Radial Force

[Primarygun]

Consider there is a car driving in a horizontal circular path, as stated in my book, a radial force exists and causes the car to accelerate towards the centre.

 

Here the questions are:

why the acceleration must be towards the centre?

with a positive acceleration in a direction perpendicular to the motion of the car, the car should be accelerate towards the centre and this results in an increase in the speed of the car but the fact indicates that is impossible, why?

 

Besides, what is centrigual force? Is it prerequisite in order to have a complete understanding of rotation?

 

This question I've tried the web links in my books, and other E-books, however, I still get a unclear mind/

[timo]

Here the questions are:

why the acceleration must be towards the centre?

I have no good explanation for it but you can see it easily from the math. Let a circular motion be described by (x' date='y)(t) = (sin(t), cos(t)). The first derivative (with respect to time t) is the velocity, the 2nd derivative is the acceleration. You will easily see that (a_x, a_y)(t) = -(x,y)(t)

 

with a positive acceleration in a direction perpendicular to the motion of the car, the car should be accelerate towards the centre ...

It does.

... and this results in an increase in the speed of the car ...

Not really. Take it this way: If the acceleration is in direction of the velocity, the speed increases. If the acceleration is in opposite direction of the velocity, the speed decreases. Seen this way it seems only natural that a perpendicular acceleration let´s the speed unchanged, doesn´t it?

... but the fact indicates that is impossible ...

I don´t get that part. What fact? And what is impossible? Hopefully, that statement becomes superfluous in the light of what I said in above.

 

Besides, what is centrigual force? Is it prerequisite in order to have a complete understanding of rotation?

No good idea how to describe it. But it´s not a prequisite, imho.

 

This question I've tried the web links in my books, and other E-books, however, I still get a unclear mind/

Circular motion -as most things in physics- is something you get used to by working with it, so don´t worry.

[Primarygun]

... and this results in an increase in the speed of the car ...

 

Not really. Take it this way: If the acceleration is in direction of the velocity, the speed increases. If the acceleration is in opposite direction of the velocity, the speed decreases. Seen this way it seems only natural that a perpendicular acceleration let´s the speed unchanged, doesn´t it?

I don't know, but there's an increase in velocity in the perpendicular way, thar is v=a *(delta t)

I suggest that as delta t is near zero, the increase of that component us also near zero.

[Rocket Man]

Consider there is a car driving in a horizontal circular path' date=' as stated in my book, a radial force exists and causes the car to accelerate towards the centre.

 

Here the questions are:

why the acceleration must be towards the centre?

with a positive acceleration in a direction perpendicular to the motion of the car, the car should be accelerate towards the centre and this results in an increase in the speed of the car but the fact indicates that is impossible, why?

 

Besides, what is centrigual force? Is it prerequisite in order to have a complete understanding of rotation?

 

This question I've tried the web links in my books, and other E-books, however, I still get a unclear mind/[/quote']

 

 

think of it in terms of the force applied to keep the car rotating against inertia, to turn inward, the wheels must apply a force perpendicular to it's motion.

without an opposing force, there is acceleration. this acceleration is limited by the cars own intertia.

 

take a weight on a string swung in a circular path; at one point, it is travelling in a direction, call it forward.

the string is applying a force perpendicular to it's travel.

it may help to take a look at pythagorases theorem,

call the forward motion, motion along the x axis, and the inward force applied on the y.

v^2 = x^2 + y^2

there is actually very little increase in total velocity unless the y compontent is comparible to the x.

so an increase in velocity seems likely untill you take a look at the motion when the string is 90 degrees from where we last took a look. the string is actually applying a force against the original direction of travel. so it also seems to be slowing it down. there is no energy change so we can say that the rate the force speeds it up is equal to the rate it slows it down. no net energy change.

 

"centrigual force" do you mean centripetal force? centripetal force is the inertia of a mass attempting to go in a straight line against the inward force.

[timo]

I don't know' date=' but there's an increase in velocity in the perpendicular way, thar is v=a *(delta t)

I suggest that as delta t is near zero, the increase of that component us also near zero.[/quote']

It was only a suggestive argument. Mathematically, let´s have a velocity v=(V,0) and an acceleration a=(0,A).

 

Now let´s use the dirty physicists dt-trick:

So v' := v + a*dt = (V, A*dt)

Calculating the magnitude of it: |v'| = sqrt(V² + A² dt²) ~= v + (A² dt²)/(2v)

where the last ~= was a Taylor expansion of the square root.

So if you calculate the differential change of the magnitude as normal by limit(dt->0) (|v'| - |v|)/dt it turns out to be zero.

If you -in contrast- had had a acceleration of a=(A,0), the result would be nonzero as then

v' = sqrt[(V+A dt )²] = V+A dt and the limit gave an A.

 

sry for the lazy formatting and maybe slightly weird formulation. i was in a hurry but wanted to give you the reply. I´ll post it in a more thought-trough and better-formatted way later if this was not understandable.

[swansont]

A force perpendicular to the displacement does no work, so there is no change in kinetic energy and thus no change in speed. Velocity is a vector, so it has changed if the direction changes, even if the magnitude is constant. A changing velocity is, by definition, an acceleration.

[Primarygun]

A force perpendicular to the displacement does no work, so there is no change in kinetic energy and thus no change in speed. Velocity is a vector, so it has changed if the direction changes, even if the magnitude is constant. A changing velocity is, by definition, an acceleration.

 

I could fully understand this statement.

But, if a change in direction is resulted in a change ( tiny change) in one of the components of velocity, right?

[swansont]

I could fully understand this statement.

But' date=' if a change in direction is resulted in a change ( tiny change) in one of the components of velocity, right?[/quote']

 

Velocity is a vector. A change in direction of motion is, by definition, a change in velocity. (and the speed, or magnitude of this velocity, does not have to change)

[J.C.MacSwell]

Velocity is a vector. A change in direction of motion is, by definition, a change in velocity. (and the speed, or magnitude of this velocity, does not have to change)

 

In fact if the speed changes then the force of the acceleration vector will not be exactly radial or perpendicular to the instantaneous velocity vector.

[timo]

In fact if the speed changes then the force of the acceleration vector will not be exactly radial or perpendicular to the instantaneous velocity vector.

In circular motion, it always will. Take the (sin(t),cos(t)) example in above. We hopefully can agree that this is circular motion. Try to find a time for which the acceleration is not perpendicular to the velocity. Try to find a time for which the magnitude of the velocity is not 1.

You cannot make a small timestep, say "now my velocity has changed direction", take the old acceleration and say "and it´s not perpendicular to the acceleration". Well you can, but you are comparing v and a for different times/positions, then. And that doens´t make too much sense.

[J.C.MacSwell]

In circular motion' date=' it always will[/b']. Take the (sin(t),cos(t)) example in above. We hopefully can agree that this is circular motion. Try to find a time for which the acceleration is not perpendicular to the velocity. Try to find a time for which the magnitude of the velocity is not 1.

You cannot make a small timestep, say "now my velocity has changed direction", take the old acceleration and say "and it´s not perpendicular to the acceleration". Well you can, but you are comparing v and a for different times/positions, then. And that doens´t make too much sense.

 

In this example it works but if you vary your speed in a circular motion the force and acceleration vector would not be directed exactly radially.

 

I agree with everything else for the special case, circular motion at constant speed.