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  • A disc of radius R is rotating with an angular speed \omega _{o} about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is \mu _{k}. (a) What was the velocity of its center of mass before being brought in contact

A disc of radius R is rotating with an angular speed \omega _{o} about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is \mu _{k}.
(a) What was the velocity of its center of mass before being brought in contact with the table?
(b) What happens to the linear velocity of a point on its rim when placed in contact with the table?
(c) What happens to the linear speed of the center of mass when disc is placed in contact with the table?
(d) Which force is responsible for the effects in (b) and (c).
(e) What condition should be satisfied for rolling to begin?
(f) Calculate the time taken for the rolling to begin.

Answers (1)

(a) Before coming in contact with the table, the disc was undergoing only rotational motion about its axis that passes through the centre. The velocity of C.O.M =0 since the point on the axis is considered at rest

(b)When placed in contact with the table, the force of friction reduces the linear velocity of a point on the rim.

(c) The linear velocity causes a change in momentum (action force). By Newton’s third law of motion wherein, every object has an equal and opposite reaction a reaction force is applied on the disc due to which it moves in the direction of reaction force, so Center of mass acquires a linear velocity.

(d) Force of friction

(e) The C.O.M acquires a velocity due to reaction force due to rotation. Therefore

v_{cm} at the start when just comes in contact with table is v_{cm}=\omega _{o}R

v_{cm}=0

v_{cm}=\omega _{o}R

F=ma

a=\frac{F}{m}=\frac{\mu _{k}mg}{m}=\mu _{k}g

Angular retardation (\alpha ) produced by frictional torque =\frac{\tau }{I}=\frac{r\times F}{I}=\frac{R\times \mu _{k}mg}{I}=\frac{R\mu _{k}mg\; \sin\; \theta}{I}

As R and F are perpendicular,

\alpha =\frac{-\mu _{k}mgR}{I}

v_{cm}=u_{cm}+a_{cm}t (for linear velocity)

v_{cm}=0+\mu _{k}gt=\mu _{k}gt

\omega =\omega _{0}-\alpha t (for rotational motion)

\omega =\omega _{0} -\frac{\mu _{k}mgR}{I}t

v_{cm}=\omega R

\omega =\frac{v_{cm}}{R}

\frac{v_{cm}}{R}=\omega _{0}-\frac{\mu _{k}mgR}{I}t

\frac{\mu _{k}gt}{R}+\frac{\mu _{k}mgRt}{I}=\omega _{0}

\omega _{0}=\mu _{k}gt\left ( \frac{1}{R}+\frac{mR}{I} \right )=\frac{\mu _{k}gt}{R}\left ( 1+\frac{mR^{2}}{I} \right )

t=\frac{R\omega _{0}}{\mu _{k}g\left ( 1+\frac{mR^{2}}{I} \right )}

Thus frictional force help in pure rolling motion without slipping.

 

Posted by

infoexpert23

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