1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Non uniform circular motion

  1. Jun 16, 2014 #1
    Here is the problem I am working on:

    A steel ball is accelerated around a circular track of radius 72 cm from rest. after 3 circuits the ball reaches a speed of 4.00 m/s, assuming tangential acceleration, [itex] a_{T} [/itex], is constant, how long after the ball starts moving is radial acceleration, [itex] a_r [/itex], equal to tangential acceleration?


    [itex] v_f = v_{\text{final}} = 4.00\, \text{m/s} [/itex]

    [itex] v_i = v_{\text{initial}} = 0.00\, \text{m/s} [/itex]

    [itex] r = 0.72\, \text{m} [/itex]

    relevant equations:

    1. [itex] C = 2\pi r [/itex]

    2. [itex] {v_f}^2 - {v_i}^2 = 2{a_{T}}d\cos (\theta) [/itex]

    3. [itex] a_r = \frac{v^2}{r} [/itex]

    4. [itex] v = a_{T}t + v_0 [/itex]


    So what I first did was solve for the total distance travelled by the ball in preparation of using 2. to solve for [itex] a_{T} [/itex];

    [itex] d = (3)C = 6\pi (0.72\, \text{m}) \approx 13.56\, \text{m} [/itex]

    So then I rearrange 2. noting that [itex] \theta = 6\pi [/itex];

    [itex] a_{T} =\frac{{v_f}^2}{2d} = \frac{(4.00 \, \text{m/s})^2}{2(13.56\, \text{m})} \approx 0.589\, \text{m}/\text{s}^2 [/itex]

    then setting [itex] a_r = a_{T} [/itex] and substituting into 3. I get:

    [itex] v = \sqrt{ra_{T}} = \sqrt{(0.72\, \text{m})(0.589\, \text{m}/\text{s}^2)} \approx 0.65 \, \text{m/s} [/itex] and thus using 4. to solve for time,

    [itex] t = \frac{v}{a_{T}} = \frac{0.65\, \text{m/s}}{0.589\, \text{m}/\text{s}^2} \approx 1.1 \, \text{s} [/itex].

    However, the answer in the back of the book says i should be getting 0.18 s. any help would be greatly appreciated!

    EDIT: I feel this would be better in the hoework section, however i do not know how to move it.
    Last edited: Jun 16, 2014
  2. jcsd
  3. Jun 16, 2014 #2


    User Avatar
    Science Advisor

    I'm more "mathematics" than "physics" but here is how I would do this problem:
    Write [itex]x= Rcos(\omega(t))[/itex], [itex]y= R sin(\omega(t))[/itex]. Here, [itex]\omega(t)[/itex] is an unknown function of t, not [itex]omega[/itex] times t as it would be for constant angular speed.

    Then the components of velocity are given by the derivatives: [itex]v_x= -Rsin(\omega(t))\omega'(t)[/itex] and [itex]v_y= R cos(\omega(t))\omega'(t)[/itex]. Note that the dot product of the position vector, [itex]Rcos(\omega(t))\vec{i}+ Rsin(\omega(t))\vec{j}[/itex] with the velocity vector, [itex]-Rsin(\omega(t))\omega'(t)\vec{i}+ Rcos(\omega(t))\omega'(t)\vec{j}[/itex], is 0. The two vectors are perpendicular.

    The speed, the length of the velocity vector, is [itex]R\omega'(t)[/itex] and we are told that "after 3 circuits the ball reaches a speed of 4.00 m/s". That is, when [itex]\omega(t)= 6\pi[/itex], [itex]R\omega'(t)= 4.00[/itex].

    The acceleration vector is given by the second derivatives: [itex]a= (-R cos(\omega(t)(\omega'(t))^2- R sin(\omega(t))\omega''(t))\vec{i}+ (-R sin(\omega(t))(\omega'(t))^2+ Rcos(\omega(t))\omega''(t))\vec{j}[tex]

    That can be split into two parts in an obvious way: [itex](-R cos((\omega(t))\vec{i}- R sin(\omega(t))\vec{j})(\omega'(t))^2[/itex], which is (anti)parallel to the position vector, and so radial, and [itex](-R sin(\omega(t))+ R cos(\omega(t))\omega''(t)[/itex]. The dot product of those vectors is 0 so those are the radial and tangential components of acceleration.

    The radial and tangential components will be equal when [itex]-R cos(\omega(t))(\omega'(t))^2= -R sin(\omega(t))\omega''(t)[/itex] and [itex]-R sin(\omega(t))(\omega'(t))^2= R cos(\omega(t))\omega''(t)[/itex].

