- #1

kostoglotov

- 234

- 6

## Homework Statement

The problem

Direct imgur link to problem: http://i.imgur.com/QCc0zj6.jpg

## Homework Equations

[tex]\alpha = \frac{\tau_{net}}{I}[/tex]

[tex] I = \frac{1}{2}MR^2[/tex]

[tex]\tau = rF = rmg[/tex]

## The Attempt at a Solution

I have figured that the moment of inertia of the entire system must be used for both torques pulling on the system, so the one I is

[tex]I = \frac{1}{2}(5.0\times (0.05)^2 + 20\times (0.10)^2)[/tex]

So shouldn't the net angular acceleration just be the net torque of the two competing weights divided by the moment of inertia of the entire system?

[tex]\frac{0.10\times 2.5\times 9.8 + (- 0.05 \times 4.0 \times 9.8)}{\frac{1}{2}(5.0\times (0.05)^2 + 20\times (0.10)^2)} \approx 4.6 \frac{rad}{s^2} [/tex]

[The second torque trying to turn CW, so therefore is negative torque.]

However, the answer from the back of the text says [itex]3.5 \frac{rad}{s^2}[/itex]

What am I doing wrong? Do I need to consider this is terms of the tensions in the ropes instead? When considering the torque applied to the system by one hanging weight, does the other hanging weight count towards the moment of the inertia of the system?