SATHEE: Rotational Motion Question 52

Rotational Motion Question 52

Question: A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height h. What is the speed of its centre of mass when the cylinder reaches its bottom?

[AIPMT 2003]

Options:

A) $\sqrt{\frac{4}{3} g h}$

B) $\sqrt{4 g h}$

C) $\sqrt{2 g h}$

D) $\sqrt{\frac{3}{4} g h}$

Show Answer

Answer:

Correct Answer: A

Solution:

Potential energy of cylinder at the top will be converted into rotational kinetic energy and translational kinetic energy.

So, energy conservation gives.

$M g h = \frac{1}{2} M v^{2} + \frac{1}{2} I ω^{2}$

$= \frac{1}{2} M v^{2} + \frac{1}{2} \frac{M R^{2}}{2} \frac{v^{2}}{R^{2}}$

$(∵ I_{c y l i n d e r} = \frac{M R^{2}}{2})$

So, $M g h = \frac{1}{2} M v^{2} + \frac{1}{4} M v^{2}$

or $M g h = \frac{3}{4} M v^{2}$

or $v^{2} = \frac{4}{3} g h$

or $v = \sqrt{\frac{4}{3} g h}$

Note: In a collision of two bodies whether it is perfectly elastic or inelastic, linear momentum is always conserved but kinetic energy need not to be conserved.

Rotational Motion Question 52

Question: A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height h. What is the speed of its centre of mass when the cylinder reaches its bottom?

Options:

Answer:

Solution:

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Rotational Motion Question 52

Question: A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of length L and height h. What is the speed of its centre of mass when the cylinder reaches its bottom?

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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