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A mass is free to move on a horizontal frictionless surface

  1. Mar 6, 2017 #1
    1. The problem statement, all variables and given/known data
    A mass m that is free to move on a horizontal frictionless surface, is attached to one end of a massless string that wraps partially around a frictionless vertical pole of radius r, as in the Figure below.You are holding onto the other end. At t=0 the mass has speed V0 in the tangential direction along the dotted circle of radius R, as shown. You pull the string in such a way that it remains in contact with the pole at all times, with the mass moving along the dotted circle. What is the speed of the mass as a function of time? Is there any special value of time that you notice? Interpret.
    Screen Shot 2017-03-06 at 6.18.47 PM.png
    2. Relevant equations
    $$\mathbf{F}=-mR\dot\phi^2\mathbf{\hat r}+mR\ddot\phi\mathbf{\hat \phi}$$

    3. The attempt at a solution
    I wasn't sure if the diagram indicated that $$tan\theta=\frac{r}{R}$$ or $$sin\theta=\frac{r}{R}$$. I assumed it meant the former. So I drew a FBD with only the tension force from the hand acting. This lead me to the equations $$T\cos\theta=mR\dot\phi^2$$ and $$-T\sin\theta=mR\ddot\phi$$
    $$\implies -tan\theta=\frac{\ddot\phi}{\dot\phi^2}$$ $$\implies -\frac{rt}{R}=-\frac{1}{\dot\phi}+C$$ When t=0, $$\dot\phi=\frac{v_0}{R}$$ $$\implies -\frac{rt}{R}=-\frac{R}{v}+\frac{R}{v_0}$$ $$\therefore v=\frac{v_0R^2}{R^2-rv_0t}$$ Is my solution and answer correct?
     
    Last edited: Mar 6, 2017
  2. jcsd
  3. Mar 6, 2017 #2

    kuruman

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    The diagram indicates that the right angle is between radius r and the string which is tangent to the pole. Which trig function is correct in that case?
    why the negative sign?
     
  4. Mar 6, 2017 #3

    BvU

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    I stay out of this, but I had the impression ##r## is drawn halfway the segment the wire is on the cylinder. Not very useful, to say the least. @kuruman 's idea is a lot better -- even if appearances are against.
     
  5. Mar 6, 2017 #4

    haruspex

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    Is the tension given to be constant? If not, you probably need to assume it anyway.

    Edit: I forgot you are given it moves in a circle, so the tension can be deduced (not constant). (Thanks @kuruman )

    As BvU notes, the diagram is badly drawn and misleading. It shows the string cutting into the pole. Redraw it properly for yourself, perhaps with it wrapping further around the pole to make things really clear.
     
    Last edited: Mar 6, 2017
  6. Mar 6, 2017 #5

    kuruman

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    I assumed that angle θ is constant, but not the tension. Anyway, here is a drawing I made for my "attempt at a solution."

    StringAroundPost.png
     
  7. Mar 7, 2017 #6
    Thanks @kuruman for the better drawn diagram. I have a new solution taking in mind what was said above. Here it is:
    First from the diagram and using the Pythagoraean Theorem we can immediately deduce that $$\tan\theta=\frac{r}{\sqrt{R^2-r^2}}$$. Upon drawing a FBD with the only force acting being tension T, we equate the tangential and radial components of the tension with the tangential and radial scalars, respectively, in the equation listed in relevant equations. I.e.
    $$\begin{align}
    T\cos\theta=mR\dot\phi^2 (1) \nonumber \\
    T\sin\theta=mR\ddot\phi (2) \nonumber \\
    \end{align}$$
    Dividing (2) by (1) we have
    $$
    \tan\theta=\frac{\ddot\phi}{\dot\phi^2}
    \implies \frac{\ddot\phi}{\phi^2}=\frac{r}{\sqrt{R^2-r^2}}
    \implies \int\frac{\ddot\phi}{\phi^2}\,dt=\int\frac{r}{\sqrt{R^2-r^2}}\,dt
    \implies -\frac{1}{\dot\phi}=\frac{rt}{\sqrt{R^2-r^2}}+C $$
    Using the initial condition that when $$t=0, \dot\phi=\frac{V_0}{R}$$ we have $$C=-\frac{R}{V_0}$$
    So after rearranging we have (assuming its correct)
    $$v(t)=\frac{V_0}{1-\frac{V_0rt}{R\sqrt{R^2-r^2}}}$$
     
  8. Mar 7, 2017 #7

    kuruman

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    That matches what I got. What about the special time?
     
  9. Mar 7, 2017 #8
    The special value of time I notice is $$t=\frac{R\sqrt{R^2-r^2}}{V_0r}$$ since as t approaches this time, the velocity approaches infinity. This corresponds to the mass m moving faster and faster as I pull on the string in order to keep it moving in a circle of constant radius. Of course, however, this is not possible in reality since at some velocity, relativistic effects start to become important and the equations we used to the derive this result break down.
     
  10. Mar 7, 2017 #9

    kuruman

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    Indeed that is a special value for the time and the equations say that at that time the speed becomes very (infinitely) large. We all know that the velocity cannot become infinitely large, so there must be a physical interpretation of what happens at the point in time when the velocity apparently becomes infinite according to the math. Hint: Examine the assumptions under which these equations were derived. Is it possible that at some instant in time the model and the motion of the object part company?
     
  11. Mar 7, 2017 #10
    Well the equations also imply (not derived directly here) that as time goes on, the tension in the string also gets larger and larger. At some point the string will break due to the tension in the rope being too high. Then the mass will initially travel at the velocity is was traveling at an instant before the rope snapped, tangentially to the dotted circle it was traveling in.
     
  12. Mar 7, 2017 #11

    kuruman

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    Not what I had in mind. Assume you have an infinitely strong string. The problem says
    How sure are you that this is the case all the way up to infinite speed?
     
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