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.

A stepped cylinder having mass 50 kg and a radius of gyration (K) of 0

1.	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.
Answer» the ANGULAR accelerationis `2.67 rads^(-2)` The force of kinetic friction is `40 N` The acceleration is `3.2 ms^(-2)` towards let. The acceleratiion is `3.2 ms^(-2)` towards right Solution :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)`