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Pinned massless stick with a concentrated mass at the other end.

  1. Jan 3, 2014 #1
    1. The problem statement, all variables and given/known data
    90sf8l.png

    A massless stick is pinned at point A. And it has a concentrated ball mass of 1kg attached to it at point B.

    Given the initial values:
    Initial angular position θ=45◦
    Initial angular velocity ω=0
    Initial angular acceleration α=0

    Show that because of the setup of the problem, the concentrated mass B follows a circular path of:
    [tex]x = \sqrt{2}cos\theta[/tex][tex]y = \sqrt{2}sin\theta[/tex]

    2. Relevant equations
    The tangential acceleration is induced by the gravitational force on the mass:
    [tex]ma_t = mgcos\theta[/tex]

    The stick reacts to the compressive force induced by the ball mass, which is also the centripetal acceleration:
    [tex]ma_c = AB = mgsin\theta[/tex]

    3. The attempt at a solution
    Find the forces acting at point B:
    [tex]\sum F_x= ma_x[/tex][tex]AB_x = ma_x[/tex][tex]ABcos\theta = ma_x[/tex][tex]mgsin\theta cos\theta = ma_x[/tex][tex]a_x = gsin\theta cos\theta[/tex]

    [tex]\sum F_y= ma_y[/tex][tex]AB_y - mg = -ma_y[/tex][tex]-ABsin\theta + mg = ma_y[/tex][tex]m(g - gsin^2\theta) = ma_y[/tex][tex]a_y = g(1 - sin^2\theta)[/tex][tex]a_y = gcos^2\theta[/tex]

    To find the x and y positions simply double integrate the respective accelerations ax and ay with respect to time.
    [tex]x = \int \int a_x = \int \int gsin\theta cos\theta dt^2[/tex][tex]y = \int \int a_y = \int \int gcos^2\theta dt^2[/tex]

    However, the problem is that θ is a function of time or θ(t). In addition, although there is a constant vertical gravitational acceleration, the angular acceleration is not constant. Therefore, the equation for constant angular acceleration is not applicable:
    [tex]\theta = \frac{1}{r}(\theta _0 + \omega _0 t + \frac{1}{2}\alpha _0 t^2)[/tex]

    So, in order to find θ, i have to use the tangential acceleration relation with the angular acceleration:
    [tex]r\alpha = a_t[/tex][tex]\alpha = \frac{a_t}{r}[/tex][tex]r\theta = \int \int \frac{a_t}{r}dt^2[/tex]

    However, because the tangential acceleration is known in terms of θ, it sets me back to my original problem that i am trying to integrate θ with respect to time:
    [tex]\theta = \frac{1}{r^2}\int \int gcos\theta dt^2[/tex]
     
  2. jcsd
  3. Jan 3, 2014 #2

    mfb

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    If you posted the full problem statement, you don't have to find the time-evolution (= θ(t)) at all.

    To get θ(t) (if you really need it), you can find a differential equation and solve this. Alternatively, use energy conservation to find ##\dot{θ}(θ)## and solve this backwards to get t(θ) and then θ(t).
     
  4. Jan 5, 2014 #3
    @mfb
    When you are talking about the energy equations, i believe you are referring to:[tex]\Delta KE = 0[/tex][tex]\frac{1}{2}I\omega _f^2 - \frac{1}{2}I\omega _i^2= 0[/tex][tex]\frac{1}{2}I(\omega _f^2 - \omega _i^2)= 0[/tex]

    However, the problem is finding the angular velocity:[tex]\omega = \int \alpha dt[/tex][tex]\omega = \int \frac{1}{r}gcos\theta dt[/tex]

    Which i am back to round one. Maybe there is a way to go from position, velocity and acceleration without going through integration/differentiation with respect to time; which i don't know. Can you please show me your method?
     
    Last edited: Jan 5, 2014
  5. Jan 5, 2014 #4

    mfb

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    That is wrong, there is a second form of energy involved in your setup.

