- #51

Chestermiller

Mentor

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No. If there were no horizontal velocity, the downward acceleration should be ##d^2r/dt^2##, not ##-d^2r/dt^2##.I don't seem to be having as much fun .

What was the mistake in the earlier analysis where we found right mass going down and left moving up ?

If origin is placed at the pulley and downwards is considered positive , shouldn't that be

$$mg-T=m\left[-\frac{d^2r}{dt^2}-r\left(\frac{d\theta}{dt}\right)^2\right]$$