    In this particular problem, we are told that the tangential acceleration, [itex]-R sin(\omega(t))\omega''(t)\vec{i}+ R cos(\omega(t))\omega''(t)[/itex] is constant.
  4. Jun 17, 2014 #3
    @Hallsofivy Albeit, I do understand the math, like where your original functions come from, and how/why you get what you do for tangential and radial acceleration, there is something that I am missing; for when I use your result to solve for tangential, I get Rw''(t).
  5. Jun 17, 2014 #4
    Hopefully someone can show us the light on this one. I am getting a third answer (roughly ##1.5\,s##) using polar coordinates and the equations

    \mathbf{r} &= r \,\mathbf{\hat r} \\
    \mathbf{\dot{r}} &= \dot r \mathbf{\hat{r}} + r \dot{\theta} \boldsymbol{\hat{\theta}} \\
    \mathbf{\ddot{r}} &= (\ddot r - r \dot{\theta}^2)\mathbf{\hat{r}} + (r\ddot{\theta} + 2 \dot{r} \dot{\theta}) \boldsymbol{\hat{\theta}}
  6. Jun 17, 2014 #5

    I just took another stab at it and ended up with $$t=\frac{2\sqrt{3\pi}}{v_\text{f}}$$ which, when i plug in the given values, gives me [itex]1.53 \text{s}[/itex].
  7. Jun 17, 2014 #6
    Sorry. I forgot to multiply by ##r## at one point. I actually get the same answers as you originally (for both ##a_T## and the final time ##1.1\,s##.

    Edit: Maybe we could get a third person to compute the solution, ideally with their own method.
    Last edited: Jun 17, 2014
  8. Jun 17, 2014 #7
    whoops, in lol my second attempt I did, I must have made the same mistake, I was off by a factor of R the answer I have is simply $$t=\frac{2r\sqrt{3\pi}}{v_f}= 1.1\, \text{s}$$

    EDIT: The book I am using, "Physics for Scientists and Engineers" by Nelson is a first edition, and I have already found numerous mistakes so this may just be another mistake in the answer key, however since I am rather new to physics and not yet all that confident in my work, I cannot say for certain that the answer in the book is wrong. Thanks a lot everybody :)
    Last edited: Jun 17, 2014
  9. Jun 17, 2014 #8


    User Avatar
    Gold Member

    Hmm... why not try using some rotational kinematic analogues?

    Translational: ##\Delta x = v_{x_{0}}t + \frac{1}{2} a_{x} t^2##
    ##\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ##& ##\Delta y = v_{y_{0}}t + \frac{1}{2} a_{y} t^2##
    Rotational: ##\ \ \ \Delta \theta = \omega_{0}t + \frac{1}{2} \alpha t^2 ##

    this allows you to solve for your angular acceleration using your initial conditions. alpha is a constant, whereas a is a function of t. I'm assuming that the question wants the time where ##|a| = \alpha ## correct? otherwise you left out a crucial piece of information: the initial displacement.

    Maybe try using this. I'll work through it and see what I get as well. Rendezvous in 5
  10. Jun 17, 2014 #9
    I wonder if it would be okay if I provided my full solution. I get the same answers as the OP and I have a hard time figuring out where (if anywhere) I could have possibly gone wrong.
  11. Jun 17, 2014 #10
    Well, this isn't for class or anything, so I don't have a problem with it. I took physics first term of last year at university, but had to drop it due to unforeseen circumstances, unfortunately. So I'm trying to learn some more during the summer before I take it again in the fall.
  12. Jun 17, 2014 #11
    I can remember a little bit about rotational stuff from high school, however the subject has not yet come up in the book that I am working through, so was just sticking with what i have read so far.
  13. Jun 17, 2014 #12


    User Avatar
    Gold Member

    So I guess my question is then what HAVE you covered in this book? I'm ~ a junior/senior level physics major (BS), so what I have in my arsenal may very well be different from what you're trying to use.
  14. Jun 17, 2014 #13
    so far 2D kinematics, projectile motion, that sort of stuff. just started circular motion and relative motion in 2 and 3 dimensions yesterday. So far there has been nothing in the book pertaining to angular velocity and such.
  15. Jun 18, 2014 #14
    So, I haven't heard any objections and since I am getting the same (wrong?) solution as the OP, I will go ahead and post my solution. I will remove it upon request.

    We use polar coordinates where we have
    \mathbf{r} &= r \,\mathbf{\hat r} \\
    \mathbf{\dot{r}} &= \dot r \mathbf{\hat{r}} + r \dot{\theta} \boldsymbol{\hat{\theta}} \\
    \mathbf{\ddot{r}} &= (\ddot r - r \dot{\theta}^2)\mathbf{\hat{r}} + (r\ddot{\theta} + 2 \dot{r} \dot{\theta}) \boldsymbol{\hat{\theta}}
    Since we have circular motion ##r## does not change, i.e., ##\dot r = \ddot r = 0## identically. Thus
    a_T &= r \ddot{\theta} \\
    a_r &= r \dot \theta^2
    We know that ##a_T## is constant, so we can integrate the first equation twice to obtain
    \dot \theta &= \frac{a_T t}{r}, \\
    \theta &= \frac{a_T t^2}{2r}
    Now, at some time ##s## we know that ##\theta(s) = 6\pi## and ##v(s) = v_T(s) = r \dot \theta(s) = 4.0\, m/s##. We get a system to solve for the unknowns ##s## and ##a_T##
    4 &= a_T s, \\
    12\pi r &= a_T s^2,
    s &= 3 \pi r\\
    a_T &= \frac{4}{3 \pi r}.
    Finally we use ##a_r = r \dot \theta^2## and ##a_r = a_T## to obtain
    a_T = a_r = r \dot \theta^2 = \frac{a_T^2 t^2}{r},
    t^2 = \frac{r}{a_T} = \frac{3}{4} \pi r^2 \approx 1.22,
    where we have used ##r = 0.72\, m##. Taking the square root gives ##t \approx 1.11\, s##. I would love to hear where this could possibly have gone wrong.

