2 cylinders with rope wrapped around them

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SUMMARY

The discussion focuses on the dynamics of two identical cylinders connected by a rope, specifically analyzing the angular acceleration of the upper cylinder when one cylinder is falling. The moment of inertia is defined as I=kmR², and the relationship between the angular acceleration (α) and the linear acceleration (a) is established, with the hint indicating that the acceleration of the lower cylinder is twice that of the rope. The final derived formula for the angular acceleration of the upper cylinder is α = g / R(k + 2), confirming the influence of tension and mass on the system's rotation.

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  • Understanding of rotational dynamics and moment of inertia
  • Familiarity with Newton's second law for rotation (M = Iα)
  • Knowledge of Steiner's theorem for parallel axis theorem
  • Basic concepts of linear and angular acceleration relationships
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  • Learn about the relationship between linear and angular acceleration in connected systems
  • Explore advanced problems involving multiple bodies and tension in ropes
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Homework Statement


2 identical cylinders of mass m, radius R and moment of inertia I=kmR2are connected with a rope wound around both, according to the drawing. One is falling.
What is the angular acceleration of the upper one?
A hint was given: the acceleration of the lower one is twice the rope's acceleration.

Homework Equations


[tex]M=I\alpha[/tex]
Steiner's theorem:
[tex]I_b=I_c+mr^2[/tex]

The Attempt at a Solution


The hint causes me to think the angular acceleration of both cylinders is the same, since the tension T, which rotates them, is the same for both.
The forces that pass through the centers of the cylinders do not have effect on their rotation since they pass through their center of mass, correct? so it leaves the tension T to have the only influence on rotation.
Sum of moments around point A, using Steiner's theorem:
[tex]mgR=(kmR^2+mR^2)\alpha[/tex]
[tex]\Rightarrow\alpha=\frac{g}{R(k+1)}[/tex]
But the solution is:
[tex]\alpha=\frac{g}{R(k+2)}[/tex]
 

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for the lower cylinder,we have...

mg-T=ma...I

from the rotation of upper cylinder,

TR=kmR^2[tex]\alpha[/tex] (torque eqn)...II


if you have understood the hint...(angular acc is equal for both,and acceleration of lower cylinder is twice that of rope)
also [tex]\alpha[/tex]R=a/2...
or,...a=2[tex]\alpha[/tex]R.....III

putting II and III in I,


mg-kmR[tex]\alpha[/tex]=2mR[tex]\alpha[/tex]...IV

IV can be solved for [tex]\alpha[/tex] to give the required solution



hope that helps...:smile:
 
Thanks!
 

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