Calculating Alpha: Magnitude of Angular Acceleration??

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The discussion focuses on calculating the angular acceleration (alpha) of a uniform solid cylinder when a block attached to a string wrapped around it descends. The block experiences a gravitational force of mg, which creates a torque on the cylinder equal to mgr. Using the moment of inertia for a solid cylinder, I = (1/2)mr^2, the relationship between torque and angular acceleration is established as alpha = torque/I. The final expression for angular acceleration is derived as alpha = 2g/r, indicating that the block's acceleration influences the cylinder's rotation. The conversation clarifies that the block's descent is not free fall but is constrained by the string.
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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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