Ohh i see! I just use Pythagorean theorem, so I square i, j, and k and take the square root and so the second part becomes 1 because everything cancels out!
so final result is L = (1/12)(Mωsinφl2)
Thanks everyone for helping!
Ok I think I finally understand this.
I first wrote r and v as:
r = [r*sin(φ)*cos(ωt)]i + [r*sin(φ)*sin(ωt)]j + [r*cosφ]k
=r[[sin(φ)*cos(ωt)]i + [sin(φ)*sin(ωt)]j + [cosφ]k]
so r*sin(φ) is the horizontal radius from the point mass to the axis of rotation and
cos(ωt) accounts for the...
Okay so this is what I imagine the problem looks like. Velocity vector V is pointing into the paper and r has an x and y component and then Torque T is perpendicular to both V and r
This is correct right?
If there is torque than L is not constant. But then you have torque on both sides of the rod so net torque is zero?
The r vector and velocity vector should be perpendicular to each other, so at the angle of 90 degrees?
Well cross product is always in the plane perpendicular to the two vectors?
Homework Statement
A thin rod of length l and mass M rotates about a vertical axis through its center with angular velocity ω. The rod makes an angle φ with the rotation axis. Determine the magnitude and direction of L (angular momentum).
So we're given: mass - M, length - l, angular velocity...