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A disc of radius R is rotating with an angular 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 being? (f) Calculate the time taken for the rolling to begin. |
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Answer» Solution :(a) The disc was in pure rotational motion before being brought in contact with the table, hence `u_(CM)=0` (b) Velocity of a point on the rim decreases when the disc is placed in contact with the table due to friction. (c ) Centre of mass ACQUIRES some LINEAR velocity when the rotating disc is placed in contact with the table due to friction. (d) Friction is responsible. (E) When rolling starts `v_(CM)=omegaR`. (f) Acceleration produced in centre of mass due to friction. `a_(CM)=(f)/(m)=(mu_(K)N)/(m)=(mu_(K)mg)/(m)=mu_(K)g` Angular retardation produced by the torque due to friction. `alpha=(tau)/(I)=(mu_(K)mgR)/(I)" "[because tau=fR=mu_(K)mgR]` `therefore v_(CM)=u_(CM)+a_(CM)t (because u_(CM)=0)` `therefore v_(CM)=mu_(K)g t` Now `omega=omega_(0)+ALPHAT` `therefore omega=omega_(0)-(mu_(K)mgR)/(I)t` For rolling without slipping, `(v_(CM))/(R)=omega` `therefore (v_(CM))/(R)=omega_(0)-(mu_(K)mgR)/(l)t` `(mu_(K)g t)/(R)=omega_(0)-(mu_(K)mgR)/(l)t` `t=(Romega_(0))/(mu_(K)g(1+(mR^(2))/(l)))` |
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