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

In summary, the equation \( a = \frac{mg \sin(\theta)}{m + \frac{I}{R^2}} \) describes the acceleration of an object on an inclined plane considering both its weight and the moment of inertia. Here, \( a \) represents linear acceleration, \( m \) is the mass of the object, \( g \) is the acceleration due to gravity, \( \theta \) is the angle of the incline, \( I \) is the moment of inertia of the object, and \( R \) is the radius at which the mass is applied. This relationship illustrates how rotational dynamics influence the linear motion of the object on the slope.
  • #1
MP97
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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:

 
  • #6
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.
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.
 
  • #7
Rishit Bhaumik said:
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.
 
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Likes Rishit Bhaumik and berkeman

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

What is rotational kinematics?

Rotational kinematics is the branch of mechanics that deals with the motion of objects that rotate about an axis. It describes the relationships between angular displacement, angular velocity, angular acceleration, and time, analogous to linear kinematics for translational motion.

What does the equation a = mg sin(theta) / (m + I/R^2) represent?

This equation represents the linear acceleration (a) of a rolling object down an inclined plane. The term mg sin(theta) is the component of gravitational force acting parallel to the incline, m is the mass of the object, I is the moment of inertia, and R is the radius of the object. The denominator m + I/R^2 accounts for both the translational and rotational inertia of the object.

How is the moment of inertia (I) related to rotational motion?

The moment of inertia (I) is a measure of an object's resistance to changes in its rotational motion. It depends on the mass distribution of the object relative to the axis of rotation. For example, a solid cylinder and a hollow cylinder of the same mass and radius will have different moments of inertia.

Why is the term I/R^2 included in the equation?

The term I/R^2 is included in the equation to account for the rotational inertia of the object. When an object rolls without slipping, its rotational motion is coupled with its translational motion. The term I/R^2 effectively converts the rotational inertia into an equivalent mass that contributes to the total resistance to acceleration.

How do you derive the equation a = mg sin(theta) / (m + I/R^2)?

The equation is derived by applying Newton's second law for both translational and rotational motion. For translational motion, the net force down the incline is mg sin(theta) - friction. For rotational motion, the torque due to friction causes angular acceleration. By combining these equations and eliminating the friction term, we arrive at the expression a = mg sin(theta) / (m + I/R^2), which simplifies the analysis of the rolling motion.

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