Chapter 8 - Rotational Motion - Problems: 27

$\Sigma \tau = mg \mathcal{l}_2-mg \mathcal{l}_1$ $clockwise$.

Work Step by Step

There is a torque due to the weight on the left that tends to accelerate the rod counterclockwise and a clockwise torque due to the weight on the right. Let’s call clockwise the positive direction. $$\Sigma \tau = mg \mathcal{l}_2-mg \mathcal{l}_1$$ The net torque is clockwise.

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