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Question

A uniform sphere of mass m is given some angular velocity about a horizontal axis through its centre and gently placed on a plank of mass m. The co-efficient of friction between the two is μ. The plank rests on a smooth horizontal surface. The initial acceleration of the centre of the sphere relative to the plank will be:


A
zero
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B
μg
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C
(75)μg
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D
2μg
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Solution

The correct option is D 2μg

Sphere is rotating clockwise, therefore friction will act rightward on the sphere.
fk=μN=μmg ...(1)
Assume a1 and a2 are the accelerations of the sphere and plank respectively w.r.t ground as shown.

For sphere: fk=ma1
Using Eq. (1) μmg=ma1a1=μg

For plank: fk=ma2
From Eq. (1) μmg=ma2
a2=μg
[-ve sign indicates direction is opposite to that assumed]
Acceleration of sphere w.r.t plank (asp)
asp=asgapg
=a1a2
=(μg)(μg)
=2μg

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