Problem with Circular Motion: Solution Attempt and Equations | Homework Help

Click For Summary

Homework Help Overview

The discussion revolves around a problem involving circular motion and an ellipse, specifically focusing on the dynamics of a man completing an elliptical path. Participants are exploring concepts related to normal acceleration, radius of curvature, and energy conservation in the context of motion along an elliptical trajectory.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants attempt to derive the radius of curvature and question the assumptions regarding the radius being 1.8m. There are discussions about using energy equations versus force equations to analyze the problem.

Discussion Status

Several participants are actively engaging with the problem, offering different perspectives and questioning the validity of certain assumptions. Some guidance has been provided regarding the use of energy conservation and the calculation of radius of curvature, but no consensus has been reached on the correct approach or final values.

Contextual Notes

There is an ongoing debate about the correct interpretation of the radius of curvature and its relation to the semi-major and semi-minor axes of the ellipse. Participants also express uncertainty about the application of certain equations and the implications of potential energy in their calculations.

  • #31
sorry what is X and Y? are they the vertical and horizontal displacements?
 
Physics news on Phys.org
  • #32
Yes.:-)
 
  • #33
Also you said,

the equation given x^2/9 + y^2/16 = 1 is distance i believe, so when you differentiate it v=2x/9 + y/8 = 0, and differentiate again would mean a=2/9+1/8=25/72. And these three equations are tangential.


This is not how you differentiate.

X^2/9+ Y^2/16=1

Differentiating
2xdx/9dt +2ydy/16dt =0

dx/dt is horizontal velocity and dy/dt is vertical velocity.

Using equation x=asin m is easier .So differentiate it.

dx/dt or Vx=acos m (dm/dt)

dm/dt is like omega
Ax= -asinm(dm/dt)^2 + a cosm(d2m/dt2)

Here -asin m(dm/dt) is x component of centripetal acceleration.

Magnitude of velocity is sqroot [(dx/dt)^2 +(dy/dt)^2]
 
  • #34
i still don't get the radius. i get everything else just not the radius. I don't get how differentiating that somehow gets me a radius. before the mistake the radius was 9/4 now what is it?
 
  • #35
so can i actually find out the velocity at the top without using a radius?
 
  • #36
See Radius of curvature at any point is defined as V^2/a where V is tangential velocity and a is centripetal acceleration( all at that point)

This formula is commonly used in uniform circular motion.(Right?)

So first we obtain tangential velocity which we get by differentiating displacement.


Secondly we obtain value of acceleration which we obtain by differentiating velocity.
One of these corresponds to centripetal acceleration and the other to tangential acceleration.

We take the centripetal acceleration and obtain its magnitude.

and thus obtain radius of curvature.

No you cannot find velocity on top without calculating radius of curvature.

All you know from the ellipse equation is what is the path traveled by the man.
You don't know how its travelled.

Here we were finding the minimum velocity so we used v^2=rg

Otherwise the velocity on top could have been different.

When you differentiate x you obtain a term dm/dt which is like w (omega)of circular motion.

This dm/dt can have any value similar to way w can have any value which decides velocity at any point.


(You may want to open standard books like Resnick Haliday and Krane and read out circular motion)
 
Last edited:
  • #37
In case you still have doubts, google circular motion and read about tangential velocity,angular velocity centripetal acceleration tangential acceleration angular acceleration.

Because questions of ellipse can only be solved when these concepts are crystal clear.

I would suggest you go through Renick Halliday and Krane.

You can also read about elliptical motion

http://opequi.free.fr/pl/icm_html/icmse17.html

This is a little differebt than the case we solved but the basic concepts are same
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K