(R²)) = mgh with h = (3v²)/(4g). 1/2(1 + (k²)/(R²)) = 3/4 ⇒ (k²)/(R²) = 1/2 k² = 1/2R² is a disc. So the answer is (b).

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Rolling Motion

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

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Answer: (b)

Energy: $\frac{1}{2}mv^2\left(1 + \frac{k^2}{R^2}\right) = mgh$ with $h = \dfrac{3v^2}{4g}$ .
$\frac{1}{2}\left(1 + \frac{k^2}{R^2}\right) = \frac{3}{4} \Rightarrow \frac{k^2}{R^2} = \frac{1}{2}$
$k^2 = \frac{1}{2}R^2$ is a disc.
So the answer is (b).

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