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## Homework Statement

A massless rope is wrapped several times around a solid cylinder of radius R = 20 cm, and mass M = 20 kg, which is at rest on a horizontal surface. Someone pulls 1 m or the rope with a constant force of 100 N, setting the cylinder in motion. Assuming that the rope neither stretches nor slips, and that the cylinder rolls without slipping, what is the final angular velocity of the cylinder of mass M and radius R. The moment of inertia of the cylinder is MR

^{2}/2.

## Homework Equations

T = F.R

L = Iw

T = dL/dt

T is torque, F force, R radius, I moment of inertia, W angular speed, L angular momentum, Y is the angle

## The Attempt at a Solution

Torque is 100x0.2 = 20 Nm. Moment of inertia is 0.4 kg m

^{2}. Now rearranging torque as a function of angular momentum, you get T = I dw/dt.

Some chain rule... dw/dt = w dw/dY. A little bit of integrating... 0.5 w

^{2}= (T/I)Y -> using Y=0 as the lower boundary.

Okay, now because 1m is pulled from the rope, and the circumference of the cylinder is 1.256m, it works out that the sphere has rotated by 5 rads. Put that, along with T=20 and I=0.4 into the above equation, you come out with 22.36 rad /s. The actual answer is 18.4 rad /s. Ugh!