Launching a projectile from a spinning wheel

[natasha13100]

Homework Statement

Two engineers are working together in a competition to launch a pumpkin as high as possible using a launching device that must be kept within a certain height. Since the engineers have a height restriction, they want to accelerate the pumpkin to a large speed before releasing it. They decide a good way to get the pumpkin moving fast is to attach the pumpkin to a vertical wheel turned by a motor.

The first engineer suggests that when the wheel is turning as fast as possible at its constant rate w, the pumpkin can be released so that it goes straight upwards.

The other engineer gets out a notebook, runs through some physics calculations, and concludes that releasing the pumpkin straight up will not achieve maximum height. If the wheel spins at a constant angular velocity w, at what angle Θ should the pumpkin be released to reach the highest distance from the ground?

In your calculations, use the small angle aproximations for the sinΘ and cosΘ, and assume the wheel is spinning quickly. Use g, R, and W in your answer. Hint: what are the necessary terms to keep in your small angle approximation?

Diagram is attached.

Homework Equations

v=Rw
a=Rα
v=v_i+at
x=x_i+v_ixt+1/2a_xt²
v_x²=v_ix²+2a_xt
x=x_i+1/2(v_x+v_ix)(t-t_i)

The Attempt at a Solution

I am not sure how to solve this without θ being 0 or H being in my equation. I also do not understand the small angle approximation thing.
Here's what I have:
v_x=vsinθ=Rwsinθ
v_y=vcosθ=Rwcosθ
0=(Rwcosθ)²+2gH

Here I do one of two things
1. (Rwcosθ)²=-2gH
Rwcosθ=√(-2gH)
cosθ=√(-2gH)/(Rw)
θ=cos^-1√(-2gH)/(Rw)
2. -(Rwcosθ)²=2gH
H=-(Rwcosθ)²/2g
g is -9.8m/s² and R and w are constant so H depends on cosθ².
For H to be maximum, cosθ would be maximum. This happens when cosθ=1.
θ=0

 

[SteamKing]

What happens to sinθ and cosθ when θ gets small? Remember, θ is in radians.

 

[verty]

I don't see why, if we are trying to find the angle that gives greatest height, we should assume the angle is small. It is conceivable that a large angle gives the greatest height. Why not check all angles?

 

[SteamKing]

I don't see why, if we are trying to find the angle that gives greatest height, we should assume the angle is small. It is conceivable that a large angle gives the greatest height. Why not check all angles?

You are saying the angle for max.height should be small and then you are saying the angle could be large. You can't have it both ways.

 

[gneill]

You are saying the angle for max.height should be small and then you are saying the angle could be large. You can't have it both ways.

This is one of those times where punctuation is important The clause between the commas is just stating the goal. It can be rearranged as:

"If we are trying to find the angle that gives greatest height, I don't see why we should assume the angle is small. It is conceivable that a large angle gives the greatest height. Why not check all angles?"​

 

[verty]

It's an odd question, is all.

 

[voko]

0=(Rwcosθ)²+2gH

This does not look right. H is positive (it is the height), and so is ## R \omega \cos^2 \theta ##, so this equation is true only when ##H = 0## and ## \theta = 90° ##, which is useless for the problem.

What is the correct formula?

Secondly, you ignore the initial elevation of the pumpkin while it is still attached to the disk.

 

[technician]

At the point of release the vertical component of the velocity = ωRCosθ and therefore the vertical distance traveled from this point is (ωRCosθ)²/2g
The point of release is at a height of RSinθ from the horizontal reference point.

 

[natasha13100]

I am getting pretty confused, but I will try to answer posts as they came.
SteamKing: Why does θ have to be in radians? Anyhow, I would think sinθ would be close to 0 and cosθ would be close to 1.

Verty: I'm sorry, but that's the question exactly as I was given it except for the "Diagram is attached" part.

Voko: I am talking about velocity in the y direction.
v²=v_i²+2ay.
At the top, v=0.
v_i is the tangential velocity from the circular motion equation (v=Rω).
since v_y=vcosθ(the angle of velocity from the horizontal is 90-θ), v_i²=(Rwcosθ)²
Okay, so the tiny initial height above θ=0 should be included in the equation which would make it 0=(Rwcosθ)²+2g(h-h_i)
h_i=Rsinθ so 0=(Rwcosθ)²+2g(H-Rsinθ)
2gH=2gRsinθ-R²w²cos²θ
H=Rsinθ-R²w²cos²θ/2g
since sinθ≈0 and cosθ=≈1, H=-R²w²/2g
could that be it?
I ignore the height from the ground to the center on purpose.

technitian: Sorry, I just realized you had already put up the answer to voko's post. Thanks. :)

 

[voko]

Voko: I am talking about velocity in the y direction.
v²=v_i²+2ay.
At the top, v=0.
v_i is the tangential velocity from the circular motion equation (v=Rω).
since v_y=vcosθ(the angle of velocity from the horizontal is 90-θ), v_i²=(Rwcosθ)²
Okay, so the tiny initial height above θ=0 should be included in the equation which would make it 0=(Rwcosθ)²+2g(h-h_i)

Again: both terms on the right hand side are positive, so their sum cannot be zero except in the trivial case ## h = h_i ## and ## \theta = \pi / 2 ##. This cannot be right.

since sinθ≈0 and cosθ=≈1, H=-R²w²/2g

And you end up with a negative height. Definitely not good. But if you ignore thesign, then the height is what you would get by launching the thing straight up. Hint: this is because ## \sin \theta \approx 0 ## is too rough an approximation; do you know a better one?

