IIT JEE , AIEEE Forum - Physics , Mechanics

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a uniform disc is given an initial linear velocity = v , and an initial angular veloctiy (in an anti clockwise sense as shown in figure) = w . Given that 2v = rw . (friction is acting on the disc)...........

in this case both linear and angular velocities become zero at a time (i.e, in same time)

then what is its motion after both the linear velocity as well as angular velocity becomes zero.

i.e, will it stop its motion, or roll in forward direction or in backward direction ?

i think that it will stop its motion.........but i am not confident about it........so waiting for the reply...........

Raghudevan

when linear and angular velocity become zero it is neither rolling or spining or moving in the forward direction or anything, zero velocity implies that the body has stopped moving.

pramod6990

evn thinkit shud stop and shudnt continue any further.........

elessar_iitkgp

Assuming the initial linear and angular velocities to be ${v}_{0}$ and ${\omega}_{0}$ , such that $2{v}_{0}=R{\omega}_{0}$

Then, initially there is forward slipping. So friction acts backward.

$ma=\mu mg\\mR^2\alpha=\mu mgR$

So,

$v={v}_{0}-\mu gt$

$R\omega=R{\omega}_{0}-\mu gt=2{v}_{0}-\mu gt$

Its evident that becomes zero first. Still due to the rotational component, the slipping is forward. Friction continues to act bacward leading to developement of velocity in the backward direction. So the disc turns back after $t=\frac{{v}_{0}}{\mu g}$ and then

$v=\mu gt'$

$R\omega=R{\omega}_{0}-\mu gt'$

where $t'=t-\frac{{v}_{0}}{g}$

The velocity of point of contact is

${v}_{PC}=v-R\omega=2\mu gt'-R{\omega}_{0}$

Let pure rolling start at , then,

${v}_{PC}(T)=0\Rightarrow T=\frac{{v}_{0}}{\mu g}$

Hence, the disc moves forward till $t=\frac{{v}_{0}}{\mu g}$ andd then it starts moving back and attains pure rolling at $t=\frac{2{v}_{0}}{\mu g}$ (just when the angular velocity becomes zero)

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