A stepped cylinder having mass `50 kg` and a radius of gyration (`K`) of `0.30 m`. The radii `R_(1)` and `R_(2)` are respectively `0.3 m` and `0.6 m`. A pull `T` equal to `200 N` is exerted on the rope attached to inner cylinder. The coefficient of static and dynamic friction between cylinder and ground are `0.1` and `0.08` respectively.

Options

A. the angular acceleration is `2.67 rads^(-2)`
B. The force of kinetic friction is `40 N`
C. The acceleration is `3.2 ms^(-2)` towards let.
D. The acceleratiion is `3.2 ms^(-2)` towards right

Answer

B, D

Explanation

There is no slipping, therefore `tau=Ialpha`
`T[R_(2)-R_(1)]=[MK^(2)+MR_(2)^(2)]alpha`
`alpha=(200(0.6-0.3))/(50(0.362+0.6^(2)))=2.67rad//s`
using Newton's IInd law
`T-f=MR_(2)alpha`
`f=(T-MR_(2)alpha)=[200-5xx0.6xx8/3]=120N`
Now `F_("max")=musmg=0.1xx50xx10=50N`
`f_(k)=mu_(s)Mg=0.08xx50xx10=40N`
`Ma=T+40impliesa=-3.2m//s^(2)`