How find moment of inertia for car in turn?

[hutchphd]

Can the OP explain what the parallel axis theorem says? I don't think so.

 

[user079622]

The car seems to be rotating so that the front left wheel is the instantaneous center of rotation (possibly you will call it "pivot"). This point is at rest, its translation and rotation velocities cancel one another.

Dont understand your view.

Center of curve is ICR not front left wheel.
Car ICR depend on acerman geometry, but that center is irrelevant for polar moment of inertia

 

[user079622]

Can the OP explain what the parallel axis theorem says? I don't think so.

Yes, for every axis that is parallel with vertical axis and perpendicular to that (in our case yaw plane) plane has moment of inertia = moment of inertia for axis that pass through c.g. + mass of object times distance^2 from c.g. to that "choosen" axis.

Also formula shows that smallest moment of inertia is when object rotate around his c.g.(axis of rotation pass through c.g)

 

[hutchphd]

So is it not obvious why one wants to know the moment of inertia about the CoM ?

 

[user079622]

So is it not obvious why one wants to know the moment of inertia about the CoM ?

If car turn front wheels 90degress to the left, and c.g. of car is in the middle, isnt obvious that axis of rotation is not pass through c.g.?

I just want to tell to find car moment of inertia we must first find out how far is axis of rotation from c.g, I bet that distance is never zero..

 

[nasu]

Dont understand your view.

Center of curve is ICR not front left wheel.
Car ICR depend on acerman geometry, but that center is irrelevant for polar moment of inertia
View attachment 334284

I did not say center of curve, did I?
Instantaneous center of rotation moves as the car moves. It's fixed just for an instant.
Do you realize that what happens here can be analized as the motion of the COM around a curved path as well as a rotation of the body around the COM? If the path of the COM is part of a circle, there is a fixed center of that path.
For the rotation around the COM, there is always an instantaneous center of rotation which is at rest at that instant. This is not the center of the curve described by the COM and may have different position on the body or outside the body, at different instants. These are basic concepts in the mechanics of rotation for rigid bodies.

 

[user079622]

Do you realize that what happens here can be analized as the motion of the COM around a curved path as well as a rotation of the body around the COM?

That implies: if COM of object move in curved path, that object can only rotate around his COM(because if not, COM will not move in curved path)

(Only when undeersteer or oversteer happened ,COM slip out from that curved path so we have complex path.)

 

[nasu]

Again, does not follow. The COM can move in a curved path no matter what. The path of the COM is determined by the external forces on the system. It is useful to consider the rotation around the COM because then the two motions can be separated (translation of COM plus rotation around the COM). This means that you can write Newton's second law separately for the two motions, you can write KE and momentum as a sum of just two terms (rotation and translation).
If he rotation is considered around any other point than COM then there are mixed terms in KE and momentum which are neither purely translation, nor purely rotation. The equatyions of motions cannot be separated anymore. Nothing wrong with this, just messy.
But the object still rotates around these not so convenient points, it does not care about how messy is our math. So when we say that "it rotates around the COM" we don't mean "it rotates only around the COM".

 

[Lnewqban]

That implies: if COM of object move in curved path, that object can only rotate around his COM(because if not, COM will not move in curved path)

(Only when undeersteer or oversteer happened ,COM slip out from that curved path so we have complex path.)

That is difficult to understand, sorry.
At high speed on a track, it will be easier to turn-in and turn-out the Porsche than the Audi, regardless the radius of the curve, or under/over steer.

The front tires of the Porsche will have less lateral force and therefore will have more grip in reserve.
The front tires of the Audi will have to apply more lateral force on the asphalt to get the car turned at the same rate, just like you and I had to apply more force on the Audi to make it spin while hanging from the crane.

As both cars should be breaking (decelerating) from the curve entry to the apex, and accelerating from the apex and out of the curve, we will have more than one type of inertia to consider.

Centralized mass is important for spinning the cars, but not for turning the cars.
Reduced total mass is important for accelerations along the curve path, but not so much for spinning the cars from North heading to West heading.

 

[user079622]

@Lnewqban

What are pros and cons of middle and rear engine position? Why people often say porsche has engine in wrong place?If engine is at rear that doesnt mean that that must be far away from c.g....

 

[Lnewqban]

@Lnewqban

What are pros and cons of middle and rear engine position? Why people often say porsche has engine in wrong place?If engine is at rear that doesnt mean that that must be far away from c.g....

That is not what the original article in post #1 explains.
It is located in the less rear position possible for a street-sport car.

 

[user079622]

I did not say center of curve, did I?
Instantaneous center of rotation moves as the car moves. It's fixed just for an instant.

So this is not ICR?

 

[user079622]

This means that you can write Newton's second law separately for the two motions, you can write KE and momentum as a sum of just two terms (rotation and translation).

Translation is tangential velocity and rotation is car spin around c.g.?

 

[nasu]

Translation means the motion of the COM, assumed to have a mass equal to the mass of the rigid object. The velocity pf the COM is always tangent to the trajectory of the COM. There is no other component. In general, velocity is tangent to the trajectory, by definition.

 

[nasu]

So this is not ICR?

View attachment 334315

It may be. Or not. How can you tell from some pictures with no context? But if they mark it as ICR it may be. It does not mean that is the same point as the center of the the trajectory described by the COM. These are distinct points in general. They may be same in some situations.

 

[user079622]

It may be. Or not. How can you tell from some pictures with no context? But if they mark it as ICR it may be. It does not mean that is the same point as the center of the the trajectory described by the COM. These are distinct points in general. They may be same in some situations.

how do you mean "with no context"?

This is center of curve, that point is at rest in space-time. Time passes but the coordinates remain the same.
Isnt it?

 

[user079622]

@Lnewqban

Does high moment of inertia drain availble tyer force so there is less for lateral direction?

 

[Lnewqban]

@Lnewqban

Does high moment of inertia drain availble tyer force so there is less for lateral direction?

Not exactly.
The lateral load trying to skid the tires is increased, but only if the car is accelerating (increasing or reducing velocity).
Therefore, there is less available traction to be used in steering and in forward acceleration or trail braking.

 

[jbriggs444]

how do you mean "with no context"?

This is center of curve, that point is at rest in space-time. Time passes but the coordinates remain the same.
Isnt it?

There is no such thing as being at rest in "space-time". Motion is always relative. Minkowski space-time offers no preferred reference frame against which to measure velocity. Nor does plain old Newtonian space-time for that matter.

The proper terminology would that the point is "at rest in our chosen frame of reference".

But which frame of reference will we choose? Probably one where the pavement is motionless. The lab frame.

But we could choose a frame of reference where the cosmic microwave background is isotropic (co-moving coordinates in cosmological terms). Then the instantaneous center of rotation could be one or two hundred kilometers away. Our drift velocity relative to the CMBR is about 369 km/sec. At a rotation rate of 0.5 radians per second, that would put the instantaneous center of rotation about 185 km away.

 

[user079622]

@jbriggs444

I understand that motion is relative term.

Flywheel rotate around center of main bearing, it is hard to me to grasp how flywheel rotate around engine mounting as well. I can't visualize that plus in reality he is not doing that...