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# What is torque, really? Can it be determined for a point on a circle that is away from the center of the rotation axis by a radius “r”?

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Let's picture a Rolling without slipping Wheel, that constantly accelerates. radius of a wheel/circle is rr.

Basically any wheel of a vehicle is rotating around its fixed axis - AoR -(axis of rotation)

Most of the times wheels have this AoR perfectly in the center of the circle/wheel/cylinder. What I want to understand is how can we describe the torque of such a wheel?

Because as I understand we could go about finding torque for the center of the circle - point O.

I am not sure if it's possible to find the torque for the point P, that is point on a line tangent to the circle on the surface level.

Edited by Classical Physicist
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Here is the theory you require.

Make sure you get you friction force acting in the correct direction (your dashed one in this case)

Also the normal reaction (again dashed) is the force ating on the wheel.
The other mg at P acts on the ground, not the wheel.

The moment provided by the contact point P about the centre = tangential force due to friction x wheel radius.
Note the frictional force is only equal to μmg when the wheel is on the point of slipping.
At other times it is less than this.

All this is detailed in the text.

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Great!, I will look into it in a moment. But first that's what I did to find acceleration of this whole wheel, is it even partially correct?

whoops my bad, not sure what happened with formatting of ω either. - reuploaded

I'm Reading the chapter You've just sent right now

Edited by Classical Physicist
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Yes, as long as that point rotates interactively with the axis of a disk or that circle

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Okay so having all of these Drawn out,  how can I find acceleration a that is at the point c which is centre of mass  of a vehicle like this? @studiot, @Lizwi

And am I Missing some important factors here? - θ is the angle that the pink/black rod moves from the equilibrium position that is on the normal line to the surface in order to accelerate a vehicle- something like a segway/ moving inverted pendulum.

Thankfully we can omit the rolling resistance - (forces that make the vehicle decelerate) But torque for a point C would have to be a sum of torques acting on this point - right?
And from here we would make some equations and transformations in order to find the acceleration a?

Edited by Classical Physicist
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2 hours ago, Classical Physicist said:

Okay so having all of these Drawn out,  how can I find acceleration a that is at the point c which is centre of mass  of a vehicle like this?

The driving torque is determined by the engine/transmission, not the conditions where the wheel meets the road.

When driven in a straight line:

For any given driving torque we have that the frictional force times the wheel radius times the number of driven wheels = the driving torque up to the point where the wheels start to spin.

Look at your first diagram.

Is the vehicle in vertical equilibrium (is it moving up or down ?)

Is the vehicle in horizontal equilibrium (is it moving left or right ?)
So what net forces are acting on the vehicle?

This must therefore be the accelerating force on the vehicle.

Can you complete this ?

In your second diagram, what does the blue arrow ma represent?

Do you know how to construct a free body diagram?

Edited just now by studiot
Edited by studiot
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38 minutes ago, studiot said:

Do you know how to construct a free body diagram?

I think I can- but it's really not that simple when there's so many parts here to consider.

38 minutes ago, studiot said:

Is the vehicle in vertical equilibrium (is it moving up or down ?)

It simply moves (accelerates uniformly with this unknown yet constant acceleration) in the right direction as the picture suggests.  So it's not moving upwards/downwards- just rolling without slipping

38 minutes ago, studiot said:

Is the vehicle in horizontal equilibrium (is it moving left or right ?)

we assume here that this motion is in the right direction of this horizontal plane. So I think we can say that the equilibrium is met here.

38 minutes ago, studiot said:

So what net forces are acting on the vehicle?

This must therefore be the accelerating force on the vehicle.

Can you complete this ?

I don't see what you tried to post here- is it incomplete? Because you talk here about "the accelerating force on the vehicle"

38 minutes ago, studiot said:

In your second diagram, what does the blue arrow ma represent?

I drew here a non-inertial case of this vehicle in motion - so ma would be the force that it takes to keep the motion going,

Edited by Classical Physicist
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18 minutes ago, Classical Physicist said:
29 minutes ago, studiot said:

Do you know how to construct a free body diagram?

I think I can- but it's really not that simple when there's so many parts here to consider.

It is that simple. Simplicity is the purpose of free body diagrams.

19 minutes ago, Classical Physicist said:
30 minutes ago, studiot said:

Is the vehicle in vertical equilibrium (is it moving up or down ?)

It simply moves (accelerates uniformly with this unknown yet constant acceleration) in the right direction as the picture suggests.  So it's not moving upwards/downwards- just rolling without slipping

You have stated no equation balancing forces here ?

20 minutes ago, Classical Physicist said:
31 minutes ago, studiot said:

Is the vehicle in horizontal equilibrium (is it moving left or right ?)

we assume here that this motion is in the right direction of this horizontal plane. So I think we can say that the equilibrium is met here.

Again no equations?

This one is the crux of the problem.

22 minutes ago, Classical Physicist said:
33 minutes ago, studiot said:

In your second diagram, what does the blue arrow ma represent?

