Acceleration of a pulley. Torque, and moment of inertia.

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SUMMARY

The discussion focuses on deriving the expression for acceleration in a pulley system involving two blocks, m1 and m2, connected by a rope over a pulley with radius R and moment of inertia I. The net force equation is established as Fnet = m2gsin(theta) - m1g, leading to the acceleration formula a = (m2gsin(theta) - m1g) / (m1 + m2 + I/R^2). The confusion arises from the inclusion of the moment of inertia I, which is necessary for accurately modeling the dynamics of the pulley. A thorough understanding of free-body diagrams and Newton's second law is essential for solving such mechanics problems.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Familiarity with free-body diagrams
  • Knowledge of moment of inertia and its role in rotational dynamics
  • Basic trigonometry, particularly involving angles and sine functions
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  • Study the concept of moment of inertia in detail, particularly I = mR^2 for rigid bodies
  • Learn how to apply Newton's second law to rotational systems
  • Explore the derivation of acceleration in pulley systems with varying masses
  • Practice solving mechanics problems involving multiple objects and forces
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Students studying physics, particularly those focusing on mechanics, as well as educators teaching concepts related to dynamics and rotational motion.

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Homework Statement


There are two blocks, one of mass m1 and the other of m2. They are connected by a rope that runs over a pully of radius R and interia I. m2 slides down a frictionless incline at an angle theta, and m1 hangs virtically while being pulled up by the tension in the rope. Find the expression for acceleration.


Homework Equations


Fnet=m2gsin(theta)-m1g
F/M=a


The Attempt at a Solution



With the previous two equations I came to this conclusion...

(m2gsin(theta)-m1g)/(m1+m2+I)

The right answer has I/R^2 instead of just I.

I figured it would be just I because the mass of an object is its inertia.

Howwoud I derive the I/R^2?
 
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How did you get an equation with an "I" when your initial equations:

Fnet=m2gsin(theta)-m1g
F/M=a

didn't have an "I" anywhere? Anyhow, for mechanics problems, you should always draw free-body diagrams for every object (including the pulley!), write down Newton's second law for each object, and then solve the resulting equations. That way, you'll see very clearly where the I/R^2 came from.
 
I got the I from thinking that I of the pulley would be the same thing as mass for the blocks because isn't mass just the inertia of an object? and I put my net force over the total mass.

I think I may have figured it out but I also may be making a mistake here...

T1 and T2 are equal to the tensions on each side, and alpha=a/R

T2R-T1R=Ialpha

R(T2-T1)=Ialpha

Tnet=I(a/R)/R

ma=Ia/(R^2)

m=I/R^2
 

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