Calculating Alpha: Magnitude of Angular Acceleration??

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SUMMARY

The discussion focuses on calculating the angular acceleration (alpha) of a uniform solid cylinder when a block is attached to a string wrapped around it. The key formula derived is alpha = 2g/r, where g represents the acceleration due to gravity and r is the radius of the cylinder. The torque acting on the cylinder is calculated as mgr, with I being the moment of inertia, expressed as I = (1/2)mr^2. This relationship establishes a clear connection between the linear acceleration of the block and the angular acceleration of the cylinder.

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calculating alpha??

A string is wrapped around a uniform solid cylinder of radius r, as shown in the figure. The cylinder can rotate freely about its axis. The loose end of the string is attached to a block. The block and cylinder each have mass m.

Find the magnitude alpha of the angular acceleration of the cylinder as the block descends.
Express your answer in terms of the cylinder's radius r and the magnitude of the acceleration due to gravity g.




Acceleration of the block is = g
Force acting on the block is = mg
The torque by this force on the cylinder about the axis = mgr
This results in accelerating the rotation of the cylinder.
mgr = I x alpha, where I = moment of inertia of the solid cylinder;
alpha = Angular acceleration of the cylinder.

I = (1/2)mr^2

alpha = torque/I = mgr/I = 2g/r
??
 
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merlos said:
Acceleration of the block is = g
Force acting on the block is = mg
That would be true if the block was in free fall. But I assume it's tied to the string that's wrapped around the cylinder.
 

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