A 4.0 kg block is suspended 1.0 m off the ground and is attached via a pulley and massless rope to a 2.0 kg mass held to the ground. The pulley is .12m in diameter and and has a mass of 2.0 kg. Friction between the axel and pulley exerts a torque or .5 Nm. How long does it take the 4.0 kg block to hit the ground.(adsbygoogle = window.adsbygoogle || []).push({});

Okay, first Happy New Year!!!!!! Why am I posting this on New Years Day? Because if I don't it will never stop bugging me!!

First I found the net Torque.

[tex] \sum \tau = \tau_{w of 4.0 kg} - \tau_{2.0 kg} - \tau_{friction}[/tex]

[tex] \sum \tau = 2.4 - 1.176 - .5 [/tex]

[tex] \sum \tau = .724 N*m [/tex]

I don't see a problem so far.

[tex] I = .5Mr^2 = .0036 kg m^2 [/tex]

[tex] \alpha = \frac{\sum \tau}{I} [/tex]

[tex] \alpha = 201.11 \frac{rad}{s^2} [/tex]

Ok now, the block falls 1.0m which is equal to 16.53 radians here.

[tex] \frac{.38 m}{2\pi rad} = \frac{1.0m}{x rad} [/tex]

[tex] x = 16.53 rad [/tex]

Then:

[tex] \theta_f = \theta_0 + \omega_0 t + .5\alpha t^2 [/tex]

If I solve this for t I get .41s, but the answer is 1.11s. Where did I go wrong? Thanks for the help.

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# Homework Help: Blocks on Pullys and Torque

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