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

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Homework Help Overview

The discussion revolves around calculating the tension in a rope while considering the mass of a pulley, specifically a pulley with a mass of 8 kg. Participants are exploring the implications of including the pulley's mass and its rotational inertia in the analysis.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are questioning whether neglecting the mass of the pulley is appropriate. There are discussions about the effect of the pulley's mass on the overall problem and the necessity of including its rotational inertia. Some participants suggest that knowing the radius of the pulley is essential for further calculations.

Discussion Status

The conversation is ongoing, with some participants affirming the correctness of certain statements while others seek clarification on the use of the moment of inertia formula. There is an exploration of different interpretations regarding the distribution of mass in the pulley and its impact on the calculations.

Contextual Notes

Participants are navigating the complexities of integrating the pulley's mass and its rotational effects into their calculations, with some expressing confusion over the assumptions made about the pulley's geometry and mass distribution.

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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