# Two reels rolling without slipping

1. May 10, 2016

### bznm

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

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

2. May 10, 2016

### BvU

Interesting, can you explain ?
The rope can roll up (or off) on the reels, I presume ?
 A bit late, but I get it. Let me sit back and think a little longer before posting

3. May 10, 2016

### bznm

Oh my gosh, you're right.. I'm totally wrong(i don't know why but I considered it as a rod) and I don't know what to do now :/

4. May 10, 2016

### BvU

But the "$\tau_1=3RF−RT$, while on the second is $\tau_2=RT$" is correct. That gives angular acceleration and therefore also (uniformly) accelerated translation

5. May 10, 2016

### bznm

I tried: (where A is friction force)
$F-A-T=M a_1 = M \alpha_1 2R$
$T-A=M a_2 = M \alpha_2 2R$

Subtracting, I get: $F-2T=M(2R)(\alpha_1-\alpha_2)$
$F-2T=M(2R)\frac{3RF-2RT}{I}$
$T=F\frac{1+6MR^2 / I}{2+4MR^2 / I}$

Is it correct?
What about the work? Is it correct my reasoning in applying kinetic energy theorem?