Solving Jumping Off a Disk Problem (d)

[Andy Resnick]

Just to be clear, this is potentially a homewowrk problem I would assign.

1. Homework Statement
This is a 2-D problem. You (mass m) are standing on the edge of a large disk (radius R, mass M). What happens when you jump off, with:

a) The disk is on a frictionless surface, you jump radially outward with velocity v? (I can do this)
b) The disk is anchored to the ground at the center and you jump radially outward with velocity v? (I can do this)
c) The disk is anchored to the ground at the center by a frictionless pivot and you jump tangential to the edge with velocity v? (I can do this)
d) the disk rests on a frictionless surface and you jump tangential to the edge with velocity v? (I am having trouble with this)

Homework Equations

Conservation of linear momentum, conservation of angular momentum, conservation of energy

The Attempt at a Solution

The problem I have with (d) is that there seems to *never* be rotation of the disk, even though I am providing an impulse J = mv at some moment arm to the center of mass of the disk, which would seem to create angular rotation of the disk, in addition to translational motion of the disk. Help? Thanks in advance!

 

[PeroK]

The Attempt at a Solution

The problem I have with (d) is that there seems to *never* be rotation of the disk, even though I am providing an impulse J = mv at some moment arm to the center of mass of the disk, which would seem to create angular rotation of the disk, in addition to translational motion of the disk. Help? Thanks in advance!

Why do you think the disk would not rotate?

 

[Andy Resnick]

Why do you think the disk would not rotate?

Because, when I set up my 3 conservation equations and start reducing them, I end up with an expression like F(d)*ω²= 0, where 'd' is the distance from the disk edge to the center of rotation. So either ω = 0 or F(d) = 0, and solving for d returns complex values; therefore ω=0. But the disk *should* rotate! I can't figure out what I did wrong: here are my 3 conservation equations-

Linear momentum: V_disk = -m/M V_you, V_disk is the velocity of the disk's center of mass.
Angular momentum: -I_diskω_disk = I_you V_you/d (I made the substitution ω_you=V_you/d). The moments of inertia are obtained with the parallel axis theorem, and are functions of 'd'.

Energy: 0 = mV_you²+ M V_disk²+ I_diskω_disk² (*)

(*): there should also be a term representing the biochemical energy required to jump, this is an input energy. I left it out... maybe that's the problem?

 

[PeroK]

Because, when I set up my 3 conservation equations and start reducing them, I end up with an expression like F(d)*ω²= 0, where 'd' is the distance from the disk edge to the center of rotation. So either ω = 0 or F(d) = 0, and solving for d returns complex values; therefore ω=0. But the disk *should* rotate! I can't figure out what I did wrong: here are my 3 conservation equations-

Linear momentum: V_disk = -m/M V_you, V_disk is the velocity of the disk's center of mass.
Angular momentum: -I_diskω_disk = I_you V_you/d (I made the substitution ω_you=V_you/d). The moments of inertia are obtained with the parallel axis theorem, and are functions of 'd'.

Energy: 0 = mV_you²+ M V_disk²+ I_diskω_disk² (*)

(*): there should also be a term representing the biochemical energy required to jump, this is an input energy. I left it out... maybe that's the problem?

Well, if you jump and create any motion that's an increase in kinetic energy of the system. So, there can't be zero kinetic energy.

 

[PeroK]

PS you need to be careful about which point you are considering AM.

 

[Andy Resnick]

Well, if you jump and create any motion that's an increase in kinetic energy of the system. So, there can't be zero kinetic energy.

Yeah, I think that's the problem. When I add an energy source term, I can then solve all three expressions. For example, I now have

ω_disk = - 2 m (m + M)/(3 m^2 + 2 m M + M^2)V_you/R, where the negative sign indicates direction, and from the energy equation:

E_in=1/2 m (1 + m (1/M + 2 M/(3 m^2 + 2 m M + M^2)))

The answers seem a little 'ugly', I'll try making some substitutions like m = M or M = αm and see if that's 'cleaner'...

