Two wheels rotating connected by a rod

[fedecolo]

Homework Statement

A cart has two cylindrical wheels connected by a weightless horizontal rod using weightless spokes and friction- less axis as shown in the figure. Each of the wheels is made of a homogeneous disc of radius R, and has a cylidrical hole of radius R/2 drilled coaxially at the distance R/3 from the centre of the wheel. The wheels are turned so that the holes point towards each other, and the cart is put into motion on a horizontal floor. What is the critical speed v by which the wheels start jumping?
Answer: $v=3\sqrt{gR}$

Homework Equations

The Attempt at a Solution

I have no idea where to start! I have to find the centre of mass of each wheel. Then I must find the forces acting on it (only gravitational attraction) and then?

 

[BvU]

I have no idea where to start

Not good enough according to the PF guidelines !

I have to find the centre of mass of each wheel

You could do that. Or you can work with R/2 disks with a negative mass ...

Then I must find the forces acting on it (only gravitational attraction)

There's the normal force too! Once that goes zero at some moment in a revolution the jumping isn't far away

PS is the connecting rod essential for answering the question / Or is it just to give the contraption a semblance of realism ?

 

[fedecolo]

There's the normal force too! Once that goes zero at some moment in a revolution the jumping isn't far away

You gave me an idea! It's correct to think (in the reference frame of the wheel) that when centrifugal force equals gravitational force, then normal force goes zero and the wheel jumps?

PS is the connecting rod essential for answering the question / Or is it just to give the contraption a semblance of realism ?

I don't know. I think that I have to use this (but I have no idea how to)

 

[BvU]

My intake is to forget the rod and look at the vertical acceleration of the c.o.m. because of the rolling motion with speed $v$. If that offsets g you're at the critical speed.

 

[fedecolo]

My intake is to forget the rod and look at the vertical acceleration of the c.o.m. because of the rolling motion with speed $v$. If that offsets g you're at the critical speed.

After some counts I came to this: the c.o.m. is $\frac{R}{9}$ distant from the centre of the cylinder, its angular speed $\omega$ is equal to the angular speed of the wheel, so $v_{c.o.m}=\frac{v}{9}$.
How can I find the vertical acceleration of the c.o.m.?

 

[BvU]

so $v_{c.o.m}=\frac{v}{9}$.

Really ? So the center of mass stays behind and after a short while it is outside of the wheel ?

 

[Dr.D]

Are both disks and the rod all in the same plane, or is this a 3D system with the rod normal to the two disks which are parallel to each other (like a rail road truck)?

 

[haruspex]

Are both disks and the rod all in the same plane, or is this a 3D system with the rod normal to the two disks which are parallel to each other (like a rail road truck)?

I think the description is quite clear. Two parallel cylinders, axles linked by a horizontal rod normal to the axles. The complication is that the masses of the cylinders are not uniformly distributed, this being represented by holes drilled parallel to their axes (not, as the text says, coaxially).

 

[haruspex]

is it just to give the contraption a semblance of realism ?

I suggest it is so that the system would keep moving the same direction at any speed. The mass centre of the whole system has constant height. Tension and compression in the horizontal rod alternates: as the mass centre of the leading cylinder descends that cylinder pulls the other along.
But since, as you say, "jumping" would mean the normal force from the ground vanishes, it seems that the forces in the rod are not relevant for that aspect.

 

[haruspex]

It's correct to think (in the reference frame of the wheel) that when centrifugal force equals gravitational force, then normal force goes zero and the wheel jumps?

Yes, that's the way to go.

the c.o.m. is $\frac{R}{9}$ distant from the centre of the cylinder

Don't deal with the c.o.m., that's unnecessarily complicated. As BvU wrote, treat the hole as a cylinder of negative mass superimposed on a solid cylinder. The solid cylinder has no centrifugal/centripetal force.

 

[fedecolo]

Don't deal with the c.o.m., that's unnecessarily complicated. As BvU wrote, treat the hole as a cylinder of negative mass superimposed on a solid cylinder. The solid cylinder has no centrifugal/centripetal force.

How can I deal with this idea of superimposed cylinder? Because maybe it's useful to find the moment of inertia but I don't think I must use this to solve the problem.

 

[haruspex]

How can I deal with this idea of superimposed cylinder? Because maybe it's useful to find the moment of inertia but I don't think I must use this to solve the problem.

No, you don't need to worry about moments of inertia here, just centripetal force.
Think of the cylinder as consisting of a uniform solid cylinder with the hole as a cylinder of equal negative density superimposed on it.
The centripetal force required is the sum of the two centripetal forces. The solid cylinder has no acceleration about the axis, so needs no centripetal force. So the centripetal force needed is for the negative cylinder only.
Now, it might not be obvious, but it turns out that the centripetal force required for a solid body is the same as for a point mass at the same mass centre. So it is very easy to find the centripetal force.
For the cylinder to become airborne, the max downward centripetal force equals the weight of the system.

