The uniform speed of a body is the same as seen from any point in the body. A light cord is wrapped around the rim of the disc and mass of 1 kg is tied to the free end. If it is released from rest,

The uniform speed of a body is the same as seen from any point in the

1.	The uniform speed of a body is the same as seen from any point in the body. A light cord is wrapped around the rim of the disc and mass of 1 kg is tied to the free end. If it is released from rest,
Answer» the TENSION in the cord is `5 N` in the first `4 s` the angular displacement of the disc is `40 rad` the work done by the TORQUE on the disc in the first `4 s` is `200J` the INCREASE in kinetic energy of the disc in the first `4 s` is `200 J` Solution :By `FBD` of particle `mg-T=ma`………i By `FBD` of the disc `TR=Ialpha=Ia/Rimplies T=(mR^2)/2a/R^2` `T=(ma)/2=a` By eqn i and ii we GET the following results `a. a=5 m//s^(2)` and `T=5N` and `alpha=a/5=5rad//s^(2)` b. For angular displacement of disc `theta=omegat+1/2alphat^(2)` `=1/2xx5xx4^2=40 rad` c. Work done by torque is `inttau d theta=tau int d theta=5xx40=200J` d. `/_\KE=/_\W=200J` `k_(2)-k_(1)=200J`