Question

An experimental system consists of the solid cylinder of mass m and radius R that is able to rotate around its axis and is attached to the ceiling as shown. Two blocks of masses m1 and m2 are connected by a light, non-stretchable rope as shown. There is no slippage of the rope on the cylinder. Find the angular acceleration 'a' of the cylinder and the ratio of tension forces T1/T2 of the vertical portions of the rope while the blocks are moving. What is this ratio if mass of the cylinder m=0?

Answer 1

here is the pic for it attached

Answer 2

-T1+m1*g= m1* a..........(i)
T1-m2*g=m2*a..........(ii) you'll get a= (m1-m2)/(m1+m2) now find the tension substituting the value of a and then
torque= force(tension)* radius= moment of inertia * angular acceleration.... from this equation you'll get angular acceleration

Answer 3

\[\tau= i *\alpha = force*r\]

Answer 4

I got \[\alpha=(2F)/(mR)\] (of the cylinder) where i'm given 'm' and 'R'. but how can i relate it to Tension from the rope?

Answer 5

here force = tension acting in the rope

Answer 6

and you can find tension from the above equation

Answer 7

thank you!!

Answer 8

anytime