A uniform, solid cylinder of mass M = 5.96 kg and radius R = 0.25 m with Icm=1/2

Question

A uniform, solid cylinder of mass M = 5.96 kg and radius R = 0.25 m with Icm=1/2MR^2 is attached at its axle to a string. The string is wrapped around a small frictionless pulley (Icm=0) and is attached to a hanging block of mass 2.35 kg as indicated in the above below. You release the objects from rest. Assume the cylinder rolls without slipping.

What is the magnitude of the acceleration of the block?  
What is the tension in the string?  

A uniform, solid cylinder of mass M = 5.96 kg and radius R = 0.25 m with Icm=1/2MR^2 is attached at its axle to a string. The string is wrapped around a small frictionless pulley (Icm=0) and is attached to a hanging block of mass 2.35 kg as indicated in the above below. You release the objects from rest. Assume the cylinder rolls without slipping. What is the magnitude of the acceleration of the block? What is the tension in the string?

Explanation / Answer

for rolling without slipping

tangential velocity of roll = linear velocity of block

at = tangential acceleration   al = linear velocity

Vt = Vl

at*t = al * t

at = al   

force required to roll without slipping

F1= (at/r) * I + M*al = al ( 0.5M + 1M) = a * 1.5 M

force required to accelerate hanging block to value acceleration a

F2=m*a

total force = a* (1.5M + m ) which is equal to gravitational force acting on hanging block.

a*(1.5M + m) = m*g

a =2.35*9.8/( 1.5 * 5.96 + 2.35) = 2.04 m/sec2

tension in the string = gravitational force acting on hanging block - ma

=2.35*9.8 - 2.04*2.35

=18.236 N

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