# Rotation of a rigid body

• Klaudia
In summary: The disk rotates about its centre2. The disk rotates about the point opposite to where the rope leaves the disk.3. The disk does not rotate because there is only one force acting on it.4. The inertia of the disk provides "a restraint" which allows the single force to cause rotation.

#### Klaudia

Homework Statement
A plane, homogeneous disk may slide on a frictionless horizontal table. A
light cord is fastened to the disk and wrapped many times around the rim of
the disk. The cord goes without friction through a small eyelet at the edge
of the table and is connected to a mass m (see the figure). The mass of the
disk is M and its radius is R. The system is started from rest and the cord
is assumed to be taut throughout the motion. Ignore the mass of the cord.

(1) Find the velocity v of the center of mass of the disk as a function of time
(before the disk hits the eyelet or the mass m hits the floor).
Relevant Equations
N = dL/dt

mgR = d(mvR + MvR + ½M(R^2)v/R)/dt
mgR = ma + Ma + ½Ma
mg = a(m + 3/2M)
v = mgt / (m+3/2M)

My answer is incorrect. The right answer is v = mgt/(3m+M), but I have no idea what I'm doing wrong.

Klaudia said:
mgR = d(mvR + MvR + ½M(R^2)v/R)/dt
What axis are you taking moments about that results in mvR+MvR?
Are you supposing m has velocity v?

If I will consider an axis passing through the center of mass, should I take into account only the momentum derived from rotation: Iω?
I assumed the velocity of m and M will be the same. If not, how should I calculate the velocity of m? I have tried to use the conservation of energy, but I don't know how to deal with calculation of potential energy of both masses.

Klaudia said:
If I will consider an axis passing through the center of mass, should I take into account only the momentum derived from rotation: Iω?
The linear velocity of the disc's centre has no moment about that centre.
Klaudia said:
I assumed the velocity of m and M will be the same. If not, how should I calculate the velocity of m? I have tried to use the conservation of energy, but I don't know how to deal with calculation of potential energy of both masses.
Since the disc is rotating, the velocity of the mass centre of the disc will not be the same as the velocity of the suspended mass. What will the relationship be?

Oh, and the disc's angular acceleration is not d(vR)/dt.

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I would write 2 dynamical equations

energy conservation and the angular momentum about the eyelet conservation

wrobel said:
I would write 2 dynamical equations

energy conservation and the angular momentum about the eyelet conservation
Sure, but angular momentum about the disc centre, coupled with linear acceleration, is fine too. All the ways I can find require two equations regarding torque/force/energy, plus one to relate the angular and linear accelerations.

Lnewqban
I just proposed how to write equations without reactions of constraints,
and yes kinematic equation is necessary anyway

haruspex said:
Sure, but angular momentum about the disc centre, coupled with linear acceleration, is fine too. All the ways I can find require two equations regarding torque/force/energy, plus one to relate the angular and linear accelerations.
Only if you have time for this questions, haruspex:
What the relevant equation N = dL/dt stands for?
Not having tried any calculation yet, it seems to me that the claimed correct answer v = mgt/(3m+M) should include R in some way.
Thank you.

haruspex said:
All the ways I can find require two equations
two degrees of freedom -- two dynamical equations

I think this can be solved by using typical Newtonian mechanics reasoning, making two equations for the movement of the two CMs, and one equation for the rotation of the disk. You ll end up calculating the tension on the string too this way along with the two linear accelerations and the angular acceleration.

The tricky part here is that the equation ##a=\alpha R## holds between the linear acceleration of the mass m and the angular acceleration of the disk.

Lnewqban
Delta2 said:
The tricky part here is that the equation holds between the linear acceleration of the mass m and the angular acceleration of the disk.
I think that is wrong

Delta2
wrobel said:
I think that is wrong
If the string unwinds by x around the disk, then the disk rotates by angle ##\phi=\frac{x}{R}## and the mass m goes down by x as well. @wrobel you got any disagreement with the above reasoning?

you do not take into account that the center of the disk moves

wrobel said:
you do not take into account that the center of the disk moves
I see now i think you are right after all, thanks for the correction.

is the correct equation $$x_m=\phi R+x_{disk}$$
then?

something like that

Delta2
I have been struggling with this as a recreational, not homework problem.

