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Homework Help: How to Add Acceleration Vector in Uniform Circular Motion

  1. Aug 27, 2014 #1
    Currently we're discussing uniform circular motion in my physics class.

    The previous unit discussed vectors and vector addition in the i, j, k format. When I try to apply the rules for vector addition to find the resultant velocity in uniform circular motion, I get an increase in the magnitude of the velocity vector.

    For example, say an object is in uniform circular motion around a circle of radius 2m with a tangential velocity of 6m/s. It's centripetal acceleration, v^2 / r, is therefore 18m/s^2 perpendicular to the direction of motion.

    So if the tangential velocity = 6i + 0j, and if the centripetal acceleration = 0i + 18j, the resultant vector would have a magnitude of sqrt(6^2 + 18^2) or about 19, which clearly isn't uniform circular motion.

    How should I think about this problem differently? Is it correct to add the initial velocity vector and the acceleration vector to find the final velocity, even though the units are different?
  2. jcsd
  3. Aug 27, 2014 #2
    Your understanding of circular motion seems just a bit off, but no worries.

    To your question about adding vectors, it's an error to add the velocity and acceleration vectors together. Like you mentioned, the units are different, so it doesn't work to add them. Just as it's an error to add length and area together, or time and mass, etc. You can always multiply two quantities together, but you can only add two quantities if they have the same units, or in other words are the same type of quantity.

    And then to your question about final velocity, think about what uniform circular motion means. Velocity could only increase if there was some tangential acceleration, so if some of the acceleration pointed along the circle (tangentially) instead of directly to the center (perpendicularly). What's the case with the problem here?
  4. Aug 27, 2014 #3
    Our book actually discusses non-uniform circular motion too, and resolves the acceleration into a radial and tangential acceleration.

    So in uniform circular motion, there is no tangential component to the acceleration, and therefore the speed is not affected.

    What I don't understand is how mathematically to go from the tangential velocity vector and the centripetal acceleration vector to the resulting tangential velocity vector at another point along the circle.

    Is not the definition of the acceleration vector the final velocity vector minus the initial velocity vector, and therefore is not the final velocity vector the initial velocity vector plus the acceleration vector?
  5. Aug 27, 2014 #4


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    That's the problem. You can't add velocity to acceleration, it just doesn't make sense.
    (Whether you're adding vectors or scalars, the units need to be the same)

    Isn't time involved? Shouldn't the new velocity be after a certain amount of time?
    The correct equation should be [itex]\vec{v_f}=\vec{v_i}+\vec{a}t[/itex] Where t is the time that passed.

    So, you could just multiply the acceleration by 1 second and then it would have units of velocity, and then you would get the same (incorrect) answer that you got. Then, why is it wrong?

    The answer involves my above question of why it is wrong to multiply the acceleration by 1 second and then add it to the velocity. Why might that give the wrong answer?
  6. Aug 27, 2014 #5
    Because by that logic we could multiply the acceleration by an arbitrary number of seconds, resolve it to the same unit as velocity, and then get an infinitely many different number of final velocities.

    With the [itex]\vec{v_f}=\vec{v_i}+\vec{a}t[/itex] definition of final velocity, it seems to me like we'd want the instantaneous acceleration, or the limit as time goes to zero, because the direction of acceleration changes at every instant along the circle. However I'm not sure how to formally approach the problem with that reasoning.
  7. Aug 27, 2014 #6


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    Exactly! :smile:

    As far as actually calculating it, I prefer to do it the other way around.
    That is, you're starting out with the centripetal acceleration equation and then trying to use it find out the velocity at another point (which should have a constant speed and a tangential direction).
    If I were you, I would start out with the fact that the speed is constant (and the path is circular) and try to derive the centripetal acceleration formula that makes that true.

    I find this to be a much more useful exercise.
    (Plus, if you ever forget the centripetal acceleration equation, you'll know what to do!)
  8. Aug 27, 2014 #7
    Well, like you said, if the situation is UCM like the example you gave, then speed won't change, and so there's no calculation to be done.

    If the situation is non-UCM, and you want find the final velocity at some point, what you first want to do is resolve the acceleration into tangential and radial components, because only the tangential component will change the velocity. After that, it's going to depend on other given information. If you have a change in time, and the components of the acceleration don't change, then you can use some kinematic formulas to find the final velocity. Otherwise, it will require some analysis of the forces at work. Because the velocity is increasing, the radius of the path will also increase, so the distance traveled will be difficult to find. Also, forces at work may change in their contributions to the centripetal and tangential accelerations over time. Do you have a specific problem involving final velocity to work through?

