Finding the tension of a rope given the mass of a pulley

AI Thread Summary
The discussion centers on calculating the tension in a rope with an 8 kg pulley, questioning whether the pulley’s mass should be considered. Participants agree that neglecting the mass of the pulley simplifies the problem, but including it requires understanding its rotational inertia, which is calculated as I = 1/2mr^2. The confusion arises over why the factor is 1/2 instead of 1, with explanations highlighting that not all parts of the pulley are equidistant from the axis of rotation. The integration of mass elements shows that the average distance from the axis leads to the 1/2 factor. Understanding these principles is crucial for accurately solving the problem.
Jacobs
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Homework Statement


PROBLEMS+24%2F142..jpg

mass of pulley is 8 kg

Homework Equations


m1a=T
m2g-T=m2a

The Attempt at a Solution


I solved question with neglecting mass of pulley but should I?
 

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How does the mass of the pulley affect the problem?
 
If you want to include the rotational inertia of the pulley, you either need to know its radius or solve for the acceleration in terms of the radius.
 
Is that right?
 

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Jacobs said:
Is that right?
That is correct.
 
But why 1/2mr^2 instead of mr^2?Yes it is cylinder but particle moves as x and y-axis not z
 
Jacobs said:
But why 1/2mr^2 instead of mr^2?Yes it is cylinder but particle moves as x and y-axis not z
Not all parts of the pulley are at distance r from the axis.
 
But r is radius of pulley
 
To find the inertia of the pulley or disk, you need to integrate r^2 dm over the dimensions of the pulley I = r^2 dm = ∫∫∫ r^2 (ρ r dr dθ dz). If you do that, you will find I = (mr^2)/2 where m = ρV = ρ π r^2 t, t = thickness of the disk.
 
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Jacobs said:
But r is radius of pulley
Yes, but only its periphery is at distance r from the axis. A mass element dm on the oeriphery does have moment of inertia dm r2 about the axis. Other parts of it are closer to the axis so have a smaller moment of inertia. The average turns out to be the same as if all parts were r/√2 from the axis. Dr dr news has posted the details of that.
 
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