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Why is the Acceleration vector perpendicular to the tangent ?

  1. Nov 29, 2011 #1

    Why is the Acceleration vector (second derivative) perpendicular to the tangent (first derivative) when velocity is constant ?

    I understand that when velocity is constant there is no acceleration but it's that vector being perpendicular to its tangent is confusing me.

    For example, if I have y=x^2 (a simple parabola), the second derivative will be y=2. Obviously Y=2 cannot be perpendicular to the tangent of every point on the parabola.

    What is it that I don't get?

  2. jcsd
  3. Nov 29, 2011 #2
    y=x^2 is not a vector function. You are heavily confused with the basics, you need to read your book again from the beginning of this chapter.
  4. Nov 29, 2011 #3


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    This is only true for circular motion or for an instant where the path is momentarily circular. For a parabola, acceleration is only perpendicular to the velocity at the peak of the parabola.
  5. Nov 30, 2011 #4
    Curl, I am confused with the basics, you're right and that's why I posted the question on this forum (I won't go and read an entire book if I can get a simple explanation by someone). I thought maybe someone could help me understand my confusion. Your comment is not fun to read, not helping at all, you're basically denigrating me. Why did you bother to answer at all if it's not to help me ?
  6. Nov 30, 2011 #5


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    When an object's velocity is constant, the object moves in a straight line at constant speed, and the acceleration is zero. There is no acceleration vector in this case. I suspect that you're confusing velocity (a vector quantity) with speed (a scalar quantity). Speed is the magnitude of velocity.

    As rcgldr noted, the acceleration vector is perpendicular to the velocity only when the object is (instantaneously at least) in circular motion at constant speed. In this case the speed is constant but not the velocity.

    Acceleration is related to change of velocity. Velocity includes both magnitude (speed) and direction. Therefore if either the speed or the direction, or both, changes, you have acceleration.
  7. Nov 30, 2011 #6
    Thank you jtbell, your answer is very helpful !!! I was asking this question because I wanted to understand how come the second derivative of a Helix parametric function is pointing towards the center of curvature. I can accept that it works mathematically but I'm still confused as the reason why this second derivative is pointing towards the center of curvature. Why is it pointing in that direction and not in another direction ?


  8. Nov 30, 2011 #7
    A Helical trajectory is a circular trajectory with motion perpendicular to the circle.

    In order to achieve constant circular motion, constant acceleration towards the centre of the circle is required. Thus, the instantaneous acceleration is at right angles to the direction of motion; in other words, the velocity vector is tangential to the circle of motion and the acceleration vector is normal to that - pointing at the centre.
  9. Nov 30, 2011 #8
    What I don't understand is specifically what you mention: "...constant acceleration towards the center of the circle is required" Why couldn't it be that what is required is a VELOCITY vector towards the center instead of ACCELERATION ?

    Why does the vector towards the center of motion "HAS" to be a vector of acceleration?

    thanks again,
  10. Nov 30, 2011 #9


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    By definition, the velocity vector points in the direction the particle moving in. In other words, it always points along the tangent of the particle's path.

    From your OP I'm guessing you are not going to be convinced by (and possibly not understand) an argument based on vector calculus, so think of it this way. Resolve the accleration into two components, along the tangent and perpendicular to it. The component along the tangent changes the speed of the particle, but not its direction of motion. The component perpendicular to the tangent changes the direction, but not the speed.

    Gonig back to your OP, an equation like y = x^2 is describing the shape of a curve. It doesn't say anything about the motion of something along the curve. To describe the motion you need two equations like
    x = f(t)
    y = g(t)
    that describe how x and y change with time. If you eliminate t from those two equations, you get the curve that the particle follows in space.

    The speed of the particle along the curve is then given by
    [tex]\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}[/tex]
    which is not the same as [itex]dy/dx[/itex].
  11. Nov 30, 2011 #10
    Hi Azoulay,

    Acceleration is a change in velocity. This can be either a change in speed or a change in direction or both.

    In order to keep an object moving in a circular path the direction of its motion has to be changed continuously which necessarily implies a constant change of velocity which implies a constant acceleration. In order to move in a circular path this acceleration must always be at 90 degrees to the current direction of travel.

    Hope this helps.
  12. Nov 30, 2011 #11
    Azoulay.... I don't think that I can answer your question in detail, I am happy with the case of circular motion but more importantly I also do not think it is very helpful to be advised to 'read your book from the beginning'.... I would hope that a forum such as this was to help and guide those who do not get everything from the text book.
  13. Nov 30, 2011 #12


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    Assume that the helix curves around the z axis as in this wiki article:


    If an object follows that helix path at constant speed, then the z component of velocity is constant, and the x and y components are circular. Acceleration is always in the direction of the x-y plane towards the z axis.

    Note that a spiral curve called involute of circle also has second derivative always perpendicular to tangent. If a moving object was attached to a string that wrapped or unwrapped around a post with no losses and infinitely thin string, the path is involute of circle, speed would be constant (until contact with the post).


    Last edited: Nov 30, 2011
  14. Nov 30, 2011 #13
    Of course mathematics will solve all of this if it's done correctly but it's very important to develop an intuition concerning this type of motion.

    Let's suppose you are driving your car in a straight line at constant speed. The car now has constant velocity also because neither the direction nor the speed are changing. There is no net force on the car in the direction of motion. (engine force balances friction forces) IF you want to change the direction of your car you must turn the steering wheel. This creates a force inward toward the center of the circle you will now travel in. The force is created by the friction on the tires. If the road is black ice and you turn the steering wheel you will continue to travel in the same straight line you were traveling in. You can try a similar experiment by twirling a ball tied to a string in a circle and then releasing the string. The ball will move tangent to the circle at the point where you released it. No first of all, no net force, no curved path.

