2
$\begingroup$

I just got started on my physics course, and already I am confused with this stuff called centripetal acceleration.

One thing I don't understand is what produces the velocity that is tangent to the circular path.

I searched on google, and got from PBS Education that there is a change in the velocity which produces the acceleration:enter image description here But my textbook tells me that there is no change in the magnitude of the velocity only that its direction is changing.

On another website, there is a proof explaining that there is an inward acceleration because the change-in-velocity vector points inward which indicated that the acceleration vector should also be in the same direction. enter image description here

If you wish to explain it in calculus, bear in mind that I don't know anything about it.

$\endgroup$
2
  • 2
    $\begingroup$ It doesn't. The velocity is already there because of some other force. $\endgroup$ – Pranav Hosangadi Dec 29 '14 at 16:32
  • $\begingroup$ @PranavHosangadi, are you saying that a charge moves in magnetic field only because I throw it into the magnetic field not because of the force the field exerts on the charge? $\endgroup$ – most venerable sir Jul 28 '15 at 20:16
3
$\begingroup$

A small addition to @Hypnosifl's answer:

Key to understanding the circular motion is the fact that at any moment in time the acceleration (force keeping the horse / object in a circular path) is exactly perpendicular to the direction of motion. We have to do "almost calculus" to explain this... but maybe you will be able to follow:

Imagine we initially have a horizontal velocity, and the acceleration is vertical (downwards). How come the new velocity isn't faster than the old one?

Mental picture:

enter image description here

When the change in velocity due to acceleration $a$ for time $\Delta t$ is added to the initial velocity $v_{init}$, the new velocity has a magnitude

$$|v_{new}| = \sqrt{v_{init}^2 + (a\Delta t)^2}$$

When $a\Delta t$ is small, we can do a Taylor expansion and get

$$|v_{new}| = v_{init}\left(1 + \frac{(a\Delta t)^2}{2 v_{init}}\right)$$

Now if we look at a shorter and shorter interval $\Delta t$ we can see that the change in magnitude in velocity, which goes with the square of the interval, will become zero.

This means that if we keep the acceleration at right angles to the velocity (which is done in circular motion) then the speed will be constant - which is of course what we observe for circular motion of things like satellites, the moon, etc (barring small effects like tides, atmospheric drag etc - which are "imperfections" and not part of the essence of the orbital motion).

$\endgroup$
1
  • 1
    $\begingroup$ Now I know calculus, I got it! $\endgroup$ – most venerable sir Oct 3 '16 at 16:32
4
$\begingroup$

The centripetal acceleration doesn't cause the tangential velocity, it has to be there to begin with in order for the body to orbit, as opposed to just fall to the surface. A good intuitive way of thinking about this is Newton's cannonball. If you place a cannon on a mountain and shoot a cannonball in the tangential direction, if the tangential velocity is small relative to the orbital velocity the cannon will fall to the surface. But as the tangential velocity gets larger, the cannon is still falling towards the surface but the Earth's surface is curving out from under it, so the distance between it and the surface shrinks more slowly despite the fact that the radial acceleration is the same, and the initial radial velocity was zero when it was fired. You can then imagine a series of cases where the tangential velocity gets larger and the fraction of the Earth's circumference traversed by the cannonball before hitting the surface gets larger, and the rate at which it approaches the surface decreases, until you reach a limit case where the rate at which it approaches the surface goes to zero and the cannonball is in a circular orbit, as depicted below:

enter image description here

There's a java applet here that allows you to play around with different initial tangential velocities for the projectile. Note that if you check the box "show full orbit" you can see what would happen if all the Earth's mass were concentrated at the center of the planet, so a cannonball fired from the same radius and with the same tangential velocity would be in some sort of elliptical orbit no matter how small the tangential velocity, an orbit that that can pass below the radius of the real Earth's surface (so in real life it would simply crash into the surface without making a full orbit). For a non-java illustration of this see the diagram in slide 23 here, attributed to Physics for the Enquiring Mind, Ch. 22, Fig. 29:

enter image description here

From the fact that all arcs of the cannonball can be understood as sections of possible elliptical orbits, you can deduce that even before the velocity needed for circular orbit, there would be a velocity that would result in an elliptical orbit that gets very close to the radius of the surface on the opposite side of the Earth from the cannon (ignoring atmospheric drag), then begins to climb in height again as it continues around the Earth, until returning to the height of the cannon at the position the cannon originally fired it.

$\endgroup$
4
  • $\begingroup$ This is very helpful. For the cannonball that makes the complete orbit, the change in velocity is always perpendicular to the current velocity which is why the magnitude doesn't change, only the direction. That might be worth adding. $\endgroup$ – Floris Dec 29 '14 at 16:50
  • $\begingroup$ @Floris - I agree the magnitude won't change for a circular orbit, but I don't understand your argument for why it won't change--if you have a velocity vector $\vec{v}$ and then you add a change in velocity $\Delta \vec{v}$, the fact that $\Delta \vec{v}$ is perpendicular to the original velocity vector isn't enough to ensure that the sum of the two vectors has the same magnitude as the original vector $\vec{v}$. I recall there's a calculus derivation of circular orbits which integrates over infinitesimal changes in the velocity due to the force, but the OP doesn't know calculus. $\endgroup$ – Hypnosifl Dec 29 '14 at 17:00
  • $\begingroup$ @Hypnosifl, if you place a charged particle in a magnetic field, will it start to move on its own in a perpendicular circle? $\endgroup$ – most venerable sir Jul 28 '15 at 19:31
  • $\begingroup$ I really like your answer! I've re-quoted some of it here, I hope you don't mind. $\endgroup$ – uhoh Feb 23 '19 at 13:31
1
$\begingroup$

Actually,the centripetal force doesn't produce the tangential velocity!