 

[haruspex]

I strongly advise not making any approximations until the equations reach a point where you need to approximate to make further progress.

 

[technician]

Don't get hung up about +ve and - ve heights!
The max height traveled has 2 parts...
1) the release height = RSinθ

2) the vertical distance from the release height. = (ωRCosθ)²/2g

I do not see any justification for having to consider "small angles"
Try any angle!...45⁰ and see what comes out !

(45⁰ is nice to try because Sinθ and Cosθ are the same ...1/√2)
You will get a 'feel' of the problem by trying things like this.
This is a nice problem...it cannot be easy because there are 3 independent variables...ω, R and θ

 

[rude man]

Don't get hung up about +ve and - ve heights!
This is a nice problem...it cannot be easy because there are 3 independent variables...ω, R and θ

There is only one independent variable, θ.
R and w are given. 2R = max allowed height of contraption.

I saw no need for approximation, got a simple answer but probably had better check my math.

 

[rude man]

I don't see why, if we are trying to find the angle that gives greatest height, we should assume the angle is small. It is conceivable that a large angle gives the greatest height. Why not check all angles?

Absolutely right.

 

[natasha13100]

Again: both terms on the right hand side are positive, so their sum cannot be zero except in the trivial case ## h = h_i ## and ## \theta = \pi / 2 ##. This cannot be right.
And you end up with a negative height. Definitely not good. But if you ignore thesign, then the height is what you would get by launching the thing straight up. Hint: this is because ## \sin \theta \approx 0 ## is too rough an approximation; do you know a better one?

Well I'm assuming g=-9.8m/s^2. I have no idea how to make a better approximation. The only thing I can think of is to approximate for cos(theta) and not sin(theta) then solve for theta. (cos(theta) is really close to 1 and doesn't make a big difference, but sin(theta) matters because anything is bigger than zero.)

 

[voko]

Then avoid making approximations just yet. If g is negative, then your equations are correct, and all you need to do is maximize ## H = R \sin \theta - \frac {R^2 \omega^2} {2g} \cos^2 \theta ##.

 

[natasha13100]

If I am thinking correctly, it will be maximized when the derivative=0. Since theta is the only unknown, I can treat it like I would normally treat an x in math. This means that 0=Rcos(theta)+R^2*w^2*2sin(theta)cos(theta)/2g and theta=arcsin(-g/Rw^2)

 

[voko]

Yes.

 

[rude man]

Then avoid making approximations just yet. If g is negative, then your equations are correct, and all you need to do is maximize ## H = R \sin \theta - \frac {R^2 \omega^2} {2g} \cos^2 \theta ##.

I think the sign before the 2nd term is wrong. Otherwise, exactly what I got.

Oh, wait, if g < 0 it's exactly what I got. An easy maximization, and no approximation necessary.

If max H is desired you need to replace Rsinθ with R(1 + sinθ). For maximization it's irrelevant of course.

 

[natasha13100]

I think the sign before the 2nd term is wrong. Otherwise, exactly what I got.

I already covered this. I'm using g=-9.8 instead of 9.8

Voko: Thank you so much. I finally got it which is good because I don't get the correct answer even after the due date because of the online system. I understand where I was going wrong (I'm used to trig-based physics and the calculus through me off. Plus I seem to always forget some small piece of information like the height above the center.) Something is still throwing me off though. The software accepted g/Rw^2 instead of arcsin(g/Rw^2) does anyone know why?

 

[voko]

The software accepted g/Rw^2 instead of arcsin(g/Rw^2) does anyone know why?

Because ## \sin \theta \approx \theta ##. You may know this in a different form: ## \lim_{x \to 0} \frac {\sin x} {x} = 1 ##.

 

[technician]

There is only one independent variable, θ.
R and w are given. 2R = max allowed height of contraption.

I saw no need for approximation, got a simple answer but probably had better check my math.

Where is it given that the max height is 2R?
Have I missed something?

 

[voko]

Where is it given that the max height is 2R?
Have I missed something?

It is given that some max height is specified. It is not given that 2R is this max height, but this is something that is reasonable to assume.

 

[technician]

It is given that some max height is specified. It is not given that 2R is this max height, but this is something that is reasonable to assume.

OK...thanks, I thought I had missed something. I like this problem !

 

[technician]

edit

 

[natasha13100]

OK...thanks, I thought I had missed something. I like this problem !

I do too now that I've gotten through it. It really makes you think. I still can't believe we were given this in first year physics. It was interesting seeing how different people thought of the problem different ways. I also posted this on yahoo answers. The answer I got there was wrong, but it looked like maybe some simple algebra went wrong somewhere. That answer actually involved kinetic energy (we are just now getting to rotational kinetic energy though so I couldn't understand where some of his work came from).