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Question

A uniform solid sphere of mass m and radius R is placed on a horizontal surface. A horizontal linear impulse P is applied at a height R2 above its center. Find the velocity of center of mass and angular velocity of sphere just after the impulse is applied.


A

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B

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C

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D

None of these

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Solution

The correct option is A


All you have to recall is

linear impulse = change In linear momentum - - - - - - (1)

angular impulse = change in angular momentum - - - - - - (2)

Here,

linear impulse = P (given)

change in linear momentum = linear momentum final - linear momentum initial

= mVCM0 [as initially the body was at rest

and let's say it moves with VCM after the impulse]

= mVCM

According to (1)

P =mVCM

VCM=Pm

Angular impulse = linear impulse × perpendicular distance of line of

application of force from the axis of rotation

Change in angular momentum =I(ωfωi)

[ωi=0,ωf=ω]

=Iω

According to (2)

PR2=Iω

PR2=25mR2ωω=5P4mR


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