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 [2007]

Question

Smart Achievers · Accepted Answer

(4)

Applying energy conservation principle of

$\frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} = m g (\frac{3 v^{2}}{4 g})$

$∴$ $\frac{1}{2} m v^{2} + \frac{1}{2} I \frac{v^{2}}{R^{2}} = \frac{3}{4} m v^{2}$ $[∵ v = R ω]$

$∴$ $\frac{1}{2} I \frac{v^{2}}{R^{2}} = \frac{3}{4} m v^{2} - \frac{1}{2} m v^{2} = \frac{1}{4} m v^{2}$ $\Rightarrow I = \frac{1}{2} m R^{2}$

This is the moment of inertia of the disc hence the object is disc.

Solution

Q.
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 [2007]

1 ring

2 solid sphere

3 hollow sphere

4 disc

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