Calculating Angular Momentum for a Rolling Sphere

AI Thread Summary
To calculate the total angular momentum of a rolling sphere in uniform circular motion, both its rotation around its axis and its motion around the center of the loop must be considered. The angular momentum due to its own rotation is calculated using L = Icmω, where Icm = (2/5)mr² and ω = v/r. For the angular momentum around the center of the loop, L = mvr is used, with r being the distance from the loop's center to the sphere's center. The total angular momentum is found by adding these two components together. Understanding how to separate and calculate these contributions simplifies the problem without needing to analyze each particle individually.
lmnt
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The problem statement
Icm=\frac{2}{5}mr2
\omega=v/r
m=mass of object

There's some things I don't really get about total angular momentum in a rigid body. Suppose a perfectly spherical object is rolling in uniform circular motion (ie. in a loop). Find the total angular momentum.

Homework Equations


L=mvr for particle, and L=Icm\omega for rigid body

3. Attempt at solution

Since it is rolling, it has angular momentum due to its own rotation around its axis, like a normal object that's just spinning right? So then L = Icm\omega and I would use \omega=v/r with r=radius of sphere

But then, it also has angular momentum around the centre of the loop-the-loop that it rolls around. Here's where i get lost. Would i consider the object as a particle and use L=mvr, or do i use L = Icm\omega again, and which radius do i use when i sub in for \omega? the radius of the loop or the radius of the mass itself?

I get that the total angular momentum for a situation like this would require adding together the separate momentums but i don't get how exactly they are seperated.
 
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actually when u are asked to calculate the angular momentum of such system, what u need to do is to apply the mv X r (with v and r being vectors) to all the particles in the rigid body, but that will be complicated
but thankfully, the calculation can be simplified (in most simple problems) by separating the angular mementum in the way you have described, they will give the same and the correct answer just like if u do it rigorously for each and every particles in the rigid body

hope that helps :)
 
Thanks, that makes it clear, i thought i was simplifying it too much, so its good to know that I don't have to do the less simplified way with all the rigorous work.
 
lmnt said:
But then, it also has angular momentum around the centre of the loop-the-loop that it rolls around. Here's where i get lost. Would i consider the object as a particle and use L=mvr, or do i use L = Icm\omega again, and which radius do i use when i sub in for \omega? the radius of the loop or the radius of the mass itself?

For the sphere's angular momentum about the center of the loop, you use mvr, where r is the distance from the loop's center to the sphere's center. For the sphere's angular momentum about its own center of mass, you'd use 2/5MR^2. For the total angular momentum you'd add them together.
 
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