Acceleration of Two Blocks Connected by a Pulley on a Frictionless Ramp

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SUMMARY

The discussion focuses on calculating the acceleration of two blocks connected by a pulley on a frictionless ramp. The blocks, each with mass M, are influenced by gravitational force and tension in the rope, while the pulley has a moment of inertia I and radius r. The derived formula for acceleration is (Mg(1-sinθ))/(2M-(I/r²)). Key concepts include the application of Newton's second law (F=ma) and the relationship between torque and tension in the system.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of torque and its calculation (Torque = Ftangential * r)
  • Familiarity with the concept of moment of inertia (I)
  • Basic principles of pulley systems and tension in ropes
NEXT STEPS
  • Study the derivation of acceleration in pulley systems with varying moment of inertia
  • Learn about the dynamics of frictionless ramps and inclined planes
  • Explore advanced concepts in rotational dynamics and torque
  • Investigate the effects of different mass distributions on pulley systems
USEFUL FOR

Students in physics, particularly those studying mechanics, as well as educators and anyone interested in understanding the dynamics of pulley systems and inclined planes.

I_Auditor
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Homework Statement


Two blocks are of the same mass M. One lies on a frictionless ramp with slope θ, while the other one, connected by a rope, hangs by a pulley with a moment of inertia of I and a radius of r. Find the acceleration of the two blocks.

I don't have an actual picture, but this one might help:
incline_wpulley.gif

Assuming m=M and that the pulley has a moment of inertia = I and a radius = r

Homework Equations


F = ma
Torque = Ftangential * r
Tension = Mg(1-sinθ)
2Ma should (?) be the total force on the blocks.

The Attempt at a Solution


I drew an FBD for the masses and got Mg(1-sinθ), but I'm unsure as to where I am supposed to go after this. I assumed that torque factored in, but I'm shaky on that aspect of the problem.

The answer given was (Mg(1-sinθ))/(2M-(I/r2)).
 
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Hello I_A, :welcome:

You sense there is more to this, but can't pinpoint it, is my impression.
I_Auditor said:
I drew an FBD for the masses and got Mg(1-sinθ), but I'm unsure as to where I am supposed to go after this. I assumed that torque factored in, but I'm shaky on that aspect of the problem.
Did you draw one for each of the two ? Draw something equivalent for the pulley too.
You may assume that the rope does not slip on the pulley.

From the m diagram you should conclude that Tension ##\ne## Mg(1-sinθ) on the left side of the pulley.
From the pulley diagram you should conclude that the Tension on the left side of the pulley is not equal to the Tension on the right side of the pulley

By the time you have digested all that, you are well under way towards the book solution. Good luck !

:smile:
 

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