I drew here a non-inertial case of this vehicle in motion - so ma would be the force that it takes to keep the motion going,

I thought and hoped you knew better than that.

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1 hour ago, studiot said:

So what net forces are acting on the vehicle?

I'd say that the only net forces acting on a point C of a vehicle are

torque (2h mg sin θ )

force that gives us acceleration of a vehicle (2h ma cos θ )

and torque of a point C that is τ=Iω

Image:

32 minutes ago, studiot said:

I thought and hoped you knew better than that.

Sorry I've been trying to figure it out for days and I still have trouble with applying these forces in order to solve this problem, so I may have a flawed understanding of it right now.

As I understand "Rolling motion" is a sum of translational motion and rotational motion.

32 minutes ago, studiot said:

You have stated no equation balancing forces here ?

Again no equations?

This one is the crux of the problem.

So you suggest starting with finding/composing some equations that will describe the motion of this vehicle? - Sounds correct. I'll see what I can do.

Edited by Classical Physicist
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A free body diagram isolates any body or part of a body from the rest of the universe.
It shows the magnitudes, positions and directions of all forces and moments exerted on the body by the rest of the universe.
It does not show any forces exerted by the body on anything.
The vector sums of these forces and moments are sufficient to maintain the body in whatever state of motion it is in.
We can establish sums of the forces in 1,2 or 3 dimensions independently.
If the sum in any dimension is zero the body is in equilibrium in that dimension.
If the sum is not zero the body is accelerating in that dimension according to Newton's second law.
If the moments in 2 or 3 dimensions are zero the body is in rotational equilibrium in that situation.

Here is a free body diagram of the car as a whole, with a couple of realistic assumptions about weight distribution.

See how simple it is ?

Following this example can you do the same for an individual wheel ?

Sorry I see that the bottom of the scan did not come out properly the equation should read

4F = W/g

Edited by studiot
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1 hour ago, studiot said:

See how simple it is ?

Following this example can you do the same for an individual wheel ?

Yes, It's easier to draw few FBD's for each of the wheels/ points, instead of making complex sketches full of informations…
Also what a cute car did You draw! - It's looks just like an Aston Martin! I will  send Wheel FBD's here shortly. After this we should go back to finally analyzing the motion, right?

Edited by Classical Physicist
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OKAY I need some serious help with it right now. What should I do to find this acceleration?

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On 11/12/2019 at 9:12 PM, Classical Physicist said:

Yes, It's easier to draw few FBD's for each of the wheels/ points, instead of making complex sketches full of informations…
Also what a cute car did You draw! - It's looks just like an Aston Martin! I will  send Wheel FBD's here shortly. After this we should go back to finally analyzing the motion, right?

1 hour ago, Classical Physicist said:

OKAY I need some serious help with it right now. What should I do to find this acceleration?

Isn't techology wonderful when it works ? Until it doesn't.

I see there was a glitch in my last post.

The scan gave you the correct equation (Newton's second Law) for the acceleration, but was cut off at the knees.

My edit/correction I see missed off the acceleration completely.

Sorry.

So here we go

4F = W/g times acceleration

or

acceleration = 4gF/W

So now we need to find a value for F.

Which is why we need a free body diagram for a wheel ?

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sure

But what forces do I need?

Torque at the point A, Rolling friction, normal force R, and …. force that is applied to a wheel?

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10 minutes ago, Classical Physicist said:

sure

But what forces do I need?

Torque at the point A, Rolling friction, normal force R, and …. force that is applied to a wheel?

The torque, yes,

What do you think the rolling friction is equal to ?

What do you think the normal force R is equal to ?

10 minutes ago, Classical Physicist said:

force that is applied to a wheel?

What force do you think that is ?

What have you learned from the FBD of the entire car?

Edited by studiot
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force applied to a wheel F=ma

rolling friction is F=-μR

normal force R=-mg

Is that right?

going back to the Two-wheeled inverted pendulum that i have to find acceleration of, by using the point c that is the centre of mass of the whole vehicle,

how do I go about finding the torque at that point, which is on a line that is displaced at exactly the center of the rod that is in the axis of rotation of a vehicle… ?

I Really need to figure it out @studiot. I can't even tell you how much It would mean to me if we figured it out!

Edited by Classical Physicist
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1 hour ago, Classical Physicist said:

I Really need to figure it out @studiot. I can't even tell you how much It would mean to me if we figured it out!

We will, but I shall be busy for a few hours now tonight.

Meanwhile,

Why do you say ?

minus mu R

rolling friction is F=-μR

Have another go at a free body diagram (google it) and only include real forces external to the body concerned.

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ok so rolling friction is just F= μR

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5 hours ago, studiot said:

acceleration = 4gF/W

So F = μR

and R = (in terms of W) ?

Can you now make the substitution?

When you do you are in for a pleasant suprise.

After that I suggest you review both your understanding of turning forces and Free Bodies.

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