    EDIT: Yes, the formula Nathanael gave above is correct for the final velocity in vector form. Finding the final velocity will also involve whether or not you want it as a vector, or just as a magnitude, and, of course, whether it's UCM or non-UCM.
    Last edited: Aug 27, 2014
  9. Aug 27, 2014 #8
    I think the basis of my understanding of velocity and acceleration is false. I thought that the acceleration vector was defined as the final velocity vector minus the initial velocity vector. My textbook has diagrams like this one that label the change in velocity vector as the acceleration vector.

    I'd love to be able to derive it myself. Can you point me in the direction about how to start the process? I'm not sure how to approach the problem.

    How can I mathematically or pictorially express the fact that some acceleration, when applied to an initial velocity, leads to a final velocity of equal magnitude but changed direction along a circular path?

    When is that the case? If there's an orbiting planet? If I was spinning an object in a circle with a rope, or driving a car around a circular track, the velocity could change without increasing the radius of the path.

    No, I just want to wrap my head around how the resulting velocity and centripetal acceleration are related.
  10. Aug 27, 2014 #9
    Indeed -- it can be difficult to confuse the change in velocity with the acceleration. Mathematically:

    ##\Delta v = at##

    Acceleration is in fact a scaled copy of the change in velocity, where the scaling factor is time. This is how the different units come into play, even though ##\stackrel{\rightarrow}{\Delta v}## has the same direction as ##\stackrel{\rightarrow}{a}##.

    Simple -- the definition of UCM is that ##a_{r} = \frac{v^{2}}{r}##. If that's true, then you've got UCM. Alternatively, you can assert that velocity is perpendicular to acceleration for a given time period. Mathematically, if:

    ##\vec{v}(t) \cdot \vec{a}(t) = 0, t_{i} \leq t \leq t_{f}##

    Then you have UCM for that interval.

    What I meant is that if velocity increases and centripetal acceleration stays the same, then the radius must increase, simply due to the formula ##a = \frac{v^{2}}{r}##. For your examples, when velocity increases, the centripetal acceleration increases through the centripetal force. Spinning an object with a rope faster and faster creates more tension in the rope (which is why ropes break when you spin something too fast), and driving faster in a circle increases the friction between the road and the tires (which is why eventually friction can't hold the car in the circle, and the car starts sliding outward).
  11. Aug 27, 2014 #10


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    Acceleration is change in velocity per change in time. (Technically this is "average acceleration")

    So acceleration is closely related to the [itex]Δ\vec{v}[/itex], but it's not the same. (You have to divide by Δt to get to the acceleration.)

    Ok, I'll give you a hint, but my hint might not help much cause I'm trying to not ruin it :smile:

    I would consider a small time Δt, and then work out what the acceleration needs to be to keep the object on the circle, then take the limit as Δt→0

    (When I first derived this formula, it took me at least several weeks! And it all started with questions similar to yours.)
    (My motivation was my passionful hate for memorizations :tongue:)

    If you can't figure it out, let it sit on the back of your mind for a while.
  12. Aug 28, 2014 #11


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    As a hint for understanding acceleration, consider this equation.

    Δv=a Δt

    A change in velocity results from acceleration acting over a small time interval. If the acceleration is changing over time then you won't be accurate if the Δt is large.

    So circular motion obviously involves the acceleration changing in direction over time. So first you start drawing a bunch of tangent vectors at time t1, then t2, then t3, and so on. You make them closer and closer together, making Δt smaller and smaller. Then you draw the acceleration vector that on average over those smaller and smaller times gives that change in velocity, Δv. The acceleration will rotate.

    But in circular motion, the tip of the velocity vector goes through a circle. So too does the tip of the acceleration vector. And with a little effort you can figure out the radius of each of those circles, and so the magnitude of the vectors.
  13. Aug 28, 2014 #12


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    The equation [itex]\vec{v}_f=\vec{v}_i+\vec{a}t[/itex] only holds for constant acceleration. I suggest you write it as [itex]\vec{v_f} \cong \vec{v_i}+\vec{a}\delta t[/itex] where ##\delta t## is small enough so that taking the acceleration as constant is a good approximation.

    If you square both sides, you get ##\vec{v}_f^2 \cong (\vec{v}_i + \vec{a}\delta t)^2##. Expand out the righthand side and keep in mind as ##\delta t \to 0## only terms up to first-order matter.
  14. Aug 28, 2014 #13


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    Is this perhaps what you are talking about? (Not the video, but the method.)

    There's also an easy way with just algebra and basic trig (and, at the end, a limit) if you assume acceleration is constant.
  15. Aug 28, 2014 #14
    Thank you guys very much for helping me head in the right direction on this.

    I'll keep looking into this on my own and wrestle with it some more and get back to you!
  16. Sep 22, 2014 #15
    Was very excited when I figured this out! Was this the direction in which you intended to point me?

    Centripetal Acceleration Derivation.jpg
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