Solving Jumping Off a Disk Problem (d)

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In summary: Is this what you were looking for?Yes, that's what I was looking for! The angular frequency seems... a little high? Is this what you were looking for?Yes, that's what I was looking for!
  • #1
Andy Resnick
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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!
 
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  • #2
Andy Resnick said:

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?
 
  • #3
PeroK said:
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: Vdisk = -m/M Vyou, Vdisk is the velocity of the disk's center of mass.
Angular momentum: -Idiskωdisk = Iyou Vyou/d (I made the substitution ωyou=Vyou/d). The moments of inertia are obtained with the parallel axis theorem, and are functions of 'd'.

Energy: 0 = mVyou2+ M Vdisk2+ Idiskωdisk2 (*)

(*): 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?
 
  • #4
Andy Resnick said:
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: Vdisk = -m/M Vyou, Vdisk is the velocity of the disk's center of mass.
Angular momentum: -Idiskωdisk = Iyou Vyou/d (I made the substitution ωyou=Vyou/d). The moments of inertia are obtained with the parallel axis theorem, and are functions of 'd'.

Energy: 0 = mVyou2+ M Vdisk2+ Idiskωdisk2 (*)

(*): 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.
 
  • #5
PS you need to be careful about which point you are considering AM.
 
  • #6
PeroK said:
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

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

Ein=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!
 
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  • #7
Andy Resnick said:
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

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

Ein=m2Vyou2(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##
 
  • #8
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.
 
  • #9
PeroK said:
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!
 
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  • #10
Andy Resnick said:
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.
 
  • #11
Orodruin said:
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.
Andy Resnick said:
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.
 
  • #12
haruspex said:
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.
 
  • #13
haruspex said:
The centre of rotation of the disk will not [...] be at the mass centre of the disc+jumper system.

Why not?
 
  • #14
Andy Resnick said:
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.
 
  • #15
Andy Resnick said:
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?
 
  • #16
haruspex said:
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.
 
  • #17
PeroK said:
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.
 
  • #18
Andy Resnick said:
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".
 
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  • #19
Andy Resnick said:
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.
 
  • #20
Andy Resnick said:
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.
 
  • #21
jbriggs444 said:
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).
 
  • #22
haruspex said:
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?
 
  • #23
Andy Resnick said:
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.
 
  • #24
Andy Resnick said:
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##
 
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  • #25
Andy Resnick said:
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.
 
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Related to Solving Jumping Off a Disk Problem (d)

What is the "Solving Jumping Off a Disk Problem (d)"?

The "Solving Jumping Off a Disk Problem (d)" refers to a mathematical problem that involves finding the maximum distance a person can jump from a rotating disk, given the disk's radius and angular velocity.

Why is the "Solving Jumping Off a Disk Problem (d)" important?

This problem has practical applications in various fields such as physics, engineering, and sports. It also helps in understanding rotational motion and projectile motion.

What are the key factors to consider when solving the "Solving Jumping Off a Disk Problem (d)"?

The key factors to consider are the radius of the disk, the angular velocity of the disk, and the initial position and velocity of the person jumping off the disk.

What is the formula for solving the "Solving Jumping Off a Disk Problem (d)"?

The formula for solving this problem is d = R(1 - cosθ), where d is the maximum distance, R is the radius of the disk, and θ is the angular displacement of the disk.

What are the steps to solve the "Solving Jumping Off a Disk Problem (d)"?

The steps to solve this problem are:

  1. Determine the values of R, ω, and initial position and velocity of the person.
  2. Use the formula d = R(1 - cosθ) to calculate the maximum distance.
  3. To find the angular displacement θ, use the equation ω = dθ/dt.
  4. Substitute the value of ω in the equation ω = dθ/dt and solve for θ.
  5. Finally, substitute the value of θ in the formula d = R(1 - cosθ) to find the maximum distance d.

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