Question

6) A uniform solid cylinder of mass M and radius R is free to rotate on frictionless horizontal axle. Two masses m hang from the two cords wrapped around the cylinder. If the system is released from rest then the tension in each cord will be -




(1) M m g 4 m + M (2) M m g 2 m + M
(3) M m g m + M (4) M m g M + 3 M

Answer 1

Answer is option (1). T = gMm/ (4m+M).

Problem clearly shown in fig. Enclosed. See.

The two masses m are hanging on the same side of the cylinder M.

mg - T = m a
2T* R. = I *alpha
a = R * alpha
Solve.

Answer 2

after solving the condition given in the image we get

a=4mg / 4m+M

therefore

T= Mmg / 4m+M