SATHEE: Rotational Motion Question 79

Rotational Motion Question 79

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 $\frac{3 υ^{2}}{4 g}$ with respect to the initial position. The object is

Options:

A) ring

B) solid sphere

C) hollow sphere

D) disc

Show Answer

Answer:

Correct Answer: D

Solution:

As $v = \sqrt{\frac{2 g h}{1 + \frac{k^{2}}{r^{2}}}}$

Given $h = \frac{3 v^{2}}{4 g}$ $v^{2} = \frac{2 g h}{1 + \frac{k^{2}}{r^{2}}} = \frac{2 g 3 v^{2}}{4 g (1 + \frac{k^{2}}{r^{2}})} = \frac{6 g v^{2}}{4 g (1 + \frac{u^{2}}{v^{2}})}$

$1 = \frac{3}{2 (1 + \frac{k^{2}}{v^{2}})}$ or $1 + \frac{k^{2}}{r^{2}} = \frac{3}{2}$ or $\frac{k^{2}}{r^{2}} = \frac{3}{2} - 1 = \frac{1}{2}$ $k^{2} = \frac{1}{2} r^{2}$

(Equation of disc) Hence, the object is disc.

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Rotational Motion Question 79

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 3υ24g with respect to the initial position. The object is

Options:

Answer:

Solution:

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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 $\frac{3 υ^{2}}{4 g}$ with respect to the initial position. The object is