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

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SUMMARY

The discussion focuses on solving a problem related to circular motion and elliptical paths, specifically analyzing the forces acting on a man completing an elliptical loop. Key equations include normal acceleration defined as (v^2)/r and the energy conservation principle. The radius of curvature is derived using the formula r = [(a^2)(sin^2(m)) + (b^2)(cos^2(m))]/sqrt[a^2cos^2(m) + (b^2)(sin^2(m))], where a and b are the semi-major and semi-minor axes of the ellipse. The conversation emphasizes the importance of understanding tangential and centripetal acceleration in relation to the motion of the man.

PREREQUISITES
  • Understanding of normal acceleration in circular motion
  • Familiarity with energy conservation principles in physics
  • Knowledge of ellipse equations and their parameters
  • Ability to differentiate functions to find velocity and acceleration
NEXT STEPS
  • Study the derivation of radius of curvature in elliptical motion
  • Learn about the conservation of energy in mechanical systems
  • Explore the relationship between tangential and centripetal acceleration
  • Review standard physics textbooks like Resnick, Halliday, and Krane for deeper insights into circular motion
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators seeking to clarify concepts related to elliptical paths and forces in motion.

  • #31
sorry what is X and Y? are they the vertical and horizontal displacements?
 
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  • #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)
 
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  • #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
 

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