Edit: yep, got it. Thanks! Problem Solved!

 

[PeroK]

Yeah, I think that's the problem. When I add an energy source term, I can then solve all three expressions. For example, I now have

ω_disk = - (2m²(m+M)/(m²M+2mM²+3M³) V_you/R, where the negative sign indicates direction, and from the energy equation:

E_in=m²V_you²(1/m + 1/M + 1/(m+M))

The angular frequency seems a little 'ugly', I may have made some errors, but the energy input seems reasonable...?

That's not what I get.

Note that if you let $M \rightarrow \infty$, then the KE of the disk will become negligible, so you should get $E \rightarrow \frac12 mv^2$. Whereas, in the limit you would get $E \rightarrow mv^2$.

I wondered why you mentioned parallel axis, as taking AM about the centre of the disk seems logical. In any case, conservation of AM should give:

$I \omega = mvR$

 

[Orodruin]

The ”collision” is inelastic so you cannot apply energy conservation. Doing so leads to an overconstrained system, as you have noticed.

You should be fine by just applying conservation of linear and angular momentum.

 

[Andy Resnick]

That's not what I get.

Note that if you let $M \rightarrow \infty$, then the KE of the disk will become negligible, so you should get $E \rightarrow \frac12 mv^2$. Whereas, in the limit you would get $E \rightarrow mv^2$.

I wondered why you mentioned parallel axis, as taking AM about the centre of the disk seems logical. In any case, conservation of AM should give:

$I \omega = mvR$

The way I understand it, the center of rotation is not at the disk center, but located at the center of mass of the system. This distance 'd' measured from the disk edge is d = M/(m+M) R. But maybe you are right... my head hurts.

I put the (I think) correct expressions up in Post #6... 3 cheers for Mathematica!

 

[PeroK]

The way I understand it, the center of rotation is not at the disk center, but located at the center of mass of the system. This distance 'd' measured from the disk edge is d = m/(m+M) R. But maybe you are right... my head hurts.

I put the (I think) correct expressions up in Post #6... 3 cheers for Mathematica!

I would just consider AM about a point on the direction of motion of the centre of the disk. That's the simplest.

 

[haruspex]

The ”collision” is inelastic so you cannot apply energy conservation.

We don't know that. You could set it up as a particle projected by a spring. As already noted, energy conservation cannot be used because there is an unknown input of energy.

the center of rotation is not at the disk center, but located at the center of mass of the system

The centre of rotation of the disk will not be at its centre, but neither will it be at the mass centre of the disc+jumper system.
It is rarely necessary to find the common mass centre or centre of rotation in such problems.
It is simpler to consider the disk's motion as the sum of a linear motion and a rotation about its centre.

 

[Orodruin]

We don't know that. You could set it up as a particle projected by a spring. As already noted, energy conservation cannot be used because there is an unknown input of energy.

Hence ”collision” in quotes.

 

[Andy Resnick]

The centre of rotation of the disk will not [...] be at the mass centre of the disc+jumper system.

Why not?

 

[PeroK]

Why not?

The motion of the disk can be decomposed into the motion of its centre of mass (linear in this case) and a rotation about its centre of mass. This is the simplest way to consider this problem.

You can, however, take any other point on the disk and decompose the motion into the motion of that point and a rotation about that point. But, the motion of any other point is not linear (unless the disk is not rotating at all).

You may be imagining the disk moving so that a point other than the centre of mass is moving linearly, but that requires an external force.

 

[haruspex]

Why not?

Let the initial positions of the person and the disk centre be P and D. The common mass centre lies between them. The instantaneous centre of rotation of the disk will lie on the opposite side of D from P, no?

 

[Andy Resnick]

Let the initial positions of the person and the disk centre be P and D. The common mass centre lies between them. The instantaneous centre of rotation of the disk will lie on the opposite side of D from P, no?

I don't understand the logic. Compound motion can be decomposed into translation of the center of mass and rotation about the center of mass.