Q1. Does the disk rotate?

Q2. If so, about which axis does it disk rotate?

Q3. And why that axis?

I can make several arguments but cannot see which is correct.

1. The disk rotates about its centre

2. The disk rotates about the point opposite to where the rope leaves the disk.

I prefer this argument. It is similar to the wheel on a car which, when viewed from a frame attached to the ground (as we have here), rotates about its point of contact with the ground. If there was friction then this is the intuitive answer suggesting that as I reduce friction to zero it will still rotate about that point.

3. The disk does not rotate because there is only one force acting on it.

4. The inertia of the disk provides "a restraint" which allows the single force to cause rotation. But, if so, how is the axis of rotation decided? Is it that about which the moment of inertia is the least - ie the centre of the disk?

I can draw a box enclosing the disk and the rope is then applying the only external force to the system.

I tried simplifying the problem to a stick on a table with friction which will then rotate about the far end of the stick and is how I came to argument 2. But why would it rotate if there was no friction - why isn't it just pulled along without rotation? Replacing the rope with a little rocket didn't help although, on second thoughts, a rocket attached off centre to a free_floating_in_space object would presumably cause it to rotate about its CofG as well as translate.

Any guidance will be welcomed. I trust that doing the calculation will be comparatively easy once I can decide what is happening.

Frodo said:
I trust that doing the calculation will be comparatively easy once I can decide what is happening.
That is a common mistake. Just very trivial problems allow you to understand what is happening directly, without analyzing equations of motion. As a rule you understand what is happening exactly by means of the equations

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Frodo said:
I have been struggling with this as a recreational, not homework problem.

Q1. Does the disk rotate?

Q2. If so, about which axis does it disk rotate?

Q3. And why that axis?
...
Any guidance will be welcomed. I trust that doing the calculation will be comparatively easy once I can decide what is happening.
This is how I see it (which can be wrong, of course):

Q1. Does the disk rotate?
Yes, the disk rotates in an accelerated way because a force is applied constantly off its center of mass.

Q2. If so, about which axis does the disk rotate?
The disk should rotate around its center of mass.
We could consider a flat disk that is free to slide and rotate over a friction free perfectly horizontal surface to behave like a body in zero-gravity conditions, at least on that horizontal plane.

If you hit or push such an object at a point and with a direction that are off its center of mass, it will naturally rotate as a result of the applied torque respect to that center.

Q3. And why that axis?
Because rotating around that axis naturally consumes the least amount of kinetic energy, by self-adopting the minimum possible moment of inertia.

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wrobel said:
two degrees of freedom -- two dynamical equations
There are sure to be two dynamical equations available, but sometimes you do not need all the available equations to answer the particular question. E.g. taking moments about a well chosen axis.

Lnewqban said:
Q2. If so, about which axis does the disk rotate?
The disk should rotate around its center of mass.
We could consider a flat disk that is free to slide and rotate over a friction free perfectly horizontal surface to behave like a body in zero-gravity conditions, at least on that horizontal plane.
It depends what you mean by rotating about an axis.
You could be asking what point is the instantaneous centre of rotation. That will not be the mass centre in this case. But it is generally simpler to think of rigid body motion as the sum of the motion of the mass centre and a rotation about that centree.

jbriggs444 and Lnewqban
Lnewqban said:
rotating around that axis naturally consumes the least amount of kinetic energy, by self-adopting the minimum possible moment of inertia.
There is no such principle (at least, not without certain provisos), and it is not generally true.
Consider a force F applied normally at one end of a rod mass m. The linear acceleration of the rod is F/m, which determines the KE gained in that mode. To minimise the total gain in KE it would not rotate.
However, if we specify the acceleration of the end of the rod instead of the force applied, its motion will minimise uptake of KE.

haruspex said:
There is no such principle (at least, not without certain provisos), and it is not generally true.
Consider a force F applied normally at one end of a rod mass m. The linear acceleration of the rod is F/m, which determines the KE gained in that mode. To minimise the total gain in KE it would not rotate.
However, if we specify the acceleration of the end of the rod instead of the force applied, its motion will minimise uptake of KE.
I got lost in this one.

