Summation(superposition) of angular momentum

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seba240698
Posts: 14
Joined: Mon Dec 18, 2006 3:04 am

Summation(superposition) of angular momentum

Post by seba240698 » Sun Jan 21, 2007 10:47 am

From the book "3000 solved problems in Physics", page 238, queston 12.48:-

"Each of the wheels on a certain four-wheel vehicle has a mass of 30kg and a radius of gyration of 30 cm. When the car is going forward and the wheels are turning at 5.0 rev/s,
a) what is the rotational kinetic energy stored in the 4 wheels?
b) what is the angular momentum of the vehicle about an axis parallel to the wheel axis and through the center of mass?
c) Is the angular momentum vector directed towards the driver's right or left?"

Answer for the book:-

a) K = 5300J (ok)

b) L = 4 I w = 340kg.m^2/s (I disagree)

c) driver's left.(ok)

I disagreed with the answer (b) of the angular momentum as the car itself is not rotating, so how to get its angular momentum?? Only the wheels are rotating, so only the wheels have angular momentum.

Am I right to say that the angular momnetum of the car, L = 0?

schmit.paul
Posts: 161
Joined: Sat Nov 04, 2006 7:48 pm

Post by schmit.paul » Sun Jan 21, 2007 3:23 pm

In classical mechanics, angular momentum is a vector quantity with two vector arguments, ie L = rxp. Once you try to project this equation into a particular coordinate system, you'll notice right away that the values you get are highly dependent on your choice of reference frame. Therefore, something that has only linear momentum in one reference frame (say, a free particle travelling in straight line through the origin of your inertial coordinate system) could have both linear and angular momentum in another reference frame (ie the same free particle travelling in a straight line but NOT through the origin) or only angular momentum (origin of coordinate system translates along with the free particle and rotates, so that particle traces out a circular path in the coordinate system). Your book said the reference axis passes through the center of mass of the car and is parallel to the axis of rotation of the wheels. The wheels retain their original angular momentum (just I*w) in this new reference frame, and you'll note that the definition of the reference axis implies that it travels with the center of mass of the car itself, so the only things moving in this reference frame are the wheels. The book is telling you the angular momentum of the car is 4*I*w, meaning the angular momentum of all 4 of the wheels, as these are the only constituents of the car that are moving in the chosen reference frame.



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