Rigid body rotation problem: Mass on a cord spinning up a wheel on an axle

[archaic]

Homework Statement

A wheel of radius R, mass M, and moment of inertia I is mounted on an axle supported in fixed bearings. A light flexible cord is wrapped around the rim of the wheel and carries a body of mass m. Friction in the bearings can be neglected. Find the final velocity.

Relevant Equations

$$\vec\tau=I\vec\alpha$$

(I know how to solve the problem, that's not what I am looking for.)
I have a problem with how I ought to understand the moment of inertia. The only torque I see applicable on the wheel is that of the tension, and so I think that $I$ should be $m_{\text{point}}R^2$, without including all the other particles. I know that it doesn't make sense to have a mass for a point since I am given a rigid body, but I frankly can't see how we are considering all the other particles as having $\vec T$ applied to them.
The derivation of the moment of inertia that I am aware of is by considering the torque at each particle then summing them (or by integration for continuous stuff).
$$\sum_i F_i\sin\theta_ir_i=\tau_{\text{total}}=\sum_i m_ir^2_i\alpha$$
I have considered that maybe because all particles have the same angular acceleration, then the force is "propagated", but that's not true since $m_ir_i\alpha\neq m_jr_j\alpha$ even if the masses are the same.
Thank you for your time.

 

[Doc Al]

I have a problem with how I ought to understand the moment of inertia. The only torque I see applicable on the wheel is that of the tension, and so I think that $I$ should be $m_{\text{point}}R^2$, without including all the other particles.

Yes, the only torque on the wheel is from the tension. To apply Newton's 2nd law to the wheel, wouldn't you use the moment of inertia of the wheel? I'm not seeing how the hanging mass enters into a calculation of the wheel's moment of inertia.

Perhaps I'm missing your point?

 

[hutchphd]

I think the issue is what is meant by a rigid body. The rigid body comprises internal forces that are very very large. Therefore the velocities and positions of various parts can be considered rigidly constrained

 

[wrobel]

. Find the final velocity.

strange

 

[Doc Al]

I know that it doesn't make sense to have a mass for a point since I am given a rigid body, but I frankly can't see how we are considering all the other particles as having $\vec T$ applied to them.

Think of the external torque as applied to the entire rigid body, not just to a point mass on its rim (I think that's what you meant, not the hanging mass as I first thought.) As @hutchphd states, there are internal forces constraining the wheel to move as a unit.

 

[Doc Al]

strange

I assume part of the problem statement has been left out.

 

[archaic]

Perhaps I'm missing your point?

It seems like you got it in your second post. :)

Think of the external torque as applied to the entire rigid body

But here the tension is at the rim, or at least how I see it.
Revising my thought once more, I have changed my understanding a bit, but it's still not such as I'd do the problem the way I am supposed to.
Here's how I see it. The tension would make the body rotate, and only the particles on the edge of a disk of some radius would feel the same force:
Consider the particles on the edge of the wheel's outermost disk, each particle can be described by $\vec F+\vec f=\vec F_{\text{net}}=m\vec a=mr\alpha\vec u_t-\frac{v^2}{r}\frac{\vec r}{r}$, where $f$ represents the sum of all internal forces on it, while the other is for external ones. Their individual torque is then $mr^2\alpha(\frac{\vec r}{r}\times\vec u_t)$, i.e each particle.
Since we only have tension as the tangential force, then $mr\alpha=T$, even particles whose velocity is not tangential to the tension but are at the same level with the outermost particles have that same tangential force since they share the angular acceleration and the radius (I am assuming uniform mass distribution). From this alone you have $2\pi r$(="number of particles") torques due to tension to consider!
...
zzZ I forgot the internal forces, they can also have tangential components.

I basically am looking for a way, a derivation, to understand how tension can be considered as the external force to be considered for the external torque.

 

[archaic]

I assume part of the problem statement has been left out.

strange

The bloc falls down $y$ units.

 

[hutchphd]

The point about a solid is that all the strong intermolecular interactions have conspired to be in their "happy place". Any attempt to move any piece in any way will result in only very small displacements which will propagate at the speed of sound through the solid and they will be negated. Unless you do something very violent (i.e. fast and large) the stuff is rigid. Whatever is necessary to accomplish this will be supplied by the huge intermolecular forces.

 

[Doc Al]

Forget the rotational issues with the wheel for a moment. Do you have the same issue with, say, pushing a rigid block of mass M along a frictionless surface? After all, you are only pushing on the outer surface, yet the block moves as a whole. (To a very good approximation, if it to be considered a "rigid" object.)

 

[etotheipi]

Forget the rotational issues with the wheel for a moment. Do you have the same issue with, say, pushing a rigid block of mass M along a frictionless surface? After all, you are only pushing on the outer surface, yet the block moves as a whole. (To a very good approximation, if it to be considered a "rigid" object.)

I think this is the crux, as in what force acts on each particle in the rigid body when a force $\vec{F}$ is applied to one side of a rigid body. I couldn't tell you, but it has to be the case that the sum of the forces on all of the particles due to this action must add up to $\vec{F}$ so we need not worry about the split when considering the rigid body as a whole system.

