Dot Physics

Basics: Centripetal Acceleration

pre-reqs: vectors, kinematics

I haven’t done a “basics” topic in quite some time. It’s odd, I have used centripetal acceleration quite often, but I never derived the expression that I use. To get to the point, the magnitude of the acceleration of an object moving in a circle is:

i-eeaccf1fe18bc08e66f237268242e7b2-cent_1.jpg

Also, the direction of this acceleration vector is always towards the center of the circle the object is moving in. This is really not too difficult to derive (but it does use at least one “trick”). Let me start with an object moving in a circle at a constant speed. I am going to show to instances of the object one and then again after a very short time.

i-034ee1c6959e4b73b53ea383496a6418-cent_2_circle_vector.jpg

To find the acceleration, I will just use the same normal definition of average acceleration:

i-919af8d6b992b3659afc1face4e1d96b-def_avg_accel.jpg

From the diagram above, I can determine the change in velocity. Let me write the vectors for v1 and v2:

i-7f802f0ad1073a0db202d0cc3c950589-v_1_and_v_2.jpg

So, the second velocity has components in both the x- and y-directions (which I didn’t label, but you get the idea). Ok, now for the first trick. This angle ? is really small. It has to be if I want to get an expression for the acceleration of an object moving in a circle. Just think, what if I took ? = 2?? In this case, I would be finding the average acceleration for the object going all the way around the circle. Clearly this would be zero since the two velocity vectors would be identical.

Since ? is small, I can say that cos(?) is about equal to 1 and sin(?) is about equal to ? (where ? must be in radians). This means I can write v2 as:

i-fae7f8a246d1484089043f87a70d32e0-v_2_new.jpg

Hopefully, you won’t feel cheated with those tricks. The first one should be clear. Look at the diagram with the vectors. The second velocity vector is moving in the x-direction essentially just as much as the first vector. The smaller ? is, the truer this is.

For the sin trick, look at the sin function. Near ? = 0, the function is really really really straightish looking. Right? If you plot y = sin(?) and y = ?, near the origin, these two functions are right on top of each other. Here is a plot showing that:

i-bb8cc6e8c9eb82ecec580bed5b74c652-theta_vs_sin.jpg

Now, I can start putting this stuff together. Let me calculate the change in velocity:

i-19c4da05e7804a638068441f14f59ba7-delta_v_with_comp.jpg

Boom. I just got the direction of the acceleration for an object moving in a circle. In this particular case, it is in the negative y-direction. If you look at the vector diagram you can see that this is towards the center of the circle. Now, for the magnitude of the acceleration, I need to first find the time that this change in velocity takes place. I know the speed (v) and I know the distance of the arc-length. This gives a time of:

i-b1287248bbe6bcd48150fb5c0700217d-delta_t.jpg

Putting the change in velocity together with the change in time (in non-vector form):

i-7eacbfafb85e7ada9e16baf454ff11c5-avg_accel_mag_derive.jpg

Done. This is the value I have been using all along. Of course, if you know the angular speed instead of the linear velocity this can be written differently. First, the relationship between linear velocity and angular velocity for something moving in a circle:

i-706f6c239decd4d41fadbf5e4b5f5b47-vwr.jpg

Substituting in for v, it is easy to see that the acceleration (magnitude) can be written as:

i-ad9ce2ee456f4b47c888216ee5cf7be2-omega_squared_r.jpg

That is it. The acceleration of an object moving in a circle.

Comments

  1. #1 Uncle Al
    July 18, 2009

    Centripetal accleration only occurs at the rim. Emit a photon from the rim toward the hub through vacuum in a compressed Harvard Tower experiment. A Beckman Coulter Optima MAX-XP Benchtop Ultracentrifuge will deliver 1,019,000 gees (MLA-130 rotor). Zero Mossbauer shift at the hub.

    What would happen to an optical photon, vacuum versus solid glass path?

    nu = nu_0[1 + gh/c^2]
    22.6 meters at 1 gee gives 4.92×10^(-15) relative. 0.0419 meters at 1.019×10^6 gees gives 9.5×10^(-13) relative. Tough experiment outside Mossbauer detection, even if high refractive index glass made a difference in pathlength.

  2. #2 Cleon Teunissen
    July 20, 2009

    There is an interesting discussion by Richard Conn Henry.
    http://msx4.pha.jhu.edu/henryDir/pubsPDFDir/circularMotion.pdf

    Around 1665 Newton had independently found a derivation for the required centripetal force in the case of uniform circular motion.
    Today one automatically uses differentation, but in Newton’s time every single proof involving what today is recognized as differentation involved taking the limit of going to infinitisimally small steps.
    A modern proof that uses goniometry will at some point invoke that in the limit of infinitissimal steps the sine function is linear.
    The two necessities to take the limit can be combined.

    From today’s perspective Newton’s derivation is an exercise in showing what the simplest means are that are sufficient for deriving centripetal force.

  3. #3 Rhett
    July 20, 2009

    @Cleon,

    Thanks, I had not seen that paper before.

  4. #4 Liz
    November 22, 2009

    Hi. Today, I came upon this web-site while researching the origin of the centripetal acceleration equation, and I have to say that I will be visiting this site more often. Thank you, whoever wrote this. Much appreciated.

    I would also like to thank Teunissen.

  5. #5 Arturo
    November 22, 2009

    Hi. Today, my best friend Liz told me about this web-site and it really helped me with researching the origin of the centripetal acceleration equation. I will probably never visit this website again however.

  6. #6 zed
    September 4, 2010

    I have a problem. I tried deriving this at home using differentiation and the formula for magnitude of vector diff. and I am getting acceleration as the product of angular velocity, constant tangential velocity and cos of half the product of angular velocity and time.
    Why am I getting that cos function? I am using differentiation, so I should get the correct answer, I suppose…

  7. #7 Ronnie
    May 18, 2011

    Many thanks !

  8. #8 Bruce Frykman
    October 2, 2011

    Great presentation but shouldn’t your phrase: “what if I took ? = 2??” instead read ? = 2PI?, since you did tell us that the angle in question was necessarily in radians?

  9. #9 Nivittina Alexander Jollykonj
    October 24, 2011

    great…..i had searched a lot for this….finally i got it from here…. ia am happy…but still i am little confused with that trick used there…any way great thanx.

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