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Motion within a circle and path kinematics

  • Thread starter MFAHH
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  • #1
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Homework Statement



John is thrown into the outer part of a circular den of radius a. At the centre of the den is a Lion. John decides his best tactic is to run round the perimeter of the den with his maximum speed u. The Lion responds by running towards John with its maximum speed U.

(i) Sketch the path taken by the Lion.
(ii) Show that the distance r of the Lion from the centre of the den satisfies:
dr/dt = sqrt [U^2 - r^2u^2/a^2]
(iii) Hence find r as a function of t.
(iv) Show that if U > u John will be caught by the Lion.
(v) What happens in the special case where u = U?


Homework Equations




The Attempt at a Solution



For i) I sketched the path of the lion to be a spiral starting from the centre and tending outwards towards the circumference of the circle.

It's on ii) that I'm not sure on how to set up the problem and the relevant equations of motion that would yield the desired equation. Any hints?

Thank you.
 

Answers and Replies

  • #2
BiGyElLoWhAt
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Have you tried drawing some triangles and invoking a dead greek?
 
  • #3
BiGyElLoWhAt
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What does U equal? Give me a magnitude in terms of r, r dot, and theta dot.
 
  • #4
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Thanks for the reply. Here's my train of thought. Essentially the lion is moving in a sort of circular motion with the radius increasing wrt time. So r=r(t). Now, U=ds/dt, where s is the distance covered by the lion. This is given by, I think, rθ. So ds/dt = (dr/dt)*θ + r*(dθ/dt) = U.

Am I on the right lines, because I have a funny feeling I'm not..
 
Last edited:
  • #5
BiGyElLoWhAt
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Almost. Think pythagorean's theorem.

Theta dot is always perpendicular to r dot, so you have a right triangle between the 2.

Check the first term again (dr/dt)*\theta
Does the magnitude change depending on which angle your facing? If you rotate the spiral, changing the starting position of Joe (? whatever his name is), does that change the magnitude of the the lions velocity?
 

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  • #6
BiGyElLoWhAt
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maybe that picture had too much going on. Try this one. Can you tell me what U is equal to now?
 

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  • #7
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maybe that picture had too much going on. Try this one. Can you tell me what U is equal to now?
Aha! U = sqrt(rdot^2+theta dot^2)?
 
  • #8
BiGyElLoWhAt
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Super close now. Check out the units. Rdot is m/s and theta dot is 1/s, what happens the the arc of a circle if you go farther away from the center?
If I travel from 0 to pi along the circumference of a circle with radius 1, then do the same thing with a radius 2, how will my total distance travelled compare?
 
  • #9
BiGyElLoWhAt
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Does the brown line or the green line have a greater magnitude? How can you relate that to what we have to work with? The angle changed in both cases will be the same, but which velocity is greater? How can you take your equation and change it (only 1 small change that needs to be made) so that it accounts for this?
 

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  • #10
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Would it be multiplying theta dot with r^2? That way the units correspond to each other on the LHS and the RHS. Also, it's the change in radius that yields the greater velocity of the green line. Would that be right?
 
  • #11
BiGyElLoWhAt
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yea yea. cool. I'm assuming we're on the same page now, with ##U^2 = \dot{r}^2 + r^2\dot{\theta}^2##
Ok next task:
The lion ALWAYS runs directly at the boy, what does this tell you about the theta dot of the lion and the theta dot of the boy?
Speaking of, what is the theta dot of the boy?
 
  • #12
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yea yea. cool. I'm assuming we're on the same page now, with ##U^2 = \dot{r}^2 + r^2\dot{\theta}^2##
Ok next task:
The lion ALWAYS runs directly at the boy, what does this tell you about the theta dot of the lion and the theta dot of the boy?
Speaking of, what is the theta dot of the boy?
Would it indicate that the theta dot of both the lion and the boy are equal. And theta dot would be u/a because the boy travels along the circumference of radius a, at linear speed u.
 
  • #13
BiGyElLoWhAt
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bada bing, bada boom. not u/r, because the radius of the circle is a, not r. r is the distance the lion is from the center. so u/a.
 
  • #14
BiGyElLoWhAt
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plug it in and solve for rdot (which is the same as dr/dt)
 
  • #15
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Haha! Excellent! Thanks a lot man I seriously appreciate all your help and patience! :)
 
  • #16
BiGyElLoWhAt
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Np bro. Take it easy.
 
  • #17
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So I got my work out of the way and tried to finish up this question. For the part where I've to find r as a function of t I'll obviously need to solve the differential. I've only been recently begun being taught about how to solve ODEs. What I'm thinking for this one is, because it has r raised to the power of 2, I'm guessing Bernoulli's equation needs to be used, and then from there calculate r(t) using the integrating factor method to solve. But I think Bernoulli can only be invoked if the ODE is in the form where there is also an r^1 term, but in this equation it's only got an r^2 term, so it can't be invoked[?].

Anyone have any suggestions? Also, the form that my diff equation is in has dr/dt all to the power of 2, is that allowed?

Thanks!
 
  • #18
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Anyone?
 
  • #19
TSny
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dr/dt = sqrt [U^2 - r^2u^2/a^2]
(iii) Hence find r as a function of t.
Can you "separate variables" and integrate?
 
  • #20
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Can you "separate variables" and integrate?
Yeah I can, but I'm not sure if I can do that here..
 
  • #21
TSny
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If you have ##\frac{dy}{dx} = f(y)##, then you can write ##\frac{dy}{f(y)} = dx##
 

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