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
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
2 answers
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.
But torque is not equal to 3g*radius but tension*radius. Do you agree with me bobpursley? Thanks.