A small object of uniform density rolls up a curved surface with an initial velocity v- It reaches up to a maximum height of 3 v 2-4 g with respect to the initialposition- The object is Assume pure rolling coditionB- Solid sphereC- Hollow sphereD- Disc

The correct option is D DiscFrom conservation of energy 12mv^2 nbsp;+12Iω^2 = mg ( 3v^24g ) For pure rolling condition, ω = vR 12IR^2=34m 12m ⇒ I =mR^22 ...

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Standard XII

Physics

Angular Velocity

A small objec...

Question

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 (Assume pure rolling codition)

Ring

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Solid sphere

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Hollow sphere

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Disc

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Solution

The correct option is D Disc
From conservation of energy
$\frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} = m g (\frac{3 v^{2}}{4 g})$
For pure rolling condition, $ω = \frac{v}{R}$
$\frac{1}{2} \frac{I}{R^{2}} = \frac{3}{4} m - \frac{1}{2} m \Rightarrow I = \frac{m R^{2}}{2}$
Object must be disc.

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A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of 3v2(4g) with respect to the initial position. The object is (Assume pure rolling codition)

The correct option is D DiscFrom conservation of energy 12mv2+12Iω2=mg(3v24g) For pure rolling condition, ω=vR 12IR2=34m−12m⇒I=mR22 Object must be 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 (Assume pure rolling codition)

The correct option is D Disc
From conservation of energy
$\frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} = m g (\frac{3 v^{2}}{4 g})$
For pure rolling condition, $ω = \frac{v}{R}$
$\frac{1}{2} \frac{I}{R^{2}} = \frac{3}{4} m - \frac{1}{2} m \Rightarrow I = \frac{m R^{2}}{2}$
Object must be disc.