# Homework Help: Coursework on Projectile Motion

1. Nov 19, 2005

### Apoorva

Projectile Motion :(

Hey guys I am doing some Coursework based on Projectile Motion. Currently I am stuck :(.

I started off my hypothesis by explaining how GPE will transfer to KE so that I could compare the error that would be created when the real equation is used which is GPE = KE + RKE.

GPE = KE where v is the subject in order to find range BUT,

for GPE = KE + RKE I am completly and utterly stuck.

Here is my current working:
GPE = KE + RKE
mgd = ½mv^2 + ½Iω^2
mgd = ½mv^2 + ½(2/5mv)ω^2
2mgd = mv^2 + 2/5mvω^2
5gd = 5/2v^2 + vω^2

GE = mgh
KE = 1/2 mv^2
RKE = 1/2Iω^2
I know that the Moment of Inertia for a Sphere is 2/5mv which can be substituted in...

Also I would prefer it if u started right from the start rather than just jump in for my sake :D.

Thanks
Apoorva

2. Nov 19, 2005

### kleinwolf

I don't know if you mean the following : I throw a sphere and once I consider it as a point particle, the other time not, so that it has a moment of inertia. I think that in this case, you get the same trajectory, because grav. force is not depending on the speed of the point it acts on (so basically, if your sphere is rotating or not, each point at its surface is feeling the same exterior grav. force, being it in rotation or not), which allows you to reduce the system to it's center of mass (gravity)(at least in classical mechanics) ...However, if you have friction forces, that either are viscous (depending on the speed, for example the air around it), or not (just depending on the force, for example on an inclinated plane, which moreover introduce rigid constraints on the movement), then it's clear that the motion will change, but you mabye need some special easy example to make the case analytically calculable......does it make any sense ?

3. Nov 19, 2005

### Apoorva

Explaining

Hmm, I think I should explain the experiment a bit more.

My aim is to find out the relationship between the Range (Horizontal Displacement) of a projectile and the vertical drop height.

To do this I need to use the Equations of Motion...

To emulate a projectile I will use a ball bearing, to give it KE I will have a curved ramp to allow the ball to gain KE from GPE and then launch at a horizontal angle. I will model the ball bearing as a particle to make life easier...

I start off with the theory that when the bearing rolls from the height it will go from having GPE = KE completely (truely it goes to KE + RKE but bare with me)

Here is my working:

GPE = KE
Hence,
mgh = 1/2 mv^2
m[gh] = m[1/2v^2]
gh = 1/2v^2
2gh = v^2
which means that the starting velocity will be the square root of 2gh.

Now we need to find the time it will take to hit the ground in order to find the range as when it is a particle the horizontal velocity will remain the same until it comes into contact with the ground.

I know that s = 1/2at^2

I know that I have s (displacement) which will be constant and I know that a = g (gravity, no number needed to be assigned for now). All I need to do is to make the subject of the equation "t":

s = 1/2at^2
2s = gt^2
t^2 = 2s/g

This means that t is equal to the square root of 2s/g

All we need to do now is to plug it into the equation s = t . v

Hence,
s = [2s/g]^1/2 . [2gh]^1/2
s = [2s/g . 2gh]^1/2
s = [2sh]^1/2

Hence the range is square root 2sh.

Now that is the incorrect derivation because GPE = KE + RKE which I am trying to do to show the difference between it as to show the error created by that small difference because as you can see from the final derivation shows that the only components that have a hand in the range is the total height and the launch height.

The equation to calculate RKE = 1/2Iw^2
I know that the moment of inertia (I) is 2/5mv for a spherical object with symmetry. So you get 1/2 [ 2/5mv]w^2 which when you simplify you get 1/5mvw^2

So lets input this into the equation:

GPE = KE + GPE
mgh = 1/2mv^2 + 1/5mvw^2
2mgh = mv^2 + 2/5mvw^2
m[2gh] = m[v^2] + m[2/5vw^2]
2gh = v^2 + 2/5vw^2

Now could you help me by where I go from here as I have to make v the subject of the equation but how do I do that???

Thanks,
Apoorva

4. Nov 19, 2005

### Astronuc

Staff Emeritus
Last edited by a moderator: May 2, 2017
5. Nov 20, 2005

### Apoorva

Yessss

Thanks, I'll have a go now :D I knew I messed up somewhere

Infact I will do the maths on paper before I reply :D...

.........

WOOT it works out to be square root of 10gh/7

PS, Sidenote, was wondering where you learned LaTeX?

6. Nov 20, 2005

### Fermat

https://www.physicsforums.com/misc/howtolatex.pdf" [Broken]

Last edited by a moderator: May 2, 2017