question_answer 56)

                  A disc of radius R is rotating with an angular speed \[{{\omega }_{0}}\] 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 centre 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 centre 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.                

Answer:

                  (a) When disc was not brought in contact with the table, it has pure rotational motion. Therefore, the velocity of C.M. is zero.                 (b) Linear velocity of a point on the rim of disc will decrease.                 (c) Linear speed of the C.M. increases.                 (d) Force of friction is responsible for the decrease of linear velocity of a point on the rim of disc and for the increase of linear speed of the C.M. of the disc                 (e) The disc will rolls over the surface without slipping, if the point of contact of the disc at any instant with the surface of die table is at rest and velocity of centre of mass, \[{{\upsilon }_{cm}}=\,R\omega \].                 (f) Force of friction acting oil the disc, \[F={{\mu }_{k}}mg\]                 \Linear acceleration of the C.M. of the disc, \[a=\,\frac{F}{m}\,=\,{{\mu }_{k}}g\]                 Torque acting oil the disc due to force of friction about the centre of mass of the disc,                 \[\tau =F\times \,R\,=\,{{m}_{k}}\,mg\,\,R,\] where R is radius of the disc                 Angular acceleration of the disc,                 \[\alpha =\,\frac{\tau }{I},\]                 where I is the moment of inertia of the disc                 \[\therefore \] \[\alpha \,=\frac{{{\mu }_{k}}\,mg\,R}{I}\]                 Let the disc begins to roll after time t.                 Linear velocity of the C.M. of the disc after time t,                 \[\upsilon =\,u+\,at=0\,\,+\,{{\mu }_{k}}\,gt\,=\,{{\mu }_{k}}\,gt\]                 Angular velocity of the disc after time t,                 \[\omega \,={{\omega }_{0}}+\,\alpha t=\,{{\omega }_{0}}-\,\frac{{{\mu }_{k}}\,mg\,R}{I}\,t\]                                                                 (Here \['\alpha '\] negative)                 Since \[\omega =\,\frac{\upsilon }{R}\]                 \[\therefore \]\[\frac{\upsilon }{R}={{\omega }_{0}}-\frac{{{\mu }_{k}}\,mg\,R}{I}\,t\]                 or \[\frac{{{\mu }_{k}}\,gt}{R}=\,{{\omega }_{0}}\,-\,\frac{{{\mu }_{0}}\,mg\,\,R}{I}\,t\]                 or \[{{\mu }_{k}}\,gt\,\,\left[ \frac{1}{R}+\,\frac{mR}{I} \right]\,=\,{{\omega }_{0}}\]                 \[{{\mu }_{k}}\,gt\,\left[ \frac{I+\,m{{R}^{2}}}{IR} \right]\,=\,{{\omega }_{0}}\]                 \[t=\frac{{{\omega }_{0}}\,IR}{{{\mu }_{k}}\,g\,[I\,+\,m{{R}^{2}}]}\]                 \[=\,\frac{{{\omega }_{0}}R}{{{\mu }_{k}}g\,\left( 1+\,\frac{m{{R}^{2}}}{I} \right)}\]                 For a disc, \[I=\,\frac{1}{2}\,m{{R}^{2}}\]                 \[\therefore \] \[t=\,\frac{{{\omega }_{0}}R}{3{{\mu }_{k}}g}\]