    There is an easier way to find the angular velocity for a given angle, you don't need to integrate anything.

    Energy conservation.


    Please post the full problem statement. It's still unclear what exactly you are supposed to do.
     
  6. Jan 5, 2014 #5
    @mfb
    That is the full problem statement because I made it up. I want to mathematically show that the path taken by the ball mass at the end of the pinned stick is:[tex]x=\sqrt{2}cos\theta[/tex][tex]y=\sqrt{2}sin\theta[/tex]Is there any unknown or missing variable you want to add to the problem?

    It is true that i can find the angular velocity of a given angle, but i want to be able to setup the equations for any angle. As for the energy equation, are you referring to this one?[tex]\Delta mgh = \Delta \frac{1}{2}mv^2 + \Delta \frac{1}{2}I \omega^2[/tex]If not, please just tell me the equation.
     
    Last edited: Jan 5, 2014
  7. Jan 5, 2014 #6

    TSny

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    From your original picture, use trigonometry to find the ##x## coordintate of the ball in terms of ##l## and ##\theta##.

    Similarly for the ##y## coordinate.
     
  8. Jan 5, 2014 #7
    Ok. Thanks to your reply TSny, i think i know what mfb meant by the problem statement being incomplete.

    So, because of the problem setup we can agree that the motion path of the ball mass will be of circular motion of the form:[tex]x=rcos\theta[/tex][tex]y=rsin\theta[/tex]Where θ is a function of time t.


    So, if the problem were to be of constant speed, the motion path would be:[tex]x=rcos(\theta_0+\omega_0 t)[/tex][tex]y=rsin(\theta_0+\omega_0 t)[/tex]

    If, the problem were to be of constant acceleration, the motion path would be:[tex]x=rcos(\theta_0+\omega_0 t+\frac{1}{2}\alpha_0 t^2)[/tex][tex]y=rsin(\theta_0+\omega_0 t+\frac{1}{2}\alpha_0 t^2)[/tex]

    So, the question for this problem should be, find exact form of the motion path. In order to do so, i have to find what is θ(t) first. I will update the original post as well.

    ------------------------------------------------------------------------------------------

    In addition, i believe that my previous energy equations are wrong. They should be:[tex]\Delta PE + \Delta KE = 0[/tex]Therefore, [tex]mg(h_f - h_i) + \frac{1}{2}I(\omega_f ^2 - \omega_i ^2) = 0[/tex]I had the kinetic energy repeated twice in my previous equation:[tex]\frac{1}{2}mv^2 = \frac{1}{2}mr^2 \omega^2 = \frac{1}{2}I \omega^2[/tex]

    Edit: How do i update the original post?
     
  9. Jan 5, 2014 #8

    TSny

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    I believe it's too late to edit your original post. But that's ok.

    Your energy equation looks good. Note that you can let h = y, which can be expressed in terms of θ.

    To find θ(t) you can try mfb's suggestion in post #2. However, I believe that it will not be possible to express θ(t) in terms of "elementary" functions.
     
  10. Jan 6, 2014 #9

    mfb

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    That is the same thing, as you can keep the "given angle" as variable.


    That is the last relevant step (and it is a step you never mentioned before as goal). Once you have θ(t), you are done.

    Looks good.

    As this is not a "real" homework question, I have the same feeling. I had a similar problem in an exam long ago, and we were supposed to (and had to) make approximations to get an answer. It was not the same problem, but it was so similar I would be surprised if that makes a difference in terms of analytic solutions.
     