Let a body be moving with constant velocity $v$ in a straight line. Then suddenly an external agent(force) acts on the body such that the body gets deflected from its straight line to follow then another path which may be curved (but need not be circular, circular makes the case simpler) . The position vector of the body gets rotated about a point but that doesn't bother the velocity $v$ of the body.

You can tell then what about the case when the tangential velocity changes.

Still, the centripetal force is not responsible for the change in tangential velocity! There may be other force responsible for the change in tangential velocity. So,velocity $v$ was present even before the centripetal force starts acting on it. Remember, centripetal force only acts to rotate the velocity vector of the body with respect to a certain point,and acts towards that point.

$\endgroup$
1
  • $\begingroup$ This seems like enough of an answer to me. $\endgroup$ – David Z Jan 1 '15 at 22:26
0
$\begingroup$

Centripetal acceleration doesn't cause the velocity, or rather its magnitude, but it is responsible for the change of its direction. Picture this scenario: the horse starts running at a constant velocity $\mathbf v$. Then it get into a bend of constant curvature $r$. The horse is then subject to a centripetal acceleration $\mathbf a$ that allows it to go on a circular motion by keeping a tangential velocity of constant magnitude. For this to be possible there must be no component of the acceleration along the tangential velocity, hence the orthogonality relation between $\mathbf v$ and $\mathbf a$.

$\endgroup$
6
  • $\begingroup$ But what is the force that makes horse go around in a circle? Isn't that the hose itself? $\endgroup$ – most venerable sir Dec 29 '14 at 15:52
  • $\begingroup$ The horse is pushing on the ground sideways and the ground is pushing (Newton's third law) the horse towards the centre of the circle through friction. $\endgroup$ – Phoenix87 Dec 29 '14 at 15:55
  • $\begingroup$ So velocity has nothing to do with the acceleration. But in the proof, the one that uses similar triangles, velocity vector 1 and velocity vector 2 and change in these two vector form a triangle. The problem with this is that change in velocity is always 0!? $\endgroup$ – most venerable sir Dec 29 '14 at 16:06
  • $\begingroup$ velocity is a vector, so it can change in both direction and magnitude. In uniform circular motion only the former is changing, while the latter stays constant in time. $\endgroup$ – Phoenix87 Dec 29 '14 at 16:14
  • $\begingroup$ So if satellite moves in a orbit. What produces its tangential velocity? $\endgroup$ – most venerable sir Dec 29 '14 at 16:25
0
$\begingroup$

What you need to know is that centripetal acceleration, $a_c=\large\frac{v^2}{r}$.
Constant Magnitude of Velocity is just as it is, as a consequence of Centripetal acceleration which result when a body trying to move in a circular motion with velocity $v$.

However, the velocity vector $\vec v$ is not a constant vector because it's direction is always changes. A vector is constant when both it's magnitude and direction are constant. And this change in direction is what is produced by the centripetal acceleration!

Also remember that, Centripetal acceleration and velocity are direct consequences of each other when a body tries to move in a circular motion with a velocity $v$ without tangential acceleration(google if you like!).

Also note that usually as you go along your physics course you will see that the centripetal force or acceleration is mostly provided by an outside agency! For e.g., Gravitational field, Magnetic field , Tension of a taut string ,etc! Which causes the body to move in a circular path with a constant magnitude of velocity $v$

And at times velocity is produced from prior sources! Before the object is subjected to circular motion. For e.g., A charge moving perpendicular to magnetic field with constant velocity! That velocity is provided beforehand by any means possible.

If you are confused about how does a change in velocity is possible when magnitude is constant.

Consider the velocity vector to be at some instant
$\vec v_1=\hat i$ (which means it's direction is entirely along $x-axis$ and magnitude 1)
And at some other instant $\vec v_2=\hat j$ (entirely along $y-axis$ and magnitude 1)
They are of same (constant magnitude but different direction).

Now, notice that $\vec v_2-\vec v_1\neq0$ but $=\hat j-\hat i$ which has a magnitude $\sqrt{2}$.

$\endgroup$
1
  • 2
    $\begingroup$ I know all you said, but where the velocity comes from is what I want to know. $\endgroup$ – most venerable sir Dec 29 '14 at 15:55
0
$\begingroup$

what produces the velocity that is tangent to the circular path.

Assuming no dissipation, e.g., friction, and uniform circular motion (the speed of the object is constant), there is nothing required to maintain the speed. This is due to the conservation of angular momentum.

If the speed is changing, the motion is not uniform circular motion and then there is tangential acceleration which produces change in tangential velocity.

So, if you're picturing an object initially at rest and finally in steady circular motion, there is a time during which there is tangential acceleration in addition to the (increasing) centripetal acceleration.

Once the tangential acceleration is zero, the speed of the object is constant and the centripetal acceleration maintains the uniform circular motion.

$\endgroup$
-1
$\begingroup$

It might seem a late answer but i was also plagued by the same issue. I know that two comppnents of perpendicular to each other doesn't affect both its magnitude and directionbut can it be said for acceleration? I looked at it this way. Since time after time a new component a x t Where a is acceleration and t is time why is the magnitude not changing? That is a valid question. However if we see the person twirling the string with a ball as the system, the total external torque acting is zero. Thus conservation of angular momentum is applicable. Now the person is at rest and the angular momentum is Mv x r where v and r are vectors keeping the position of the person as origin. Thus r is zero. Thus total angular momentum magnitude reduces to mvr of the ball which must be constant as momentum is conserved.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.