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## AP®︎/College Physics 1

### Course: AP®︎/College Physics 1>Unit 5

Lesson 5: Angular momentum and torque

# Angular momentum of an extended object

In this video David derives and shows how to use the formula for the angular momentum of an extended object. Created by David SantoPietro.

## Want to join the conversation?

• Angular momentum of a rod rotating about one end: David did not complete the calculation for momentum because he didn't explain how m(rsqr'd) became 1/3m(rsqr'd). Was he trying to avoid the use of integrals? • Short Answer - Yes he was trying to avoid the use of integrals and you can choose to just trust him or...

Long Answer - I answered a similar question to this on another Moments, Torque and Angular Momentum Video

We know that the total moment of inertia for a set of point masses is the sum of each of their masses multiplied by the square of their corresponding distance from the pivot point.
Total Moment of Inertia = Σmr²
We can consider a rod of uniform density as an infinite number of infinitesimally small point masses and thus find a similar sum to the one above. We can represent this as a definite integral. If we take the moment of inertia to be from the end of a rod of length L and total mass M we can calculate this moment of inertia.
Total Moment of Inertia = (From 0 to M)∫ r² dm
This effectively takes the sum of an infinite number of point masses in the rod multiplied by the square of their distances from the pivot point. However we can't integrate r² with respect to m. We can use the formula
m = (r/L)M
r/L represents the ratio of the current distance from the pivot point to the total length of the rod. We can multiply this by the total mass of the rod M to give us the total mass of the rod up to the distance r away from the pivot. This formula only works because we assume that the rod is of a uniform density.
m=(r/L)M = (M/L)r Now we can differentiate this equation with respect to r
dm/dr = M/L
Total Moment of Inertia = (From 0 to M)∫ r² dm
Now using the substitution dm = (M/L)dr and changing the limits of our integral accordingly our Total Moment of Inertia becomes
(From 0 to L)∫ (M/L)r² dr = [(M/L)(1/3)r³] (From 0 to L) = (M/L)(1/3)(L)³ - (M/L)(1/3)(0)³ therefore
Total Moment of Inertia = (1/3)ML² I hope that explains it and I didn't over complicate things!
• At ,
when the units were multiplied ,where did the radian go.
kgX(m^2)/sec • Where did the 1/3 come from when doing the example? • So, if you're rolling coins into one of those hyperbolic funnels that are used for donations. What physics and associated mathematics could you use to determine what coins would go down the funnel faster? I'm just stumped as to how you could apply this content to that situation, but I'd assume it's similar to orbits, but with a component of rotating disks. • Can angular momentum be conserved about any other point inside or outside the body, other than the center of mass of a body?? • Is the angular velocity constant, because all the small infinitesimal masses rotate through the same angle theta , in the same period of time? • Guys please help, perhaps it is a ridiculous thought, but it just came to my mind that we may consider a ball of mass, traveling along a straight line as rotating around an infinitely large circle. In such case R would be infinite. Does it mean that a translating ball has an infinitely large angular momentum! So please point out where is a mistake here. • my intuition goes to the similar direction as yours. and that says yes

1. if R gets close to infinity, angular momentum goes to infinity too. but what's more interesting is angular velocity as well as acceleration goes to 0 in such a case. in other words, the mass would move in a "straight" line perpendicular to the invisible bar connecting it and its orbital center of mass

2. so i think it's not because it travels along a straight line thus it has an infinite R. but the other way around

3. if the distance of a body from the orbital centre goes to infinity, it has an incredibly small (if not at all) angular velocity and acceleartion. thus it can have a straight travel

and the opposite edge seems also interesting. if a body get closer to the orbital centre, say R=infinitely small or ~0, then its angular velocity and acceleration grows so big that it has only one motion, spin on the same spot   