About centripetal acceleration

[Aitor Bracho]

I've been thinking about centripetal force and its effects on motion in uniform circular motion. I've actually found it difficult to accept that velocity magnitude can ever be maintained constant. Here is why:

if this is our velocity vector, v, at the top of the circle: →
Then the centripetal acceleration vector must then point downward (perpendicular to the velocity vector.
However, acceleration that is perpendicular to a component has no effect on that component. In other words, if you have acceleration in the y axis, this has no effect on velocity on the x axis. So here is my question: how is it possible that centripetal acceleration can change the direction of the velocity at the top of the circle without affecting its magnitude? After all, the x component of the above vector, v (→) is the only component of said vector. If acceleration is point perpendicular to this vector, how, then, can the next vector for velocity (which would point slightly diagonal) have the same magnitude as the original vector v? Would it not have an x component whose magnitude would be equivalent to that of vector v PLUS a y component, thus summing up to a larger magnitude?

 

[jbriggs444]

I've been thinking about centripetal force and its effects on motion in uniform circular motion. I've actually found it difficult to accept that velocity magnitude can ever be maintained constant. Here is why:

if this is our velocity vector, v, at the top of the circle: →
Then the centripetal acceleration vector must then point downward (perpendicular to the velocity vector.
However, acceleration that is perpendicular to a component has no effect on that component. In other words, if you have acceleration in the y axis, this has no effect on velocity on the x axis. So here is my question: how is it possible that centripetal acceleration can change the direction of the velocity at the top of the circle without affecting its magnitude? After all, the x component of the above vector, v (→) is the only component of said vector. If acceleration is point perpendicular to this vector, how, then, can the next vector for velocity (which would point slightly diagonal) have the same magnitude as the original vector v? Would it not have an x component whose magnitude would be equivalent to that of vector v PLUS a y component, thus summing up to a larger magnitude?

There is no such thing as a "next vector". There is a continuum of vectors.

For any "next vector" that you can point to, no matter how close to the first, the acceleration vector will have continuously changed directions on the way from the first vector to the next vector, always remaining at right angles to the path. There will never be a non-perpendicular component to add.

 

[Aitor Bracho]

I can understand that, however, the issue is that at moments when the velocity vector is strictly horizontal or strictly vertical, the radial acceleration should not affect that component of velocity and therefore would create a larger magnitude of velocity, meaning that it cannot be constant. Am I wrong?

 

[Nugatory]

I can understand that, however, the issue is that at moments when the velocity vector is strictly horizontal or strictly vertical, the radial acceleration should not affect that component of velocity and therefore would create a larger magnitude of velocity, meaning that it cannot be constant. Am I wrong?

There's nothing special about the moment when the velocity vector is strictly horizontal or vertical. We could draw our x and y axes so that they are at 45 degree angles instead of horizontal and vertical, and the physics wouldn't change.

 

[sophiecentaur]

I can understand that, however, the issue is that at moments when the velocity vector is strictly horizontal or strictly vertical, the radial acceleration should not affect that component of velocity and therefore would create a larger magnitude of velocity, meaning that it cannot be constant. Am I wrong?

Your use of the terms Vertical and Horizontal, implies rotation in a vertical plane. The forces involved in that situation will not be straightforward because the tension in the string will be constantly modified by the Weight Force, that always acts downward. Left to itself, a mass on a string, rotating in a vertical plane will keep changing its magnitude of velocity on the way round.
Can you clean up your question a bit so we can see what you are actually bothered about?

 

[PeroK]

I've been thinking about centripetal force and its effects on motion in uniform circular motion. I've actually found it difficult to accept that velocity magnitude can ever be maintained constant. Here is why:

Suppose a car is moving round a circular race track. At what points on the track does it speed up and at what points does it slow down?

If it always slows down, how far can it go before it stops?

If it always speeds up, what happens when it reaches the car's maximum speed?

 

[Voltageisntreal]

Consider throwing a ball. If you throw the ball, disregarding air resistance, it will not slow down in the direction you throw it in- Newtons first law.

Yet, gravity will pull it down to the Earth. so in the direction you threw the ball in, it will not slow down- it will just speed up towards the Earth.

Hence this answers how you can change the direction of something without changing it's magnitude of speed in the original direction with an analogy.

Why is this possible?

A resultant force or acceleration does not necessarily have to result in a change in speed, since simply changing the velocity (direction of travel) of something also takes a force and an acceleration.

sometimes, explicitly regarding centri-petal forces now, the force is not enough to overcome the velocity of the particle to keep it in orbit, and that particle zips off into nowhere.
sometimes, the force is just enough to keep the particle where it is traveling at the velocity it is travelling.
sometimes, the force is too much and creates a velocity towards the centre and causes a collision.

The middle ground is actually rare without controlled systems, most gravitational inteactions result in collisions, it's just that they've already happened and our own vicinity is rather tame right now.

Therefore, you have to consider that the force changes the direction of the particle JUST the right amount to keep it in orbit at the same distance away from the centre it is in right now but not move it. It's a strange, even unusual balance when you think about it, hence it breaks intuituion since usually we think of a resultant force getting work done.

