Question

A uniform disc of radius R is rotating about its axis with angular velocity mo and then it is gently
placed on a rough horizontal surface. After what time its Rotational motion ceases ?
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Answer 1

Dear Student,

◆ Answer -

t = Rω0 / 2μg

● Explanation -

Let μ be the coefficient of friction for rough surface & wheel.

F = μmg

Moment of inertia of solid disc is -

I = ½ mR²

Angular decceleration of the disc is thus calculated as -

α = τ / I

α = F.R / I

α = μmg.R / ½mR²

α = 2μg/R

Let t be the time at which the rotation ceases. i.e. ω=0.

Using laws of rotational motion -

ω = ω0 - αt

0 = ω0 - αt

t = ω0/α

t = ω0 / (2μg/R)

t = Rω0 / 2μg

Thus, the rotation will cease after Rω0/2μg .

Thanks dear...