Inclined plane and 2 wheels problem

[TheColector]

Suppose we have an inclined plane and 2 wheels. Both wheels have the same mass M but different radius R.
So the only difference is that one is bigger than another. We put both wheels on the top of the inclined plane. They are in the same position, and then we let them go. I know that the wheel with bigger radius will reach the bottom of the inclined plane faster than the smaller one. It also accelerates faster.

Here I need a little of your help. The thing I don't know/understand is why does the bigger radius wheel accelerates faster, why does it reach the bottom of the inclined plane faster ? Should both wheels reach the same max velocity ?
Now tell me if I'm wrong. In my opinion both wheels have the same force F1 where F1 = mg * sinα - force which pulls down.
By the definition of Torque which is

. we have the same F1 force and different Radiuses.
In conclusion the bigger wheel has larger Torque value than small wheel, and both wheels need the same force to move but bigger radius means larger Torque value, so it's easier to gain acceleration and starts moving faster. It's one of my thoughts. Sorry for long post, I'm new here, hope you'll understand

 

[HallsofIvy]

Have you looked at the "moment of inertia" of the two wheels? For circular motion the formula corresponding to "F= ma" is T= I\alpha where T is the torque, \alpha is the angular acceleration (second derivative of angle with respect to time), and I is the moment of inertial. For a "wheel" ("thin circular hoop") of radius r and mass m the moment of inertia, about its axis, is \frac{mr^2}{2}.
(https://en.wikipedia.org/wiki/List_of_moments_of_inertia).

The moment of inertia increases as the square of the radius.

 

[Doc Al]

I know that the wheel with bigger radius will reach the bottom of the inclined plane faster than the smaller one. It also accelerates faster.

How do you know this?

Model the wheel as a disk or a hoop (your choice) and figure out what the acceleration depends on, assuming rolling without slipping.

 

[Doc Al]

For a "wheel" ("thin circular hoop") of radius r and mass m the moment of inertia, about its axis, is \frac{mr^2}{2}.

That's the moment of inertia of a uniform disk; that of a hoop would be mr^2.

 

[nasu]

How do you know that the bigger one arrives first?

Edit.
Sorry, I did not refresh in a while. Asked same question as Doc Al. :)

 

[TheColector]

I know this because I perceived that during experiment.

 

[nasu]

Both wheels have the same geometry? Are they full cylinders?

 

[Doc Al]

I know this because I perceived that during experiment.

The first thing to do is figure out (using Newton's laws) the behavior you would expect.

 

[TheColector]

@nasu I mean smth like 2 bike wheels

 

[nasu]

You mean with spokes?

 

[TheColector]

T=Iα, => α= T/I => α= F*R/M*R^2 , so the angular acceleration is inversely proportional to the Radius, it means smaller acceleration with bigger radius. What's wrong ?

 

[nasu]

I is not just mr^2. It depends on the specific distribution of mass. If your objects are like bicycle wheels it will be quite tricky to find I theoretically.
The shape is essential.

 

[TheColector]

Sure but we have here as simplest situation as can only be, equal distribution of mass, perfecr circle etc. So tell me what is the answer ?

 

[nasu]

You said you did an experiment, didn't you? And you said your objects are like bicycle wheels.
So what are you after? An ideal situation with some simple geometrical shapes or your experiment with real objects? Bicycle wheels are not a simple case.
You have to decide what problem are you solving before starting to look for solution.

 

[TheColector]

@nasu I see your point. My bad, but I'd love to know what couses the bigger wheel(with larger radius) accelerating faster, reaching the bottom faster. Let's assume the simple situation for the simplicity of calculations.

 

[Doc Al]

T=Iα, => α= T/I => α= F*R/M*R^2 , so the angular acceleration is inversely proportional to the Radius, it means smaller acceleration with bigger radius. What's wrong ?

1) How is angular acceleration related to linear acceleration?
2) Don't assume that the torque-creating force (F, the friction) is the same.

 

[TheColector]

α=a/R and then a= F*R\M, I think it's the answer, Relation to the radius is correct. I thought about it at first but I got distracted or smth and rejected this option. I get it know I think(If it's the answer). Thanks to all of you

 

[nasu]

@nasu I see your point. My bad, but I'd love to know what couses the bigger wheel(with larger radius) accelerating faster, reaching the bottom faster. Let's assume the simple situation for the simplicity of calculations.

Well, in the simplest case, of uniform cylindrical (or disk shaped) objects, they should get down with the same speed, at the same time.
The larger torque on the big one is compensated by a larger moment of inertia. You need to solve two equations, one for linear and one for angular acceleration to see this. And use the relationship between angular and linear accelerations.

But you claim you have seen something else so this means your experiment does not satisfy the requirements of the simplest case. And you said that they are like bicycle wheels.