# 2 Accelerating Rotational Points in One System

by Mechanics
Tags: accelerating, points, rotational
 P: 7 A 2 dimensional circle of radius "r" and mass "m" is attached through the center of the circle by a rigid, massless rod to a fixed point of rotation a distance "l" away. A massless rocket is attached to the outside of the circle a height of 0 away from the circle's surface and "r" away from the circle's center. The rocket exerts a constant force "F" tangentially to the circle. The system is ideal. How would I go about finding the total system energy with respect to time?
 P: 7 I realize it's mechanics, but does this really belong in introductory physics?
P: 335
 Quote by Mechanics A 2 dimensional circle of radius "r" and mass "m" is attached through the center of the circle by a rigid, massless rod to a fixed point of rotation a distance "l" away. A massless rocket is attached to the outside of the circle a height of 0 away from the circle's surface and "r" away from the circle's center. The rocket exerts a constant force "F" tangentially to the circle. The system is ideal. How would I go about finding the total system energy with respect to time?
If I'm visualizing this correctly, we have two axes of rotation at right angles. You should be able to work out the rotational kinetic energy for each rotation individually. How do you add the kinetic energies in this case? (Hint: Is kinetic energy a scalar or a vector?)

-Dan

 P: 7 2 Accelerating Rotational Points in One System Dan, the axes of rotation are not at right angles to each other. The circle is free to rotate around its center and the fixed point at the end of the rod a distance "l" from the center of the circle. The rocket will also at times be slowing down the system as the periods about each axes of rotation is not necessarily the same so the force will at times be in the direction opposite to the motion.
P: 335
 Quote by Mechanics Dan, the axes of rotation are not at right angles to each other. The circle is free to rotate around its center and the fixed point at the end of the rod a distance "l" from the center of the circle. The rocket will also at times be slowing down the system as the periods about each axes of rotation is not necessarily the same so the force will at times be in the direction opposite to the motion.
Hmmm...not seeing it. I must not be visualizing it right. Ah well, that was never one of my strong points...I guess someone else better take it from here.

-Dan
 P: 7 If no one here can help me solve it, can this thread be moved back to Classical Physics where I posted it?

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