Falling Block Pulling Rolling Cylinder

1. The problem statement, all variables and given/known data

A uniform, solid cylinder of mass M = 4.22 kg and radius R = 0.36 m with I=1/2(M*R^2) is attached at its axle to a string. The string is wrapped around a small frictionless pulley (I=0) and is attached to a hanging block of mass 1.69 kg. You release the objects from rest. Assume the cylinder rolls without slipping. What is the acceleration of the block

2. Relevant equations


3. The attempt at a solution


∑X: Tension-ƒs=M(c)*a
∑Torque (C.O.M in center of cylinder): ƒsR+M(c)g2R=I[itex]\alpha[/itex]

I’m assuming, on the cylinder, normal and weight offset, so N=M(c)g, and because the normal makes contact at the bottom of the cylinder, the distance is the diameter (or 2R in the torque calculation).


Back to Torque, subbing for tension and static friction, inertia and angular acceleration in terms of linear acceleration (and canceling out on the I[itex]\alpha[/itex] side):

M(b)gR + M(c)g2R = a((M(c)/2) + M(b)R + M(c)R)

So, substituting in the given numbers, I get:

((1.69*9.81*0.36) + (4.22*9.81*0.72)) / ((0.5*4.22) + (1.69*0.36) + (4.22*0.36)) = a

Basically, I solve for tension using the block. Then I solve for static friction using the sum of forces in the X direction of the cylinder. Then I solve for linear acceleration using torques of the cylinder, which is just the normal (with a radius of 2R) and static friction. Answers I’ve gotten somehow:

1 Incorrect. (Try 1) 0.36 m/s/s
2 Incorrect. (Try 2) 36.96 m/s/s
3 Incorrect. (Try 3) 2.62 m/s/s
4 Incorrect. (Try 4) 1.41 m/s/s
5 Incorrect. (Try 5) 2.13 m/s/s
6 Incorrect. (Try 6) 2.67 m/s/s
7 Incorrect. (Try 7) 8.44 m/s/s


Leave a Reply

Name *
Email *