    Last edited: Jan 6, 2014
  11. Jan 7, 2014 #10
    I have tried various things and they don't seem to come out quite right. Maybe you guys can help me spot any mistakes.[tex]\alpha=\frac{a_t}{r}=\frac{gcos\theta}{r}[/tex]
    If:[tex]x=rcos\theta[/tex][tex]a_x=gcos\theta sin\theta[/tex]So:[tex]rcos\theta=\int\int gcos\theta\sin\theta dt^2[/tex][tex]\frac{d^2}{dt^2}(rcos\theta)=gcos\theta\sin\theta[/tex][tex]\frac{d}{dt}(-rsin\theta\frac{d\theta}{dt})=gcos\theta\sin\theta[/tex][tex]-rcos\theta(\frac{d\theta}{dt})^2-rsin\theta\frac{d^2\theta}{dt^2}=gcos\theta\sin\theta[/tex][tex]-rcos\theta\omega^2-rsin\theta\alpha=gcos\theta\sin\theta[/tex][tex]-rcos\theta\omega^2=gcos\theta\sin\theta+rsin\theta(\frac{gcos\theta}{r})[/tex][tex]-rcos\theta\omega^2=2gcos\theta\sin\theta[/tex][tex]\omega^2=\frac{-2gsin\theta}{r}[/tex]
    But if i try the same with:[tex]y=rsin\theta[/tex][tex]a_y=gcos^2\theta[/tex]I get:
    [tex]rsin\theta=\int\int gcos^2\theta dt^2[/tex][tex]\frac{d^2}{dt^2}(rsin\theta)=gcos^2\theta[/tex][tex]\frac{d}{dt}(rcos\theta\frac{d\theta}{dt})=gcos^2\theta[/tex][tex]-rsin\theta(\frac{d\theta}{dt})^2+rcos\theta\frac{d^2\theta}{dt^2}=gcos^2\theta[/tex][tex]-rsin\theta\omega^2+rcos\theta\alpha=gcos^2\theta[/tex][tex]-rsin\theta\omega^2=gcos^2\theta-rcos\theta(\frac{gcos\theta}{r})[/tex][tex]-rsin\theta\omega^2=0[/tex][tex]\omega^2=0[/tex]

    I don't get the same answer. So, there is a possibility i left a minus sign somewhere. So let's assume this instead:[tex]-rsin\theta\omega^2=gcos^2\theta+rcos\theta(\frac{gcos\theta}{r})[/tex][tex]-rsin\theta\omega^2=2gcos^2\theta[/tex][tex]\omega^2=\frac{2gcos^2\theta}{-rsin\theta}\frac{sin\theta}{sin\theta}[/tex][tex]\omega^2=\frac{-2gsin\theta}{r}cot^2\theta[/tex]Which is still different by cot2θ.
     
  12. Jan 7, 2014 #11

    TSny

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    You have assumed that ax comes entirely from the x-component of the tangential acceleration at. But the ball also has centripetal acceleration that will contribute an additional part to ax. If you include that and go through a similar calculation to what you did, I think you will just end up with an identity: 0 = 0.
     
  13. Jan 7, 2014 #12
    Actually, if you review my opening post, ax comes from the x-component of the centripetal acceleration ac. The centripetal force AB is the mass times the centripetal acceleration ac.

    The reason why i put the angular acceleration relationship with the tangent acceleration at is because it is later on useful to substitute:[tex]\frac{d^2\theta}{dt^2}=\alpha=\frac{gcos\theta}{r}[/tex]

    On the other hand, the y-component acceleration is derived by the centripetal reaction plus the gravitational force acting on the mass.

    Also, using the energy equation i get similar results, unless i did something wrong:[tex]mg\Delta h+\frac{1}{2}I\Delta\omega^2=0[/tex][tex]\Delta\omega^2=\frac{-2mg\Delta h}{I}[/tex][tex]\omega^2=\frac{-2mgrsin\theta}{mr^2}[/tex][tex]\omega^2=\frac{-2gsin\theta}{r}[/tex]
    So I think there is something wrong on the y and ay side instead.

