Answer:
Explanation:
mass, m = 48 kg
radius, r = 0.59 m
acceleration of center of mass, a = 4.2 m/s2
The free body diagram of cylinder is as follows:
Since the cylinder is rolling without slipping, therefore,a=αr,
which gives, angular acceleration, α = a/r = (4.2/0.59)rad/s2 = 7.12 rad/s2
Moment of inertia of cylinder about center of mass C, IC= (1/2)mr2
So, by parallel axis theorem, moment of inertia about point A,IA = (1/2)mr2 + mr2 =(3/2)mr2 = (3/2)*48*0.592 kg.m2 =25.06 kg.m2
Now, taking the moment of forces about point A, T =IAα
which gives, B*r = IAα, and we get, B =(25.06*7.12)/0.59 N = 302.42 N