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

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SUMMARY

The discussion centers on calculating the tension in a rope given an 8 kg pulley and the effects of the pulley’s mass on the problem. Participants clarify that neglecting the mass of the pulley is incorrect, as it influences the system's dynamics. The moment of inertia for a solid cylinder is established as I = (1/2)mr², derived from integrating the mass distribution. Understanding the rotational inertia is crucial for accurately solving the tension equations.

PREREQUISITES
  • Understanding of Newton's second law (m1a = T, m2g - T = m2a)
  • Knowledge of moment of inertia and its calculation for solid cylinders
  • Familiarity with rotational dynamics and integration techniques
  • Basic principles of mechanics involving pulleys and tension
NEXT STEPS
  • Study the derivation of moment of inertia for various shapes, focusing on solid cylinders
  • Learn about the dynamics of pulley systems in physics
  • Explore the integration of mass distribution in three-dimensional objects
  • Investigate the effects of rotational inertia on tension and acceleration in mechanical systems
USEFUL FOR

Students in physics, particularly those studying mechanics, as well as educators and anyone involved in solving problems related to pulleys and rotational dynamics.

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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