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A sphere having radius 5 cm rolling purely on a horizontal surface with angular velocity 3 rads-1, then what is the linear velocity for bottom-most point?

A sphere having radius 5 cm rolling purely on a horizontal surface with angular velocity 3 rads-1, then what is the linear velocity for bottom-most point? Correct Answer 0 ms<sup>-1</sup>

CONCEPT:

  • Rolling motion: Rolling motion can be regarded as a combination of pure rotation and pure translation.
  • In pure rolling motion, the point of contact (bottom-most point) moves with velocity Vp = velocity of point P = V
  • Vp = (V’ - ωR) = V
  • V’ = Vp – ωR

Where V’ = linear velocity of solid sphere

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EXPLANATION:

  • When an object is rolling on a plane without slipping, the point of contact of the object with the plane does not move.
  • In pure rolling motion, there is no relative motion between body & floor on which it is rolling (No sleep).

 

Vbottom point = Vlinear + (- Vrotational) = 0           

  • Hence, net linear velocity for the bottom-most point in pure rolling is zero. Therefore option 4 is correct.

NOTE:

If there is no pure rolling then two types of sleep occur

1. Forward slip

2. Backward slip

1. Forward slip

Linear velocity ˃ linear velocity (rotation)

So,  Vnet = V – (Rꙍ)rotation

2. Forward slip

Linear velocity ˂ linear velocity (rotation)

So,  Vnet = (Rꙍ)rotation – V 

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