Due to slipping, points $A$ and $B$ on the rim of the disc have the velocities shown. Distance between centre of disc and point on disc which is having zero velocity.

0.04

m

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0.05

m

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0.06

m

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None

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Solution

The correct option is A 0.04 $m$
Angular velocity with respect to $\frac{v_{1}}{r}$
$= \frac{4.5}{0.4}$
If $A$ be at rest let a point at distance $x$ from
$A$ has zero velocity . Then
$\begin{matrix} V_{A} + r q = 0 1.5 = - k (\frac{4.5}{0.4}) - \frac{0.4}{3} = k k = - 0.133 \end{matrix}$
Distance of this point from center $= 0.2 + 0.133$ m
$= 0.333$ m

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Due to slipping, points A and B on the rim of the disc have the velocities shown. Distance between centre of disc and point on disc which is having zero velocity.

The correct option is A 0.04mAngular velocity with respect to v1r=4.50.4If A be at rest let a point at distance x fromA has zero velocity . ThenVA+rq=01.5=−k(4.50.4)−0.43=kk=−0.133Distance of this point from center =0.2+0.133m=0.333m

Due to slipping, points $A$ and $B$ on the rim of the disc have the velocities shown. Distance between centre of disc and point on disc which is having zero velocity.