    Since the force to cause a mass to move in a curved path must be in toward the center of the curve at that point and since force is proportional to acceleration, the acceleration must also be in toward the center of the curve also. It's much easier to visualize force than acceleration so any time you're interested in the direction of the acceleration, just think about the direction of the net force on the object to create the give motion and since mass is always positive, the acceleration must be in the direction of the net force.

    When an object moves, it's velocity is always in the direction of motion so if you visualize the motion, you have the velocity direction.

    Remember that velocity is a vector meaning it has both magnitude and direction or speed and direction. To change velocity you may change the speed, the direction or both and the force and the acceleration necessary to do this depend on what is happening at that point in time.

    In the case of a parabola the force will be variable so there will not be a constant speed, nor will there be a constantly changing direction so this is more complicated than the object which travels in a circle which will have both constant speed and constantly changing direction which produces a constant acceleration produced by a constant force toward the center of the circle.

    So consider the force producing the motion, the acceleration produced by the force and the speed of the object.

    and remember your definitions - acceleration, time rate of change of velocity and velocity, time rate of change of displacement, all of which are vector quantities.

    The object on a string twirling around a post is particularly interesting because there is no force in the direction of motion so the speed of the object can't change but there is obviously in inward force because the object gets closer and closer to the center pole and because the radius of the circle gets smaller the object makes more and more revolutions per second as it gets tighter.

    The individual who had both physical intuition and good mathematics skills always has a second way to check his/her results because if the answer does not agree with what you know is physically correct, you made a math error very likely.
    Last edited: Nov 30, 2011
  15. Nov 30, 2011 #14

    D H

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    There's nothing special about velocity and acceleration here. The derivative of any constant length vector is either zero or is perpendicular to that constant length vector.

    Let [itex]\mathbf q[/itex] be some vector with a constant but non-zero length. This vector can be written as the product of a constant scalar and a time varying unit vector: [itex]\mathbf q(t) = q \hat u(t)[/itex]. The magnitude of this vector, q, is constant, and thus so is q2. Another way to compute the square of the magnitude is to take the inner product with itself: [itex]q^2=\mathbf q(t)\cdot\mathbf q(t)[/itex]. Differentiating with respect to time,
    [tex]\frac {dq^2}{dt}
    = \dot{\mathbf q}(t)\cdot\mathbf q(t) + \mathbf q(t)\cdot\dot{\mathbf q}(t)
    = 2\,\mathbf q(t)\cdot\dot{\mathbf q}(t)[/tex]
    As q2 is constant, this time derivative must be zero:
    [tex]\mathbf q(t)\cdot\dot{\mathbf q}(t) = 0[/tex].
    Given that q is not zero, this inner product will be zero only if dq/dt is zero or if dq/dt is orthogonal to q.
  16. Dec 1, 2011 #15
    Well thank you all for all these GREAT explanations, I couldn't sleep after reading them all :-) it makes a lot more sense now.

    Last question promise:
    I made a relation between 2 concpets and I'm still unsure as if that relationship is good: first concept: acceleration pointing towards the center of motion and second concept: second derivative of a parametric helix equation pointing towards the center of curvature. Or those coincidences or are they related (since acceleration is also a second derivative)?
  17. Dec 1, 2011 #16
    If an object is accelerating, its acceleration is in the direction of the change of velocity, final velocity - initial velocity, the vector difference. If the object is accelerating in a straight line the acceleration will be in the direction of motion but if it is moving in a curved path it will be in toward the center of curvature of the path of the motion. In the direction of the force needed to create the curved path.
  18. Dec 1, 2011 #17


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    The second derivative of a parametric equation depends on the relation ship to t. For a helix defined by x(t) = cos(t), y(t) = sin(t), z(t) = t, it is the equivalent of an object moving at constant speed while circling the z axis, the second derivative is perpendicular to the first derivative, and the acceleration is perpendicular to the velocity (tangent). However if the helix equation is x(t) = cos(t2), y(t) = sin(t2), z(t) = t2, the path is the same, but the speed increases over time, and the acceleration would not be perpendicular to velocity.

    If acceleration is perpendicular to velocity, then speed is constant, and only direction changes. The path could be just about any shape, for example an object following the parabolic path y = 1/2 x2 at constant speed c, x component of velocity Vx = c / sqrt(1+x2), and y component of velocity Vy = c x / sqrt(1+x2) (the parametric form of this is messy).
    Last edited: Dec 1, 2011
  19. Dec 1, 2011 #18
    The Newtonian explanation (in regards to force) is now well :eek:) understood.

    This second derivative (of a helix equation), in terms of pure mathematics, is a vector towards the center of curvature.

    Are both notions related or simply coincidence?

  20. Dec 2, 2011 #19


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    In my previous post, I mentioned a parametric equation for a helix that would result in the second derivative not being a vector towards the center of curvature, because the speed along the path increases by a factor of 2t. Using vector notation:

    Code (Text):

    position     = {      cos(t^2),       sin(t^2),  t^2}
    velocity     = {  -2t sin(t^2),    2t cos(t^2),  2t}
    acceleration = {-4t^2 cos(t^2), -4t^2 sin(t^2),  2}
    I also mentioned an equation for a parabola (y = 1/2 x^2) where the second derivative would be a vector towards the center of curvature, although it's probably not possible to express this as a normal parametric equation:

    velocity = {c / sqrt(x^2 + 1), c x / sqrt(x^2 + 1) }

    The key factor is that the speed along the path of any curve must be constant in order for acceleration to always be perpendicular to velocity. It doesn't matter what the path is, as long as the path is reasonably smooth (no sharp corners).
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