Question

A bucket containing cement has a mass of 3kg. One end of a rope is attached to the handle of the bucket and the other end is wound around a horizontally mounted cylinder on frictionless bearings. The mass of cylinder is 6kg and its radius is 0.10m. The moment of inertia of the cylinder is given by I=1/2MR^2, where M is its mass and R its radius. given that the bucket is released from rest , calculate

(i) the moment of inertia of the cylinder

I=1/2MR^2
I=0.03

(ii) the acceleration of the bucket
(iii) the tension in rope
(iv) rotational ke of cylinder after 5s

Answer 1

the acceleration of the bucket is
a=angular acceleartion of cylnder*radius
torque=I*angacceleration
3g*radius=1/2 m r^2*alpha
solve for angacceleration.
a= angacceleration*radiuscyclinder

Tension=mg-ma

KE=1/2 I w^2
where w=angacceleration*time

Answer 2

I agree that torque=I*ang_acc.
But torque is not equal to 3g*radius but tension*radius. Do you agree with me bobpursley? Thanks.