- #1

bznm

- 184

- 0

## Homework Statement

I have two cylindric reels laying on the ground. They are identical: mass M, composed by a cylinder of radius R and two side cylinders of radius 2R. I have to find angular acceleration and work done by F in delta t.

**2. The attempt at a solution**

On the first cylinder the torque is ##\tau_1=3RF-RT##, while on the second is ##\tau_2=RT##

as the rope isn't extensible the two cylinders should have the same angular acceleration.

So ##\alpha_1=\alpha_2##

##3RF-3RT=RT##

##T=-\frac{3}{2} F##

##\alpha=R/I *3/2 F=\frac{3RF}{2I}##

Angular velocity of cylinder is ##\omega = \alpha t = \frac{3RF}{2I} t##.

Kinetic energy of the first cylinder is ##1/2 I \omega^2= \frac{9R^2F^2t^2}{8I}## and this is also the work done by F plus the work done by T on the first cylinder. But as ##T/F=-3/2## and they are applied in the same way with respect to the displacement I expect ##W_T/W_F=-3/2## so ##W_F=2/5 \Delta K##. Is it correct?

Thanks a lot