 

[fedecolo]

Now, it might not be obvious, but it turns out that the centripetal force required for a solid body is the same as for a point mass at the same mass centre. So it is very easy to find the centripetal force.

What is the radius to put in the centripetal formula? Is $\frac{R}{3}$ (but the answer is not correct) (or $\frac{R}{3} + \frac{R}{9}$

 

[BvU]

Why don't you do some work an write it down in a post ? If you derail, someone will help you back onto the right track.

 

[fedecolo]

Why don't you do some work an write it down in a post ? If you derail, someone will help you back onto the right track.

Ok I'm sorry. My work is this

No, I'm wrong. Instead of M I have to put $(M-m)$. So the final equation becomes $\frac{R^2}{4} \frac{3v^2}{R} \geq \frac{3}{4} R^2 g$
And I think that the radius for the centripetal force I'm taking is wrong.

 

[haruspex]

Ok I'm sorry. My work is thisView attachment 204213

No, I'm wrong. Instead of M I have to put $(M-m)$. So the final equation becomes $\frac{R^2}{4} \frac{3v^2}{R} \geq \frac{3}{4} R^2 g$
And I think that the radius for the centripetal force I'm taking is wrong.

Is that v the speed of the system or the instantaneous speed of the centre of a hole?

 

[fedecolo]

Is that v the speed of the system or the instantaneous speed of the centre of a hole?

I assumed v as the speed of the system.

 

[zwierz]

I would like to add a comment which does not have direct relation to the problem under consideration but it seems that it would also be useful here. For simplicity assume that a rigid body rotates about a fixed axis with constant angular velocity. What can one say about centrifugal forces relative to a body fixed frame? Very common wrong answer is as follows: the centrifugal force is applied to the center of mass.

 

[fedecolo]

I would like to add a comment which does not have direct relation to the problem under consideration but it seems that it would also be useful here. For simplicity assume that a rigid body rotates about a fixed axis with constant angular velocity. What can one say about centrifugal forces relative to a body fixed frame? Very common wrong answer is as follows: the centrifugal force is applied to the center of mass.

I can state that here the centre of rotation is not the centre of mass. Is it possible to use this fact to get the solution?

 

[haruspex]

I assumed v as the speed of the system.

But in your centripetal force calculation you seem to have taken it as the tangential speed of the centre of the hole relative to the system.

 

[fedecolo]

But in your centripetal force calculation you seem to have taken it as the tangential speed of the centre of the hole relative to the system.

Ok, but also as well is wrong[emoji24]

 

[haruspex]

Ok, but also as well is wrong[emoji24]View attachment 204260

If v is the velocity of the system, what is the tangential velocity of the centre of the hole relative to the system?
What is the radius of rotation of the centre of the hole?
Get those two things right at the same time and you will be there.

 

[zwierz]

-

 

[fedecolo]

If v is the velocity of the system, what is the tangential velocity of the centre of the hole relative to the system?
What is the radius of rotation of the centre of the hole?
Get those two things right at the same time and you will be there.

Maybe I got it: because $\omega$ is constant, $v_{hole}=\frac{v}{3}$, then the radius of rotation is R/3, so I have that $\frac{R^2}{4} \frac{\frac{v^2}{9}}{\frac{R}{3}} \geq \frac{3}{4} R^2 g$ that gives the solution. Thank you all!

 

[haruspex]

Maybe I got it: because $\omega$ is constant, $v_{hole}=\frac{v}{3}$, then the radius of rotation is R/3, so I have that $\frac{R^2}{4} \frac{\frac{v^2}{9}}{\frac{R}{3}} \geq \frac{3}{4} R^2 g$ that gives the solution. Thank you all!

That's it.
But there is a subtlety in that which might have you tripped you up, and is worth being aware of.
The velocity of the hole, v/3, you (correctly) used is its tangential velocity relative to the centre of the solid cylinder, not relative to the ground. Just wanted to make sure you realized that.

 

[fedecolo]

That's it.
But there is a subtlety in that which might have you tripped you up, and is worth being aware of.
The velocity of the hole, v/3, you (correctly) used is its tangential velocity relative to the centre of the solid cylinder, not relative to the ground. Just wanted to make sure you realized that.

Yes because the tangential velocity of a homogeneous wheel is 0 at its lowest point (in contact with the ground), isn't it? Is it the same for a wheel with an hole?

 

[haruspex]

Yes because the tangential velocity of a homogeneous wheel is 0 at its lowest point (in contact with the ground), isn't it? Is it the same for a wheel with an hole?

No, I think you are missing the point.
The cylinder wil start to jump when the hole is at its lowest. At that position, the hole's centre is moving backwards at v/3 relative to the system, so its velocity relative to the ground is 2v/3 forwards. But its centre of rotation is moving forward with the system, so for the centripetal force calculation it is the relative velocity of v/3 that must be used.