Calculate Maximum Velocity of Girl on Swing

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Homework Help Overview

The discussion revolves around calculating the maximum velocity of a girl swinging on a swing, with specific details about the swing's height and length. The problem involves concepts from mechanics, particularly potential and kinetic energy, as well as the motion of pendulums.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore whether the maximum velocity occurs at the lowest point of the swing and discuss the relationship between potential and kinetic energy. Some express confusion about the role of mass in energy calculations and the implications of gravitational acceleration. Others question the relevance of certain equations and concepts, such as the period of a pendulum and the nature of the swing's motion.

Discussion Status

The discussion is active, with participants sharing various thoughts and equations related to energy conservation and motion. Some guidance has been offered regarding the use of energy principles, but there is still a lack of consensus on the best approach to take, and several interpretations are being explored.

Contextual Notes

Participants note constraints such as the absence of mass in calculations and the need for clarity on height references in energy equations. There is also mention of specific angles and trigonometric considerations that may not be necessary for solving the problem.

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

Answering you questions in order: yes, no, yes. [Use energy]
 
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.
 
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?
 
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...
 
how do you find the period of a pendulum?
 
okay, then the other hting I can think of is using 2piR

2 x pi x 10m = 62.83m ?

Heading in the right direction?
 
bionut said:
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...

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
The only issue i have using an energy formula is I don't 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
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
bionut said:
The only issue i have using an energy formula is I don't 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...

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
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
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
Rayquesto said:
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?

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
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
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
bionut said:
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

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
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
Rayquesto said:
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...

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.
 
  • #21
@timacho yes that's right for smaller angles. I remember my high school book said any angles from the vertical 15 degrees and under. I was told there was a taylor series proof, but I'm not that far in calculus just yet to exactly research it and understand it completely.

And about finding the velocity, well, I think the method I was thinking of was wrong. I was thinking you could take a quarter of the period, and then find the acceleration of the parabola then multiply that acceleration times the quarter of the period to get the velocity at the amount of time it takes to go to max velocity. how would you find the acceleration? would it simply be 9.81m/s^2 or some costheta time 9.8 or finding it a different way? I have to think about that.
 
  • #22
sorry not parabola, but hahahaha pendulum!
 
  • #23
Okay, if PE = 9.81 J @ 2m above the ground... and energy is conserved, @1m above the ground all energy will become KE?; KE=9.81J?

if I plug that into KE=1/2mv(sqr) I still can't comput an answer... I don't have mass

(sorry... this is fustraing the ! out of me lol)... I understand concepts of energy conservation but isn't this imposssible to solve without a mass??
 
  • #24
but they give you height so I think the velocity is just v=sqrt2gh
 
  • #25
rayquesto ... but this would make here velocity at the highest point faster than the lowest point... My undersatdning is velocity is at it/s max when d=1m...
 
  • #26
okay well.. the answer is 4.43 m/s... so v=sqrt (2 x 9.81 x 2) = 4.43m/s... aff yeas on reading the question it say's her velocity at highest point...

Thanks for all your help... I undersatnd how to getthe answer ..but confused how it can be solved without a mass...
 
  • #27
v=sqrt19.62 then since the height is 1meter. it's the concept of conservation of potential energy to kinetic energy.

techinically, mv^2/2 final mv^2/2initiall=mghfinal- mghinitial

and the initial mgh is high, final mgh is zero, initial mv^2/s is zero and the final mv^2/2 is high. high=high...
 
  • #28
So then this indicates that the max velcoity is at the higest point? I though max velcoity is at the lowest point??... but her velocity @ 1m using v= sqrt 2gh also = 4.43 m/s... so is the velcoity then constant?
 
  • #29
bionut said:
Okay, if PE = 9.81 J @ 2m above the ground... and energy is conserved, @1m above the ground all energy will become KE?; KE=9.81J?

if I plug that into KE=1/2mv(sqr) I still can't comput an answer... I don't have mass

(sorry... this is fustraing the ! out of me lol)... I understand concepts of energy conservation but isn't this imposssible to solve without a mass??

It CAN be solved without knowing the mass

At the highest point we have mgh + 0 = total energy [not moving so zero KE]
At the lowest point we have 0 + 1/2 mv2 = total energy [no PE as lowest point]

So numerically mgh = 1/2 mv2

There goes the mass [is cancels out]

gh = 1/2 v2

so v2 = 2gh

so v = √(2gh)

but h = 1 and g = 9.81

so v = √19.62

so v is between 4 and 5 [because I know √16 = 4 and √25 = 5]
 
  • #30
its the velocity at the lowest point. That's what you originally asked right? the velocity is not constant, but energy is conserved from point a(highest point) to b(lowest point).
 

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