    Edit: The magnitude of the resultant x and y accelerations is equals to the tangential acceleration:[tex]\sqrt{a_x^2 + a_y^2}=a_t[/tex][tex]\sqrt{g^2cos^2\theta sin^2\theta + g^2cos^4\theta}=gcos\theta[/tex][tex]\sqrt{g^2cos^2\theta(sin^2\theta + cos^2\theta)}=gcos\theta[/tex][tex]\sqrt{g^2cos^2\theta(1)}=gcos\theta[/tex][tex]gcos=gcos\theta[/tex]
     
    Last edited: Jan 7, 2014
  14. Jan 7, 2014 #13

    TSny

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    Oh, I misinterpreted how you got ##a_x##. I thought you took the x-component of ##a_t## which would be ##g\cos\theta\cdot\sin\theta## which is the same as the expression you wrote for ##a_x##.

    Looking back at the original post, I don’t understand your analysis of the centripetal acceleration, ##a_c##. There are two separate forces acting on the ball that produce the centripetal acceleration: (1) the centripetal component of the force of gravity (acting toward the center) and (2) the force that the stick exerts on the ball (which could act away from the center or toward the center depending on where the ball is located in its fall.)

    Force (1) is ##mg\sin\theta## and force (2) I'll write as ##F_{\rm stick}##.

    Newton’s second law for the centripetal direction is ##\sum F_c = ma_c##.

    This gives ##mg\sin\theta + F_{\rm stick} = ma_c##, where ##F_{\rm stick}## is positive if it acts toward the center and negative if it acts away from the center.

    So, it is not true that ##ma_c = mg\sin\theta## (except at the one point during the fall where ##F_{\rm stick}=0##).
     
  15. Jan 7, 2014 #14
    By doing it your way, if you look only at the centripetal forces, the sum of all the centripetal forces must equal to zero. That is because the resultant does not move into or away from the distance of the stick. So using your formula:

    ##mgsin\theta + F_{stick} = 0##
    ##F_{rm stick} = -mgsin\theta##

    In other words, the sticks reacts on a opposite force to the centripetal component of the gravitational force of the mass.

    The way i originally did it was to project the gravitational acceleration of the mass onto a centripetal and tangent component. That is because the mass causes both the centripetal and tangential force. Without the mass, the massless stick would not have any tangential or centripetal force. If the mass were to be alone by itself, it would freefall. The combination of both creates this mass falling on a circular motion kind of fashion.
     
  16. Jan 7, 2014 #15

    TSny

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    If the sum of the centripetal forces equals zero, how could there be a nonzero centripetal acceleration?
     
  17. Jan 7, 2014 #16
    I guess, zero tangential acceleration would be a non-zero centripetal acceleration. I mean, i know it is weird that there is no centripetal acceleration and i am with you.

    Edit: Maybe we should a similar example with a pendulum. The only difference is that there is going to be tension instead of compression for the massless string.

    1zg51nb.png

    The sum of the centripetal force using Newton's second law.

    ##\sum F_c = ma_c##
    ##mgcos\theta + T_{string} = 0##
    ##T_{string} = -mgcos\theta##
    ##ma_c = -mgcos\theta##
    ##a_c = -gcos\theta##
     
    Last edited: Jan 7, 2014
  18. Jan 7, 2014 #17

    TSny

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    Yes, it's similar to a pendulum.

    You know from kinematics that centripetal acceleration for circular motion is related to the speed and radius: ##a_c = \frac{v^2}{r}##. So, there will be centripetal acceleration as long as the ball has some speed.
     
  19. Jan 7, 2014 #18
    Yes but the centripetal acceleration of ##a_c = \frac{v^2}{r}## is for constant tangential velocity and no tangential acceleration. In the case of the pendulum, there is tangential acceleration.
     
  20. Jan 7, 2014 #19

    TSny

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    ##a_c = \frac{v^2}{r}## is valid even if there is tangential acceleration.
     
  21. Jan 7, 2014 #20

    mfb

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    Do you assume that the mass starts at the highest point here?


    I don't see the relevance of centripetal acceleration, and the thread begins to get messy with all those calculation fragments.

    For a clean way:

    Can you find a differential equation for the angular acceleration as function of the angle? This is identical to an inclined plane, as locally the mass is like on such a plane.

    Alternatively, write the angular velocity as function of the angle.
     
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