Question

A projectile of mass m is launched with an initial velocity i making an angle θ with the horizontal as shown below. The projectile moves in the gravitational field of the Earth. Find the angular momentum of the projectile about the origin when the particle is at the following locations. (Use the following as necessary: vi, θ, m, and g for the acceleration due to gravity.) a) at its highest point of trajectory? B) Just before it hits the ground.

Answer 1

can you draw up the diagram or post it?

Answer 2

angular momentum = inertia of object X angular velocity
and momentum is 'conserved' then we can set up both sides of the 'before event' and 'after event' for an equatoin

Answer 3

http://www.webassign.net/serpse8/11-p-019.gif

Answer 4

yikes ^_^ @AllTehMaffs

Answer 5

Were you able to see the diagram?

Answer 6

ya, this question is over my head, i messaged somone smatert than me to come and look

Answer 7

OOOOo okay. This is WEIRD. Okay. one sec.

Answer 8

I am totally not smarter than you, shush yoself

Answer 9

thats what I said.
but if you can solve the lenghts of a messed up triangle that only has one length given, i know you can handle this, haha

Answer 10

^_^

Answer 11

@gkendoyan Do you know the notation

\[ \frac{dx}{dt}=\dot x\]

? If no it's not a problem. It just makes things prettier ^_^

Answer 12

@AllTehMaffs No i don't

Answer 13

no worries.

Okay, so I think this isn't as complicated as it seems. It's just weird.

What's the equation for angular momentum?

Answer 14

L=IW

Answer 15

yah (also hold on a sec. This is harder than anticipated :P)

Have you gotten anywhere with this yet,, or is just one of those "I have no idea" questions

Answer 16

@AllTehMaffs thats what I was thinking, this problem is weird

Answer 17

Well i started but couldn't continue.. so basically its one of those i have no idea questions

Answer 18

where'd ya get to? this gets complicated real quick...

Answer 19

hmmm :/

Answer 20

\[t_{tot}=2\frac{v_0 \sin \theta}{g}\]
\[R = v_x t_{tot} = \frac{2v_0^2 \cos \theta \sin \theta}{g}\]

\[ h=\frac{(v_0 \sin \theta)^2}{2g}\]
\[| \textbf r_{max \ height}|=\sqrt{\Big( \frac{R}{2} \Big)^2+h^2} \\ \ \\ \ \\ \ = \sqrt{\Bigg (\frac{2v_0^2 \cos \theta \sin \theta}{2g}\Bigg)^2+\Bigg( \frac{(v_0 \sin \theta)^2}{2g} \Bigg)^2} \]

" t_{tot}=2\frac{v_0 \sin \theta}{g}

R = v_x t_{tot} = \frac{2v_0^2 \cos \theta \sin \theta}{g}

h=\frac{(v_0 \sin \theta)^2}{2g}

| \textbf r_{max \ height}|=\sqrt{\Big( \frac{R}{2} \Big)^2+h^2} \\ \ \\ \ \\ \ = \sqrt{\Bigg (\frac{2v_0^2 \cos \theta \sin \theta}{2g}\Bigg)^2+\Bigg( \frac{(v_0 \sin \theta)^2}{2g} \Bigg)^2} "

Answer 21

well I guess first, do you see where I got all of those values?

Answer 22

(I didn't actually write down anything to get to that point. SOrry :P)

Answer 23

thats fine. I'm trying to follow

Answer 24

The height is from the fourth equation of motion

\[\cancel{v_{fy}^2}^0 = v_{fi}^2+2ah\]

and the time is from

\[a=\frac{\Delta v_y}{\Delta t}=\frac{\cancel{v_{yf}}^0-v_{yi}}{(1/2) \ t_{tot}}\]

Answer 25

Angular momnetum is

\[L = I \omega\]

but also \[\textbf L =\textbf r \times m \textbf v\]

Have you studied cross products?

Answer 26

\[ \textbf r_1 =\frac{R}{2}\hat{ \textbf x } + h \hat{ \textbf y }
\\ \
\\ \
\\ \ = \Bigg( \frac{v_0^2 \cos \theta \sin \theta}{g}\Bigg) \hat{ \textbf x } + \Bigg(\frac{(v_0 \sin \theta)^2}{2g} \Bigg) \hat{ \textbf y } \]
\[\textbf v = v_{x} \hat {\textbf x} + v_y \hat{ \textbf y}\]

Does any of this look familiar? (I'm going at a snail's pace because my brain is slowz)

Answer 27

Is this looking like it's going in a useful direction @DemolisionWolf ?

Answer 28

Yeah, I think you need to cross product. Have you done that before?

Answer 29

http://en.wikipedia.org/wiki/Cross_product

\[ \textbf L_1 = \textbf r_1 \times m \textbf v_1 = \left|\begin{matrix}\hat{ \textbf x } & \hat{ \textbf y } & \hat{ \textbf k }\\ R/2 & h & 0 \\v_{xi} & 0 & 0 \end{matrix}\right|\]

The angular momentum will be about the k direction for both L1 and L2. I gotta run, but I'll be back later. Tell me if this is making sense to you or ^_^

Answer 30

cross product gives us that at the maximum height, the angular momentum is

\[ \textbf L_1 = -hv_{xi} \hat{ \textbf k }\]

\[ \textbf L_1 = -\frac{v_0^3 \sin^2 \theta \cos \theta}{2g} \hat{ \textbf k }\]

angular momentum is in the z axis pointing into the screen. Makes sense using the right hand rule!

Answer 31

still need to find L2?