The concepts of torque and angular momentum are unnecessary when analyzing the motion of a projectile; however, using these concepts for an analysis of projectile motion does show that the concepts are perfectly general, and are not restricted purely to the analysis of rotation. You should regard this problem as chiefly a mathematical exercise in using the cross product definitions of torque and angular momentum.
A restored Civil War cannon is being tested using cannonballs of various mass and using various loads of gunpowder. In one test, the cannon fires a cannonball at a speed of 107 m/s in a direction which is 48 degrees Up from East. The cannonball is intended to land on a mesa which is to the east of the cannon. The near edge of the mesa is a vertical cliff which is 153 meters high; the base of the cliff is 186 meters to the east of the cannon.
In answering the questions below, use an xy coordinate system with East as the positive x direction and Up as the positive y direction.
(g) Use the definition of angular momentum, in three dimensions, along with your equations for the position of the cannonball (as measured from the base of the cliff) and the velocity of the cannonball, to find an equation for the angular momentum of the cannonball as a function of time, as measured with respect to the point at the base of the cliff. Use that equation to determine the angular momentum of the cannonball at the following times. Because your position and velocity vectors are all in the xy plane, your angular momentum vectors will always be purely in the positive or negative z directions. Take South as the positive z direction, and indicate the direction of angular momentum by the appropriate algebraic sign. Mass of cannonball 5.5kg
Determine the z component of angular momentum for the following times:
At t = 2.0 s, Lz = ________ kg * m2/s
Answers to previous questions
At t = 2.0 s, x = 143.19 m and y = 139.43 m
At t = 4.0 s, x = 286.39 m and y = 239.67 m
At t = 6.0 s, x = 429.58 m and y = 300.7 m
At t = 2.0 s, vy = 59.92 m/s.
At t = 4.0 s, vy = 40.32 m/s.
At t = 6.0 s, vy = 20.72 m/s.
When passing over the cliff edge, vy = 54.14 m/s.
Just before crashing into the ground, vy = -57.88 m/s.
X position w.r.t the base of cliff
At t = 2.0 s, xC =-42.81 m
At t = 4.0 s, xC = 100.39 m
At t = 6.0 s, xC =243.58 m
Just before crashing into the ground, xC = 817.79 m
L = r x p
L = rmvsinθ
The Attempt at a Solution
My attempt was to use pythagoras to find r vector from base of cliff to the cannonball at t = 2 s
sqrt(139.432+(-42.81)2) = 145.85m
Then I needed an angle between the position and and origin so at t = 2s
arcttan(139.43/42.81) = 72.93°
I also need a velocity so I used pythagoras once again to get actual velocity at t = 2 s
then plug everything into equation r X p
L = 145.85m(5.5kg)(93.36m/s)sin(72.93°0
= 71596.533 kg*m2/s
The vector should be negative by RHR.
I believe I may be looking at the wrong angle, any kind of guidance would be appreciated.