How Does Torque Relate to Different Types of Acceleration?

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SUMMARY

This discussion focuses on the relationship between torque and acceleration in the context of rotational motion. The key equation presented is τ = Iα, where τ represents torque, I is the moment of inertia, and α is angular acceleration. The moment of inertia for a disk about the axle is defined as I_z = (1/2)MR². The participants clarify that linear acceleration a can be derived from torque using the formula a = (2τ)/(MR³), emphasizing the distinction between linear and angular acceleration.

PREREQUISITES
  • Understanding of rotational dynamics and torque
  • Familiarity with the moment of inertia concept
  • Knowledge of angular acceleration and its relationship to linear acceleration
  • Basic grasp of Newton's second law (F = ma)
NEXT STEPS
  • Study the derivation of the moment of inertia for various shapes
  • Learn about the relationship between linear and angular quantities in rotational motion
  • Explore the implications of rolling without slipping in rotational dynamics
  • Investigate the effects of torque on different physical systems
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Students studying physics, particularly those focusing on mechanics, as well as educators and anyone interested in understanding the principles of torque and acceleration in rotational systems.

saltine
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Homework Statement


3044245373_1001682a5c.jpg

Find acceleration a given the torque, T, about the axle.

Homework Equations


[tex]\tau = I\frac{d\omega}{dt} = I\alpha[/tex]

Moment of Inertia of a disk (about the axle):
[tex]I_z = \frac{1}{2}MR^2[/tex]

The Attempt at a Solution



[tex]\alpha = Ra[/tex] in this context, so
[tex]\tau = I Ra[/tex]
[tex]a = \frac{\tau}{IR} = \frac{2\tau}{MR^3}[/tex]
I think this is correct, however:

[tex]\tau = R \times F[/tex], since F = ma, [tex]a = \frac{\tau}{mR}[/tex].
What does this a represent? Does it represent the linear acceleration of a particle of mass m that happens to be on the edge of the disk?

- Thanks
 
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Welcome to PF!

Hi saltine! Welcome to PF! :smile:

(i assume the wheel rolls without slipping?)
saltine said:
[tex]\alpha = Ra[/tex]

No, [itex]R\alpha = a[/itex] … just consider the dimensions! :redface:
[tex]\tau = R \times F[/tex], since F = ma …

No, I don't understand that :confused: … tau doesn't come from a force at a distance R … it probably comes from two forces very close together, on either side of the axle. :wink:
 

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