 

[Andy Resnick]

The motion of the disk can be decomposed into the motion of its centre of mass (linear in this case) and a rotation about its centre of mass. This is the simplest way to consider this problem.

Maybe this is the source of my confusion: the disk motion can't 'honestly' be separated from motion of the system, at least when using the conservation laws (e.g. Ef = Ei + Ein - Eout). The correct solution should consider the motion of the system center of mass and rotation about the system center of mass.

Sure, if M>>m, then we can separately consider the disk motion. But that's a limiting case.

 

[jbriggs444]

The correct solution should consider the motion of the system center of mass and rotation about the system center of mass.

What entities are part of the "system" and what entities are considered to be outside it?

1. If the system contains both disk and jumper then no you cannot decompose the motion of the system into motion of the center of mass and rotation about the center of mass. Because this "system" is not rigid.

2. If a sub-system contains disk alone then yes, we can consider translation of the disk and rotation of the disk about its center. But for some reason you are objecting to this as "dishonest".

 

[PeroK]

Maybe this is the source of my confusion: the disk motion can't 'honestly' be separated from motion of the system, at least when using the conservation laws (e.g. Ef = Ei + Ein - Eout). The correct solution should consider the motion of the system center of mass and rotation about the system center of mass.

Sure, if M>>m, then we can separately consider the disk motion. But that's a limiting case.

Once the disk and the jumper are separated, they move with the linear and, in the case of the disk, rotational angular momentum they gained during the joint impulse.

The system cannot be rotating about its centre of mass as there were no external forces.

You seem to be confusing rotation with angular momentum.

 

[haruspex]

and rotation about the system center of mass.

Let's make this easy and do the algebra.
Let the original position of the disk centre be O, that of mass m be at the top of the picture, and the common mass centre distance x above O. Set the moment of inertia of the disk about its centre to be I.
Mass m leaves to the left at speed u, the disk centre moves to the right at speed v and rotates clockwise at rate ω. The instantaneous centre of rotation of the disk is y above O.
By linear momentum mu=Mv
By angular momentum about O: muR=Iω
For the instantaneous centre of rotation: yω+v=0
Solving: $y=-\frac I{MR}$, $x=\frac{mR}{M+m}$
If we assume I=½MR²:
y=-R/2.
In particular, as remarked, the common mass centre and instantaneous centre of rotation of the disk are on opposite sides of O. In the special case of a uniform disk and m=M, the two are equidistant from O.

 

[Andy Resnick]

What entities are part of the "system" and what entities are considered to be outside it?

1. If the system contains both disk and jumper then no you cannot decompose the motion of the system into motion of the center of mass and rotation about the center of mass. Because this "system" is not rigid.

I don't understand this- for example, a common homework problem is an exploding firework, where the center of mass is said to move along the original projectile trajectory (air resistance is neglected).

 

[Andy Resnick]

Let's make this easy and do the algebra.

If we assume I=½MR²:

But can you assume that when you already said the disk rotates about a point 'y' away from the center?

 

[jbriggs444]

I don't understand this- for example, a common homework problem is an exploding firework, where the center of mass is said to move along the original projectile trajectory (air resistance is neglected).

An exploding firework cannot be completely modeled as a system that merely translates and rotates. It also expands.

 

[PeroK]

But can you assume that when you already said the disk rotates about a point 'y' away from the center?

Here is a complete analysis. The jumper moves in a straight line with speed $v$. The centre of mass of the disk moves in a straight line in the opposite direction with speed $V = \frac{mv}{M}$.

The disk rotates about its centre of mass with angular speed $\omega$. Taking angular momentum about the initial centre of the disk, we see that the linear motion of the disk contributes zero AM about this point. That leaves:

$I\omega = mvR$

 

[haruspex]

But can you assume that when you already said the disk rotates about a point 'y' away from the center?

I took moments about O, the initial position of the disk centre, so I must use the moment of inertia about that point.

You could consider a much simpler system, if it helps: replace the disk with two point masses joined by a light rod.