Rotational Kinematics -- questions about a=mg sin(theta) / (m+I/R^2)

  • #1
MP97
4
0
Hi, I am learning. Rotational Kinematics and I was given this formula in class: a=mgsin(theta)/(m+I/R^2); however, I couldn't understand the professor's explanation of where it comes from. Could someone provide some insights about it?

I appreciate any help you can provide.
 
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  • #2
Do you at least have an idea of what physical situation it is supposed to describe? Is the formula you posted all there is in your notes? It looks like you have a mass sliding down a frictionless incline of angle ##\theta##. A massless string is tied to the mass and wrapped around a pulley of radius ##R## and moment of inertia ##I##. The expression you posted describes the linear acceleration of the mass down the incline.

It comes from the application of Newton's second law for linear motion and for rotations. If you want to see its derivation, I am sure you can find it on the web now that you know what it (probably) is.
 
  • #3
Correct it describes an object rolling down an inclined plane without slipping. Thank you!
 
  • #4
MP97 said:
Correct it describes an object rolling down an inclined plane without slipping. Thank you!
That's not what I said, but it describes that too. You are welcome.
 
  • #5
MP97 said:
Hi, I am learning. Rotational Kinematics and I was given this formula in class: a=mgsin(theta)/(m+I/R^2); however, I couldn't understand the professor's explanation of where it comes from. Could someone provide some insights about it?

I appreciate any help you can provide.
For some insights, you could try this:

 

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