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 radii `s^(2)`
B. The force of kinetic friction is 40N
C. The acceleration is `-3.2p m//s^(2)`
D. None of the above

Answer

B, C

Explanation

Let F is the force applied to the rope and assuming the cylinder rolls without siding.

`rArr F- f=ma rArr a=(f(1-R_(1)//R_(2))/(m(1+R^(2)//R_(2)^(2))))`
`fR_(2) -FR_(1) = mk^(2) alpha`
`a = R_(2)alpha`
N=mg & `f = (F[K^(2) + R_(1)R_(2)])/(K^(2) + R_(2)^(2)]`
And to avoid sliding `f le N mu`
`rArr (F [R^(2) + R_(1)R_(2)])/(K^(2) + R_(2)^(2)) le mg mu_(s) rArr F le (mgmu_(S)[R^(2) + R_(2)^(2)])/(K^(2) + R_(1)R_(2))`
`rArr F le 50 xx 10 xx 0.1(0.3^(2) + 0.6^(2))/(0.3^(2) + (0.3) (0.6)) rArr F le (250)/3`N
But applied force is 200 N `rArr` rolling with sliding and `f=f_(g) = Nmu_(K) = 50 xx 10 xx 0.08 = 40 N`
`a=(F-f)/m` towards left `=(20 -40)/50 = 3.2 m//s^(2)`