 

[Jamison Lahman]

However, acceleration that is perpendicular to a component has no effect on that component. In other words, if you have acceleration in the y axis, this has no effect on velocity on the x axis.

Is there constant acceleration in the y-direction?

 

[Jamison Lahman]

Think of it like this: for each position on the circle, there is point where the acceleration is exactly equal and opposite. For example, at the top the acceleration is -a_y while at the bottom it is a_y. If you were to sum all of these pairs up, it is easy to see that the total net acceleration is 0 for one, full revolution in the x and y directions. By definition, since there is no net acceleration, there is no change in velocity's magnitude.

 

[jbriggs444]

I can understand that, however, the issue is that at moments when the velocity vector is strictly horizontal or strictly vertical, the radial acceleration should not affect that component of velocity and therefore would create a larger magnitude of velocity, meaning that it cannot be constant. Am I wrong?

The velocity is strictly horizontal for an instant of zero duration. It is meaningless to ask whether it changes during that instant.

A constant acceleration applied over an interval of non-zero duration at right angles to an object's starting velocity would indeed increase that object's velocity. But in uniform circular motion the acceleration is no more constant than velocity or position. All three are continually changing direction. Hence @Jamison Lahman's hint

 

[Khashishi]

This has to do with the mathematics of infinitesimals. If you expand out the change in speed over some infinitesimal step in time, what you get is that the change in speed is second order infinitesimal. E.g.
At time 0, the velocity is (1,0), and the acceleration is (0, -1)
You can approximate the velocity at time dt as
v(dt) = v(0) + a(0)*dt
v(dt) = (1,0) + (0,-1)dt = (1,-dt)
This is called Euler integration, and the accuracy is first order in dt.
The new speed is |v(dt)| = sqrt(1^2 + dt^2) ≅ 1 + 0.5*dt^2 + higher order terms.
The new speed differs from the old speed by 0.5*dt^2, which is second order in dt. But Euler integration is only accurate to the first order in dt, so the new speed is equal to the old speed, within error. You can choose dt as small as you want, and you find that instantaneously, the speed is not changing.

 

[Jamison Lahman]

This has to do with the mathematics of infinitesimals. If you expand out the change in speed over some infinitesimal step in time, what you get is that the change in speed is second order infinitesimal. E.g.
At time 0, the velocity is (1,0), and the acceleration is (0, -1)
You can approximate the velocity at time dt as
v(dt) = v(0) + a(0)*dt
v(dt) = (1,0) + (0,-1)dt = (1,-dt)
This is called Euler integration, and the accuracy is first order in dt.
The new speed is |v(dt)| = sqrt(1^2 + dt^2) ≅ 1 + 0.5*dt^2 + higher order terms.
The new speed differs from the old speed by 0.5*dt^2, which is second order in dt. But Euler integration is only accurate to the first order in dt, so the new speed is equal to the old speed, within error. You can choose dt as small as you want, and you find that instantaneously, the speed is not changing.

This approximation should only be used in the case of constant acceleration which is not the case in circular motion. In my opinion, without constant acceleration, this explanation is no better than Zeno's Paradoxes.

 

[sophiecentaur]

This approximation should only be used in the case of constant acceleration which is not the case in circular motion. In my opinion, without constant acceleration, this explanation is no better than Zeno's Paradoxes.

The radial acceleration is constant.

 

[jbriggs444]

The radial acceleration is constant.

True, but arguably just word play. The radial component of the acceleration is constant and the tangential component of the acceleration is also constant. But using those components means adopting the rotating frame -- which is non-inertial. In any inertial frame, the acceleration is not constant.

 

[Jamison Lahman]

The radial acceleration is constant.

Not only is it constant, it is constantly zero.
$$
\omega = \omega_0 + \alpha t \\
\alpha = 0 \\
\omega = \omega_0 \\
$$
QED

 

[nasu]

∝ in that formula is not radial acceleration.

 

[PeterO]

Suppose a car is moving round a circular race track. At what points on the track does it speed up and at what points does it slow down?

If it always slows down, how far can it go before it stops?

If it always speeds up, what happens when it reaches the car's maximum speed?

Driving around a horizontal, circular track at constant speed is probably easier to understand than a vertical circle - you can bank the track to make it easier for the driver if you like. By managing the accelerator pedal carefully, constant speed (not constant velocity since the direction is changing) is quite easy to attain - and secondly, the magnitude is anything you like from as slow as you can drive the car up to the fastest the car can manage (without spinning out - a banked track can help there).
If you were to attempt to drive a car, at constant speed, inside a vertical, circular track, firstly the minimum speed is limited by the speed necessary to pass through the top of the loop without falling off the track - and the manipulation of accelerator and brakes to actually maintain constant speed is probably way beyond the capabilities of most humans. It does not, however, stop us imagining and analysing the situation when that was magically achieved.

 

[sophiecentaur]

Not only is it constant, it is constantly zero.
$$
\omega = \omega_0 + \alpha t \\
\alpha = 0 \\
\omega = \omega_0 \\
$$
QED

That is angular acceleration not radial.