    Edit: typo.
    Last edited: Jun 18, 2014
  16. Jun 18, 2014 #15


    User Avatar
    Gold Member

    When you plugged r double dot = 0 into your r double dot equation, how exactly did you get that a_T = r theta double dot? did you take the magnitude? one (negative) r theta double dot is attached to rhat, and 2 r theta double dot is attached to theta [edit] hat[]. You can't just neglect the unit vectors, you can take the magnitude of the vector, but you can't just let them disappear.

    also, in your equation before your final solution, where did the a_T^2 come from in (a_T^2)*(t^2)/r; should it not just be a_T? Working from your final equalities to your final solution, I get r thetadot /a_T = t, not t^2, and I'm assuming the lack of theta dot in your final solution is a typo, as you used the substitution r theta dot = 4 to get your answer.
  17. Jun 18, 2014 #16


    User Avatar
    Gold Member

    2. vf2−vi2=2aTdcos(θ)

    where did you get this equation matinee? I'm not familiar with it, and I don't like it. That cos theta really messes stuff up i.m.o.

    Lets look at our situation, but let the ball progress around the track another pi radians.

    Theta is now 7 pi, v_i still equals 0 so v_f^2 = 2a_T*d*cos(7pi)

    v_f^2 (now a scalar, and always positive) = 2a_T*d*(-1)

    2 is positive and d is positive, which means that either v_f at 7pi is imaginary, or a_T is negative, but if a_T is negative, that means that it is a vector, and you have a scalar equal to a vector. Maybe I'm being obtuse about this, and if so, could someone enlighten me? Is a_T a signed scalar? what is the sign of a_T at multiples of pi/4 ?
  18. Jun 18, 2014 #17
    ##\mathbf{\ddot{r}}## is the acceleration vector. The terms in front of ##\mathbf{\hat{r}}## in the RHS constitute the radial acceleration and the terms in front of ##\boldsymbol{\hat{\theta}}## constitute the tangential acceleration. Some of them vanish because ##r## is constant. Also note that anything not in bold is scalar-valued.

    To answer your latter question, I used the equation ##\dot \theta = \frac{a_T t}{r}## from above.

    Edit: Maybe I should have written ##|a_T| = |a_r|## to avoid problems with the sign.
    Last edited: Jun 18, 2014
  19. Jun 18, 2014 #18


    User Avatar
    Gold Member

    ahhh I see what you did there. Misread it the first time, I thought you were trying to solve for theta dot and theta double dot. Sorry about that.

    ##r \dot{\theta}^2 = \frac{a_{T}^2t^2}{r}## =>
    ##\sqrt{r^2 \dot{\theta}^2} = \sqrt{a_{T}^2t^2}##
    ##r\dot{\theta} = a_{T}t##
    ##\frac{r\dot{\theta}}{a_{T}} = t = \frac{4 \frac{m}{s}}{\frac{4}{3 \pi r} \frac{m}{s^2}} = 3 \pi r\ s \approx 6.78s \approx t##
    Last edited: Jun 18, 2014
  20. Jun 18, 2014 #19
    At this point the constant ##a_T## has been computed and is known and I am looking for the time ##t## such that ##|a_T| = |a_r|##. I know the expression ##a_r = r \dot \theta^2## and I also know ##\dot \theta = \frac{a_T t}{r}## from above. Thus ##a_r = \frac{a_T^2 t^2}{r} ## which I was supposed to equate with ##a_T## and find ##t##.
    Last edited: Jun 18, 2014
  21. Jun 18, 2014 #20
    That equation comes from starting with [itex]d = v_{i}t+\frac{1}{2}at^2 [/itex] and [itex]v_f = v_i + a t [/itex], then using the second to solve for time, ie [itex] t = \frac{v_f - f_i}{a} [/itex] and substituting that into the first equation. If we are using vectors we end up with [itex]\widehat{v}_f-\widehat{v}_i = 2\widehat{a}_T\bullet \widehat{d}[/itex]

    so I just used the magnitude version of the dot product since i am not dealing with vectors. However your example kind of makes me think it's not right?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Threads - uniform circular motion Date
Uniform Circular Motion Problem Nov 8, 2017
Uniform Circular Motion, Acceleration problem May 9, 2017