# Homework Help: Maximum Velocity

1. Oct 2, 2011

### bionut

(Q) A girl is swinging back and forth on a swing suspended from a rope 10m long. At the lowest point of her swing she is 1m from the ground, and at the highest point she is 2m above the ground. What will be her maximum velocity?

Would her maximum velocity be at the lowest point d = 1m above the ground?

Is so v= d /t .... ???

Or is this related to her potentional and kinetic energy?

2. Oct 2, 2011

### Rayquesto

"Would her maximum velocity be at the lowest point d = 1m(from the ground) above the ground?"
yes!

3. Oct 2, 2011

### PeterO

Answering you questions in order: yes, no, yes. [Use energy]

4. Oct 2, 2011

### Rayquesto

it has to do with like an inverted projectile motion. I don't know what you'd call that and I don't think it's smart to call anything an "inverted projectile motion," but you can think of it as so.

5. Oct 2, 2011

### Rayquesto

it will be constantly affected by gravity. that's why. I can tell you a method for figuring this out, but first, I got to ask, have you learned how to determine the period of a pendulum?

6. Oct 2, 2011

### bionut

So if her maximum velocity is at d-1m...

Then??

I can only think of:

acc due to gravity = 9.81 ms/sqr

im stuck with how to approach the next step...

7. Oct 2, 2011

### Rayquesto

how do you find the period of a pendulum?

8. Oct 2, 2011

### bionut

okay, then the other hting I can think of is using 2piR

2 x pi x 10m = 62.83m ???

9. Oct 2, 2011

### PeterO

This is definitely NOT inverted projectile motion. projectiles follow a parabolic path - a swing follows a circular path.

USE ENERGY as I said in post #3

10. Oct 2, 2011

### bionut

The only issue i have using an energy formula is I dont have the mass?

othwise I would use:

PE + KE = C
wt x h x 1/2 mv(sqr) = c

then I would sub in C into:

wt x h x 1/2 mv(sqr) = c, to find V....

11. Oct 2, 2011

### Rayquesto

I'm a bit stumped then. I remember some equation where v= sqrt 2glength(1-costheta)
but I'm trying to think about it in a simpler way. if the pendulum is constantly affected by gravity, then doesn't the velocity have anything to do with an angular change? well that's kinda self explanatory with the equation I gave. hm...

12. Oct 2, 2011

### PeterO

You have two separate equations.

At ALL times mgh + 1/2 mv2 = Constant

At maximum height, 1/2 mv2 = 0
At minimum height, mgh = 0

It is those two extreme positions that give you the two equations. [and mass will cancel out in the final steps]

13. Oct 2, 2011

### bionut

All I can think of is the realtionship between energy conservation... So at 2m (max height ) Her PE = max and at 1m (min) her KE=max where her velocity is max and her PE = min/0j)

If I use PE=mgh at 1 m; 0=m x 9.81 x 1 = 9.81 kg...

Then KE at 2m = 1/2mv(sqr) = 0=4.91v(sqr) = 2.2m/s

now...im confused

14. Oct 2, 2011

### Rayquesto

thanks for clarifying though. if this is circular, then isn't the dx from one point to the next is also the same as dy? so, dx=1. if you know that, then you know that the line between the motion of the pendulum when it started and the bottom is sqrt2. If you think about it, at 90 degrees between the max velocity and the point it starts, the line will be sqrt2 times the length of itself dy=10 and dx=10. if the pendulum was dropped from an angle that contains a dy of 1 which is 1/10 of 10(90 degrees of drop), and dy is also the same, then the angle should also be 1/10 of 90degress?

15. Oct 2, 2011

### PeterO

You have some very incorrect ideas in this.

The original post included values for vertical position only. You could calculate a whole lot of horizontal displacements if you like, but they are totally unnecessary for the solution of this problem.

The length of the swing is irrelevant for the solution of this problem, but would have great significance for all these dx values you are trying to find. Not sure why you want them.
When this swing is in a position where the seat is 2m from the ground [1m above the low point] the swing is about 26o to the side - much, much more than 1/10 of 90o .

16. Oct 2, 2011

### timacho

to;rayquesto

period pendulum=2pie surd L/g
=2pie surd m/k ,im i right???

what should we do if we have the period???
i do think that max velo is at the lowest point..

17. Oct 2, 2011

### Rayquesto

v= sqrt 2glength(1-costheta)

I think this is right because this is saying:

mv^2/2=mgL-mgLcostheta
since h=L-Lcostheta

its hard for me to understand why h=L-Lcostheta though hm.....

18. Oct 2, 2011

### PeterO

You need to draw yourself a picture, and decide which height is you reference value - i would suggest the low point.

This line makes absolutely no mathematical sense??
Then KE at 2m = 1/2mv(sqr) = 0=4.91v(sqr) = 2.2m/s

If I use PE=mgh at 1 m; 0=m x 9.81 x 1 = 9.81 kg...

Where did the m disappear to, PE should have been 9.81m Joules

And when you said PE at 1 metre ... 1 metre above where?

Does it help to say: The loss in potential energy when moving from 2m above the ground to 1m above the ground, will equal the gain in kinetic energy when it moves from a position 2m above the ground [at the extreme position] to the position 1m above the ground [at the bottom of the arc].

19. Oct 2, 2011

### bionut

Hi all thanks for all you help...

PeterO... the equations 1/2mv(sqr)= 0 = KE @ 2m? and mgh=0 = PE @ 1m?

so.. PE + KE = C
would be: mgh + 1/2mv(sqr) = c?
im really confused... I don't think the question is intended to be this complicated... I think its trying to gte me to think about PE and KE min and max...

I understand how 0=mgh at lowest point and how o=1/2mv(sqR0 at highest point .... and your saying velocity is constant...? so v(sqr) = 0.5 x 9.81 x 1 = 2.21m/s???

20. Oct 2, 2011

### PeterO

Rayquesto,

You are muddying the waters!!!

its hard for me to understand why h=L-Lcostheta though hm

It is hard for me to understand why you are doing this. We were told that at maximum point, the swing was 2m above the ground. At the lowest point the swing was 1m above the ground. Therefore your h = 1. Why all this trigonometry???? You only need that if you were told how far to the side the swing was moved between extreme and mean positions.