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A cylinder of mass mand radius R is rolling without slipping on a horizontal surface with angular velocityω0. The velocity of the centre of the mass cylinder isωoR. The cylinder comes across a step of heightR/4. Then the angular velocity of the cylinder just after the collision is (assume cylinder remains in contact and no slipping occurs on the edge of the step)

1303843

A

5ω06

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B

ω0

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C

2ω0

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D

6ω0

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Solution

The correct option is A

5ω06


Step 1: Given data

Mass of cylinder=m

Radius of cylinder=R

Angular velocity of the cylinder before the collision=ω0

The velocity of the centre of mass, v0=ω0R

Height, h=R4

The radius of the centre of mass, R0=R-R4=3R4

Step 2: Assumptions

Initial angular momentum=Li

Final angular momentum=Lf

Velocity of the after collision=v

Angular velocity of the cylinder after the collision=ω

Step 3: Calculation of the angular velocity of the cylinder just after the collision

The cylinder remains in contact with the step after the collision and hence the collision is “in-elastic”.

After the collision, torque about point “P” is zero as depicted by the diagram and hence the law of conservation of angular momentum can be applied.

According to the basic crux of the law of conservation of linear momentum, “the initial angular momentum must be equal to the final angular momentum”.

Li=Lf ………………………….(a)

Step 3.1: Calculation of the initial angular momentum

Initial angular momentum, Li=mvR0+12mR2ω0

Li=mv03R4+12mR2ω0

Li=34mv0R+12mR2ω0 ………………………………(b)

Step 3.2: Calculation of the final angular momentum and the angular velocity of the cylinder after collision

Final angular momentum, Lf=12mR2ω+mR2ω

Lf=32mR2ω……………………….(c)

Using equations (b) and (c) in equation (a), we get

32mR2ω=34mv0R+12mR2ω0

Using, ω0=v0R

32mR2ω=34mv0R+12mR2v0R32Rω=34v0+12v032Rω=3v0+2v0432Rω=5v04ω=56×v0Rω=56ω0

Hence, option A is correct.


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