Solving Torque on a Pulley: 0.5Nm, 4kg Block, 12cm Diameter

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The discussion focuses on calculating the time it takes for a 4 kg block to reach the floor when connected to a pulley with a diameter of 12 cm and a mass of 2 kg, experiencing a torque of 0.5 Nm due to friction. The user initially applied the equation for torque, Ʃτ = τleft-weight - τtension - τright-weight, but encountered discrepancies with the expected answer. Key insights include recognizing that the tension in the ropes, rather than the weights of the hanging blocks, exerts the torque on the pulley, necessitating additional equations for accurate acceleration calculations.

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cjavier
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The problem is: The two blocks in the figure are connected by a massless rope that passes over a pulley. The pulley is 12cm in diameter and has a mass of 2kg. As the pulley turns, friction at the axle exerts a torque of magnitude 0.5Nm. If the blocks are released from rest, how long does it take the 4 kg block to reach the floor?
Here is the picture:
pulley3.png


My approach was as follows:
Ʃτ = τleft-weight - τtension - τright-weight
Then, when I found total torque...
Ʃτ = I\alpha(\alpha is angular acceleration)
Then finally...
atangential = \alphar

When I found a, however, it was different than the answer on chegg. What am I doing wrong?

Thanks To Everyone on this site
 
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cjavier said:
My approach was as follows:
Ʃτ = τleft-weight - τtension - τright-weight
What exerts the torque on the pulley is the tension in the ropes, not the weights of the hanging.

Of course, those tensions depend on the weights. You'll need additional equations for each mass to solve for the acceleration.
 

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