In the context of the problem, the total torque of internal torques has to be zero so all that remains to be shown is that the sum of the external torques due to the cord on all of the particles in the rigid body equals the torque of the cord on the rigid body as a whole. And I'm guessing this follows from the same sort of logic.

I could be totally off, if so please do ignore what I've said!

 

[archaic]

The point about a solid is that all the strong intermolecular interactions have conspired to be in their "happy place". Any attempt to move any piece in any way will result in only very small displacements which will propagate at the speed of sound through the solid and they will be negated. Unless you do something very violent (i.e. fast and large) the stuff is rigid. Whatever is necessary to accomplish this will be supplied by the huge intermolecular forces.

Forget the rotational issues with the wheel for a moment. Do you have the same issue with, say, pushing a rigid block of mass M along a frictionless surface? After all, you are only pushing on the outer surface, yet the block moves as a whole. (To a very good approximation, if it to be considered a "rigid" object.)

Well, no, but this is different. In a rotating body, even though the angular acceleration is the same, I don't see how the force is the same since two points at different radii will move in the same time different angular displacements.

 

[archaic]

all that remains to be shown is that the sum of the external torques due to the cord on all of the particles in the rigid body equals the torque of the cord on the rigid body as a whole.

Yes, this is what's annoying me. I don't see how it adds up to that.

 

[etotheipi]

Yes, this is what's annoying me. I don't see how it adds up to that.

I see what you mean, I wonder if this line of reasoning gets anywhere. In the linear case, the total external force equals the rate of change of the total momentum

$\vec{F}_{ext} = \frac{d\vec{P}}{dt}$

The total rate of change in momentum equals the sum of the rate of change of the individual momenta,

$\vec{F}_{ext} = \sum_i \frac{dp_{i}}{dt} = \sum_i \vec{f}_i$

where $\vec{f}_i$ are the resultant forces on each of the particles. The internal forces will also cancel out in the summation, and we can conclude that the total external force equals the sum of the external forces on each individual particle.

I believe the same might hold true if we replace $\vec{F}$ and $\vec{f}_{i}$ with $\vec{\tau}$ and $\vec{\tau}_{i}$ and also $\vec{P}$ and $\vec{p_{i}}$ with $\vec{L}$ and $\vec{L_{i}}$, nothing is jumping out at me to suggest that it wouldn't.

So if that constitutes a proof that the total external torque equals the sum of the external torques on each particle, that appears to make the maths work out. This might still not satisfy you since I think you've been pondering this much more deeply than I have but hopefully it's a start!

 

[Doc Al]

Well, no, but this is different. In a rotating body, even though the angular acceleration is the same, I don't see how the force is the same since two points at different radii will move in the same time different angular displacements.

The net force on any particular element of the wheel is not the same.

I'll make a few comments, essentially rephrasing what I think @etotheipi was getting at above.

While the rope exerts a net torque $TR$ on the wheel (where $T$ and $R$ are the tension in the rope and radius of the wheel), the net force on those bits of the wheel in contact with the rope is not the tension in the rope -- other parts of the wheel exert constraint forces to keep the wheel in one piece. Since we assume the wheel manages to keep its shape, we can deduce that the net torque on any piece of the wheel $m_i$ must be such to maintain the same angular acceleration $\alpha$. Since each small mass element has a moment of inertial of $I_i = m_ir^2$, that means the net torque on a given piece must be $I_i\alpha$. And that angular acceleration must satisfy $TR = I\alpha$, where $I$ is the moment of inertia of the entire wheel.

That's why I suggested looking at the simpler case of pushing a block along a frictionless surface. The force you exert is the net force on the block as a whole, but any piece of the block also experiences constraint forces from its neighbors.

Not sure if this helps or gets at your issue, but worth a shot.

 

[etotheipi]

While the rope exerts a net torque $TR$ on the wheel (where $T$ and $R$ are the tension in the rope and radius of the wheel), the net force on those bits of the wheel in contact with the rope is not the tension in the rope -- other parts of the wheel exert constraint forces to keep the wheel in one piece. Since we assume the wheel manages to keep its shape, we can deduce that the net torque on any piece of the wheel $m_i$ must be such to maintain the same angular acceleration $\alpha$. Since each small mass element has a moment of inertial of $I_i = m_ir^2$, that means the net torque on a given piece must be $I_i\alpha$. And that angular acceleration must satisfy $TR = I\alpha$, where $I$ is the moment of inertia of the entire wheel.

That's a good explanation, each mass element is acted upon by an external torque $\vec{{\tau_{t}}}_{i}$ due to tension and an internal torque due to constraint forces, $\vec{{\tau_{c}}}_{i}$, and we have

$\vec{{\tau_{t}}}_{i} + \vec{{\tau_{c}}}_{i} = m_{i}{r_{i}}^{2} \alpha$

And the other constraint is that $\sum_{i} \vec{{\tau_{t}}}_{i} = \vec{\Gamma}$ if $\vec{\Gamma}$ is the total torque due to the tension force.

At least that's how I think it fits together.