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

[MP97]

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.

 

[kuruman]

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.

 

[MP97]

Correct it describes an object rolling down an inclined plane without slipping. Thank you!

 

[kuruman]

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.

 

[PeroK]

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:

 

[Rishit Bhaumik]

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.

Hey!
Let's solve your issues.
Consider a block of mass 'm' attached to a fixed, pulley of radius 'R', and moment of inertia 'I' by a rope having constant tension 'T'. Assuming there is no slipping between the pulley and the rope, we are asked to find the acceleration of the block when released under gravity.

By Newton's 2nd law,
F(net) = ma
=> mg - T = ma .....(i)

Now for the pulley, it will tend to rotate due to the torque produced by the tension in the rope.
=> Tau = RT = I(alpha) = Ia/R [Since no slipping, acceleration of rope is same as that of pulley, and a = R(alpha)]

=> T = Ia/R² ....... (ii)
Substituting this value in (i) and solving for a, we get
a = mg/(m+I/R²)

I hope this helps,
Signing off,
Rishit.

 

[kuruman]

I hope this helps,

I hope so too but note that this thread is more than a year old and the original poster has not been seen since January 2024.