Question

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,
A. the tension in the cord is `5 N`
B. in the first `4 s` the angular displacement of the disc is `40 rad`
C. the work done by the torque on the disc in the first `4 s` is `200J`
D. the increase in kinetic energy of the disc in the first `4 s` is `200 J`

Answer 1

Correct Answer - A::B::C::D
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`