Why Does the Denominator Include mR^2 in the Angular Acceleration Formula?

[physics noob]

A mass (m) is connected to a pulley (rotational inertia I and radius R) which sits atop an incline with angle theta find an expression for angular acceleration in terms of M I R and theta... incline is frictionless


using T= aI

i get Rmgsin(theta) = aI so a = Rmgsin(theta) over I

but alas this isn't correct and I am not sure why

the correct answer has the same numerator but the denominator is I + mR^2 where is the mR^2 coming from? THANKS for any help

 

[siddharth]

Drawing the FBD of the mass, it is seen that the force acting down the plane is mgSin(Theta) and the force acting up the plane is Tension(T). From Newton's second law,

mgSin(Theta)-T=ma.

and drawing the FBD of the pulley, The torque about the center of the pulley is due to the Tension of the rope only. So,

RT=I(alpha)

also,
a=R(alpha) [Can you see why this is so?]
From these three equations solve for alpha.

 

[thepatientmental]

!The correct answer takes into account the moment of inertia of the mass connected to the pulley. In rotational motion, the moment of inertia is a measure of an object's resistance to changes in its rotational motion. In this case, the mass attached to the pulley also has a moment of inertia, which is represented by the term mR^2 in the denominator. This term is necessary to accurately calculate the angular acceleration of the system.

To understand why this term is included, let's look at the equation for rotational motion:

τ = Iα

Where τ is the torque applied to an object, I is the moment of inertia, and α is the angular acceleration. In this case, the torque is provided by the force of gravity acting on the mass, and the angle of the incline affects the direction and magnitude of this torque.

When we solve for α, we get:

α = τ/I

So, in order to accurately calculate the angular acceleration, we need to know both the torque and the moment of inertia. The torque is given by Rmgsin(theta), but we also need to consider the moment of inertia of the mass attached to the pulley, which is represented by mR^2.

Therefore, the correct expression for angular acceleration is:

a = Rmgsin(theta) / (I + mR^2)

I hope this helps clarify the issue. Good luck with your problem!