Linear Momentum of Projectile

Homework Statement

Background
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.

Specifically
(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

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

sqrt(Vx2+Vy2)
=sqrt(71.5972+59.922)
=93.36 m/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.

That is just too difficult and error-prone. Dealing with angles in such a way is always messy. Use vectors throughout, it is much simpler. You know that the x-component of the velocity vector is some constant ## v_x ##; the y-component is ## v_y - gt ##; the x-component of the displacement vector is ## v_x t ## and its y-component is ## v_y t - \frac {gt^2} 2 ##. The angular momentum is ## \vec r \times \vec p ##, using the properties of the cross product, its z-component is ## m (v_x t (v_y - gt) - (v_y t - \frac {gt^2} 2) v_x) = - m \frac {g t^2} 2 v_x ##, so all you really need to do is find ## v_x ##, the x-component of the initial velocity.

Sorry for the delay, so I understand what your saying about using just vectors. The answer I got using the equation was -7718.15 kg*m2/s. Webassign said it was wrong, but I'm going to go back and use stored values since it is real sensitive about sig figs.

No, wait. I missed that the problem said the origin of the coordinate system must be at the base of the cliff, and placed it at the cannon. That changes the displacement components by adding some fixed offsets. I will let you do the algebra this time.

Okay so I used (RXP)z=rxpy-rypx
I ended up with
(Vi,xt)[(Vi,y-gt)m]-(Vit+.5at2)(Vxm) = mt[Vi,xVi,y-Vi,xgt-Vi,yVx+(Vxat)/2]

All of the Vx's and Vi's started jumbling together towards the end so I may have done the algebra wrong hence why I got the wrong answer, I'm going to redo it right now

Here is another trick. Let's say that the initial position of the ball is ## \vec r_0 ##. Then at any time ## \vec r = \vec r_0 + \vec r' ##, where ## \vec r' ## is the displacement from the cannon. Now, angular momentum is ## \vec r \times \vec p = \vec r_0 \times \vec p + \vec r' \times \vec p ##. We have already found that the z-component of ## \vec r' \times \vec p ## is ## -m \frac {gt^2} 2 v_x ##; the z-component of ## \vec r_0 \times \vec p ## is ## m(r_x(v_y - gt) - r_y v_x) ##, where ## r_x ## and ## r_y ## are components of ## \vec r_0 ##; so the full z-component is $$m \left[ r_x(v_y - gt) - \left( r_y + \frac {gt^2} 2 \right) v_x \right]$$ Far messier than before, but still manageble with a calculator.

1 person
I appreciate all of the help, I only have a certain amount of tries and I have almost exhausted all of them for this part. I am going to talk with my professor tomorrow at this point, I will post the solution tomorrow after speaking with him or the TA's at the school. Thanks again

So, we were on the right track with using properties of cross products but it should have been like this.
L = r X p
= (xi+yj) X m(Vxi + Vyj)

after simplifying we get,

L = m[xVyk + yVx(-k)]

x being distance from origin @ specific time, and y being height @ specific time.

He did say I was making it to complicated with how I was trying to do it at first.