Delta2
Thank you for your responses as the fog is beginning to lift.

I am still surprised how difficult this problem is for me which is because it uses techniques I haven't been taught so it's back to school for me. Can anyone recommend something I should read or will searching on dynamical equations find what I need?

I shared it with another engineer, a maths graduate and a physics teacher and all were as confused as me.

Delta2
by the way , are there any countries where such problems are considered at high school ?

Frodo said:
shared it with another engineer, a maths graduate and a physics teacher and all were as confused as me.
that is strange this is a very simple problem ,not of high school level as I understand, but very simple

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Frodo said:
1. The disk rotates about its centre

2. The disk rotates about the point opposite to where the rope leaves the disk.

I prefer this argument. It is similar to the wheel on a car which, when viewed from a frame attached to the ground (as we have here), rotates about its point of contact with the ground. If there was friction then this is the intuitive answer suggesting that as I reduce friction to zero it will still rotate about that point.

3. The disk does not rotate because there is only one force acting on it.

4. The inertia of the disk provides "a restraint" which allows the single force to cause rotation. But, if so, how is the axis of rotation decided? Is it that about which the moment of inertia is the least - ie the centre of the disk?
I assume here you are referring to the point that is stationary, i.e. the instantaneous centre of rotation.
It is hard to guess this in advance. As I posted, it is best to consider the disc's motion as the sum of a linear acceleration of the mass centre and an angular acceleration about that centre.
If the tension in the string is T then, that being the only force acting horizontally on the disc, the linear acceleration is a=T/M.
The torque the string exerts about the disc's centre is RT, so the angular acceleration is given by ##I\alpha=RT##. With ##I=\frac 12 MR^2##, we have ##M\alpha R=2T##.
We can get the corresponding velocities at time t by multiplying each by t.
Suppose the instantaneous centre of rotation is distance X from the disc's centre, distance X+R from where the string leaves the disc. We have ##X\alpha=a##, so ##2TX=MX\alpha R=MaR=TR##, whence X=R/2.

Frodo said:
shared it with another engineer, a maths graduate and a physics teacher and all were as confused as me.
We are electronic and electrical engineers respectively so mechanics was not a large part of the syllabus.
wrobel said:
that is strange this is a very simple problem, not of high school level as I understand, but very simple
The only methodology we have been taught is (it was 50 years ago!)
1. work out what is happening and how it moves by examination
2. once you know what happens calculate the values.

We have never been taught the approach of setting up the equations and calculating the motion; nor Lagrangians, Hamiltonians or the principle of least action. I am now reading Equations of motion and, in a separate project, looking at least action as it is used in quantum theory.

Sorry for offtop. Perhaps it would be of some interest. I remember such a problem. On a smooth horizontal table there is a homogeneous disk of radius ##r## and mass ##M##. The disk is pivoted by its center ##O## and can rotate freely about the point ##O##. A weightless string is wrapped around the disk such that initially the tail of the string has a length ##b##. A particle of mass ##m## is attached to the end of the tail. Initially all the system is at rest and suddenly the mass ##m## is pushed with initial velocity ##\boldsymbol u## perpendicular to the tail.
Question: will the particle reach the disk?

Thank you. I will have a go at that by applying the equation(s).

Frodo said:
1. work out what is happening and how it moves by examination
2. once you know what happens calculate the values.
That's reasonable, but I would change it to "how it might move", and keep an open mind about the possibilities. As I showed in post #26, you cannot determine the instantaneous centre of rotation by inspection.

The more I read about Newton and what he did the more impressed I am. He had the critical insight on this problem as Definition III in Principia.
III in Principia.

Lnewqban