# Solve the Juggler Problem: Find Vo for Rod to Make Integer Number of Rotations

• miew
In summary, a juggler is trying to catch a rotating rod by making it make an integer number of rotations. He needs to find the time it takes to make one rotation, and the rod goes up and down so the total time is also the time it takes to go up and down.
miew
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?

L=rxp
torque=dL/dt

## The Attempt at a Solution

I don't even know how to start...

hi miew!

(have an omega: ω and try using the X2 icon just above the Reply box )
miew said:
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?

think this,
lest rod goes up by some distance l(doesn't matter what's 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

So, for the linear motion:
V=V0-gt

And for the rotational motion:

$$\omega$$= $$\omega$$ 0+$$\alpha$$t

Cupid.callin what do you mean by lest rod goes up by some distance l?

miew said:
Cupid.callin what do you mean by lest rod goes up by some distance l?

i meant Let

hi miew!

(what happened to that ω i gave you? )
miew said:
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

(didn't I use omega :/ ? )

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

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

Am I in the right track ? :)

hi miew!

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

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

(easier to type and to read if you use ω and π instead of the LaTeX versions )

yup! (gt2, of course) …

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

ok, now find an equation relating v0 and ω0, and then solve for v0 !

(I can't find your omegas and pi...:( )

Okay, so this is what I got.

t=2npi/$$\omega$$

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/$$\omega$$

Is that right ? :)

hi miew!

(you could have copied-and-pasted the ω and π i used … 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

(alternatively, 0 = v0t - gt2/2 give the same result without having to know that v = -v0)

Oh you are right !

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?

hi diligence!
diligence said:
… 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

Thanks Tim. So I guess it's basically inherent in the definition of center of mass? Yes, that's now obvious in hindsight. Thanks!

## 1. What is the Juggler Problem?

The Juggler Problem is a mathematical puzzle that involves finding the velocity (Vo) needed for a rod to complete an integer number of rotations when thrown into the air and caught at the same position.

## 2. How is the Juggler Problem solved?

The Juggler Problem can be solved using a mathematical formula that takes into account the initial velocity (Vo), the height of the throw (H), and the gravitational constant (g). The formula is Vo = sqrt(2gH).

## 3. What is the significance of solving the Juggler Problem?

The Juggler Problem is a challenging mathematical puzzle that requires critical thinking and problem-solving skills. It also has real-world applications in sports and entertainment, such as juggling and other types of ball throwing performances.

## 4. What are some tips for solving the Juggler Problem?

Some tips for solving the Juggler Problem include breaking down the problem into smaller parts, using trial and error, and checking your calculations for accuracy. It is also helpful to have a basic understanding of physics and kinematics.

## 5. Are there any variations of the Juggler Problem?

Yes, there are several variations of the Juggler Problem, such as finding the optimal angle for the throw or solving for the time it takes for the rod to complete an integer number of rotations. These variations require a different approach and may have different formulas or solutions.

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