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Juggler problem

  • Thread starter miew
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  • #1
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b]1. Homework Statement [/b]

A juggler is juggling a uniform rod one end of which is coated in tar and burning. He is holding the rod by the opposite end and throws it up so that, at the moment of release, it is horizontal, its Center of Mass is traveling vertically up at speed vo and it is rotating with angular velocity wo. To catch it, he wants to arrange that when it returns to his hand it will have made an integer number of complete rotations. What should vo be, if the rod is the have made exactly n rotations when it returns to his hand?

Homework Equations



L=rxp
torque=dL/dt

The Attempt at a Solution


I don't even know how to start...
 

Answers and Replies

  • #2
tiny-tim
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hi miew! :smile:

(have an omega: ω and try using the X2 icon just above the Reply box :wink:)
I don't even know how to start...
ok, the linear motion and the rotational motion are completely independent, so start by writing out the two equations for them (one each), as a function of t …

what are they? :smile:
 
  • #3
1,137
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think this,
lest rod goes up by some distance l(doesn't matter whats l)
you can calc. what time will it take to come to same horizontal level
in that time the rod should make n rotations
find time for 1 rotation using wo
now time of flight = time for n complete rotations
 
  • #4
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So, for the linear motion:
V=V0-gt

And for the rotational motion:

[tex]\omega[/tex]= [tex]\omega[/tex] 0+[tex]\alpha[/tex]t

Cupid.callin what do you mean by lest rod goes up by some distance l?
 
  • #5
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Cupid.callin what do you mean by lest rod goes up by some distance l?
i meant Let
 
  • #6
tiny-tim
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hi miew! :smile:

(what happened to that ω i gave you? :confused:)
So, for the linear motion:
V=V0-gt

And for the rotational motion:

ω = ω0 + αt
Yes, that's correct, but you really need equations for s and θ, not ω and α :smile:
 
  • #7
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(didn't I use omega :/ ? )

So the time to make one rotation is t= 2 [tex]\pi[/tex] / [tex]\omega[/tex]0 right ?

And in that time it goes up a distance x= Vot - [tex]\frac{1}{2}[/tex]gt where t is the one above ?

Am I in the right track ? :)
 
  • #8
tiny-tim
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hi miew! :smile:

(just got up :zzz: …)
So the time to make one rotation is t= 2 [tex]\pi[/tex] / [tex]\omega[/tex]0 right ?

And in that time it goes up a distance x= Vot - [tex]\frac{1}{2}[/tex]gt where t is the one above ?
(easier to type and to read if you use ω and π instead of the LaTeX versions :wink:)

yup! :biggrin: (gt2, of course) …

except the question says he's a really good juggler, and you have to use nt instead of t :wink:

ok, now find an equation relating v0 and ω0, and then solve for v0 ! :smile:
 
  • #9
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(I can't find your omegas and pi...:( )

Okay, so this is what I got.

t=2npi/[tex]\omega[/tex]

Vf=v0-gt and since Vf=0, v0=gt
but it goes up and down, so the total time is

t=2vo/g.

And then, v0=gnpi/[tex]\omega[/tex]

Is that right ? :)
 
  • #10
tiny-tim
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hi miew! :smile:

(you could have copied-and-pasted the ω and π i used :wink: … see a fuller list below)

yes that's fine …

but you could have done it slightly quicker by saying that when it returns, v will be minus v0, so v0 - (-v0) = gt :wink:

(alternatively, 0 = v0t - gt2/2 give the same result without having to know that v = -v0)
 
  • #11
27
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Oh you are right !:smile:

THank you so much for your help, you were really helpful :smile:
 
  • #12
144
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Hi, bringing up this old thread because I was stuck on the same problem for a little bit, though for a different reason. I understand the derivation as it is, but the thing that held me back was accounting for the torque due to gravity. Is the torque on each infinitesimal rod element canceled by the torque on element opposite the center of mass?
 
  • #13
tiny-tim
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hi diligence! :smile:
… the torque due to gravity. Is the torque on each infinitesimal rod element canceled by the torque on element opposite the center of mass?
yes (even for an irregular shape) …

∫ ρ (r - rc.o.m) x g dxdydz

= {∫ ρ (r - rc.o.m) dxdydz} x g

= 0 x g :wink:
 
  • #14
144
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Thanks Tim. So I guess it's basically inherent in the definition of center of mass? Yes, that's now obvious in hindsight. Thanks!
 

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