How Do You Derive the Tension in a Classical Physics Pulley Problem?

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SUMMARY

The tension in a classical physics pulley problem involving two hoops with masses M1 and M2 and radii R1 and R2 is derived as T = gM1M2/(M1+M2). The discussion highlights the importance of considering both linear and rotational acceleration, particularly the moment of inertia of the hoops. Participants emphasize the need for a free body diagram and the application of Newton's second law (F=ma) to derive the correct tension. The distinction between point masses and hoops is crucial for accurate calculations, as it affects the factor of tension derived.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of moment of inertia for rotating bodies
  • Familiarity with free body diagrams
  • Basic concepts of gravitational force in classical mechanics
NEXT STEPS
  • Study the derivation of tension in systems involving rotational dynamics
  • Learn about the moment of inertia for various shapes, including hoops and rings
  • Explore advanced applications of free body diagrams in multi-body systems
  • Investigate the differences between point masses and extended bodies in physics problems
USEFUL FOR

Students and educators in physics, particularly those focusing on classical mechanics, as well as anyone solving pulley problems involving rotational dynamics and tension calculations.

danny271828
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A massless string is placed over a massless pulley, and each end is wound around and fastened to a vertical hoop. The hoops have masses M1 and M2 and radii R1 and R2. The apparatus is placed in a uniform gravitation field g and released with each end of the string aligned along the field.

I have to show that the tension is T = gM1M2/(M1+M2)

I can sort of solve the problem by just saying that (M1M2)/(M1+M2) is the reduced mass. Then all we have to do is say the tension is balanced with the weight, so that T = Mg = gM1M2/(M1+M2). But then doesn't this imply that the hoop isn't moving? I think I'm missing something here...
 
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You don't know anything about whether they are moving or not. What you do know is that the acceleration of each hoop is equal (one up and one down). Write down a free body diagram, F=ma for each hoop and eliminate the acceleration to solve for the tension.
 
I realize that this is an old post, but I have the same exact problem.

If we assume that the rings are point masses, we see that

F1_net = T - (m1)g = (m1)a
F2_net = T - (m2)g = -(m2)a

where the right hand side must be of different sign (one mass accelerates upward, the other downward). Rearranging the equations so that each is equal to "a" and then equating gives

T = 2g(m1)(m2)/(m1 + m2)

but this is not the answer--it is off by a factor of 2. Of course, my work assumes point particles, so what is the distinction between point particles and rings to obtain this factor difference?
 
Last edited:
I was assuming the problem was to be read that the rings were in rotational acceleration as well as linear acceleration. The moment of inertia should make some difference. Try doing it that way. If you don't get it let me know and I'll travel two years back in time...
 
Last edited:
No need for time travel. Your right about the rotational acceleration. Thanks. Sometimes understanding what the problem is itself can be the hardest part of solving a problem.
 

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