3
$\begingroup$

I have the following problem :

A point $P$ of mass $m$ is constrained to move on a frictionless circle of radius $l$, centred at $O$, in a vertical plane in a gravitational field with acceleration $g$. If the particle is released from rest when $OP$ makes an angle $\theta_0$ with the downward vertical, calculate the speed at which the particle passes the lowest point.

I think I solved it with energy. Setting the zero of potential energy at its lowest point I simply get that at the point of release I have $E = mg(l-l\cos\theta_0 )$ (because the cosine is negative in the upper half and positive in the lower half). Also at the lowest point I have $E=\frac{1}{2}mv^2$ hence I have $v^2 = 2gl(1-\cos(\theta_0))$

I wanted to solve this using Newton's laws, so I tried polar coordinates.

Setting first the usual Cartesian axes, I wrote $\underline{W} = -mg\underline{j}$ to be the weight and I used the formula $\underline{j} = \cos(\theta)\underline{e_\theta}+\sin(\theta)\underline{e_r}$ I get $\underline{W} = -mg\cos(\theta)\underline{e_\theta}-mg\sin(\theta)\underline{e_r}$. The only component that counts towards the velocity in the motion is the tangential one, hence $\underline{e_\theta}$.

Hence I get $m\ddot{\theta} = -mg\cos(\theta)$ from Newton's Second Law. However I think this is the usual pendulum equation which can only be solved for small oscillations using Taylors. So I must have done something wrong? Certainly I can find a solution for any oscillation (By oscillation I mean by releasing the particle at any point of the circle, maybe excluding the highest point). The last step is the one I am most doubtful about but I really can't see how to get the same result as above.

  • 1
    what are $\underline{e_r}$ and $\underline{e_\theta}$. Are they along the tangential and radial direction, cause I think they ought to be that way. – 2017-01-03
  • 0
    @user350331 they are the (non-constant) standard unit vectors for polar coordinates! We normally have $\underline{e_r}=\cos(\theta)\underline{i}+\sin(\theta)\underline{j}$ and $\underline{e_\theta} = -\sin(\theta)\underline{i}+\cos(\theta)\underline{j}$ – 2017-01-03
  • 0
    Well it would be much easier using the usual unit vectors $\hat{i}, \hat{j}, \hat{k}$. – 2017-01-03
  • 0
    @user350331 would it though? I tried, but it doesn't seem to work. Indeed those vectors are fixed and don't move with respect to the position. I mean you have to decompose the weight $\underline{W} = -mg\underline{j}$ into what contributes to the acceleration (and hence the final velocity) and what doesn't. I think (not sure though!) these are exactly the tangential and the radial components. But I accept also the solution in Cartesian coordinates! I'd just like to see another way of solving it and checking that my result is correct. Cartesian is fine, if you can – 2017-01-03
  • 0
    From what I can see, I believe you are not taking into account the fact that the vectors $e_\theta$ and $e_r$ \textbf{rotate} over time - therefore they are accelerated and the system is not an inertial system. Therefore Newton's equations don't hold unless you add the fictitious force term – 2017-01-03
  • 0
    Check out this pdf https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-notes/MIT16_07F09_Lec05.pdf – 2017-01-03
  • 0
    @shost71 oh right. Although there's something it is not clear to me about what you said. As the point falls, yes the polar unit vectors will accelerate, but does that imply that the system is not inertial? Because I think the system itself is not accelerating, just the unit vectors. – 2017-01-03
  • 0
    @shost71 indeed, as you can see at page 10 of that pdf, in the pendulum example, also there the unit vectors accelerate, but they can still use them as in a inertial frame, I think – 2017-01-03
  • 1
    The pendulum itself is not accelerating; perhaps it would be clearer to say that the frame of reference is not inertial - the problem is not with the pendulum but rather with the "point of view" from which you are measuring the distances - by using polar coordinates you are "riding along" with the ball of the pendulum, so you feel the centripetal acceleration that the ball feels. So from your frame of reference you would see things move according to a (fictitious) centrifugal force. – 2017-01-03
  • 1
    In page 10 of the pdf, when they write the equations of motion they put "Newton's" force on the LHS of the equations and the fictitious force on the RHS – 2017-01-03
  • 0
    @shost71 It makes sense until "you feel the centripetal acceleration that the ball feels". Okay I understand that. But how does this become a centrifugal force in my frame of reference? – 2017-01-03
  • 0
    The way you "feel" it is this: the vectors $e_r$ and $e_\theta$ are moving in circular motion, so when you get the derivative, e.g., for $r$ you must also take the derivative of these unit vectors. When you derive again to obtain the accelerations, you end up with new terms that account for the rotation – 2017-01-03
  • 0
    @shost71 yeah I normally get $\dot{\underline{r}} = \dot{r}\underline{e_r}+r\dot{\theta}\underline{e_\theta}$ and $\ddot{\underline{r}} = (\ddot{r}-r\dot{\theta}^2)\underline{e_r}+(r\ddot{\theta}+2\dot{r}\dot{\theta})\underline{e_\theta}$ but then as the radius doesn't change it simplifies as in the question, once you compare the components, I think – 2017-01-03
  • 0
    @shost71 , well putting $r=l$ as in the question, and multiplying by the mass by N2 I should have $ml\ddot{\theta} = -mg\cos(\theta)$ which gives me $\ddot{\theta}+\frac{g}{l}\cos(\theta) = 0$, but then again I think this is not solvable in a closed form, unless you consider small angles, I think at least – 2017-01-03
  • 0
    @shost71 hold on. So when comparing, (first of course $\ddot{r} = 0$ and $2\dot{r}\dot{\theta} = 0$ as radius is constant), I should compare $-r\dot{\theta}^2\underline{e_r} + r\ddot{\theta}\underline{e_\theta} = -mg\cos(\theta)\underline{e_\theta}$ which gives the one in my previous comment or what should I compare? I got confused – 2017-01-03
  • 0
    I apologize because I myself got mixed up. That's OK, that's the correct equation. The pendulum equation is not linear (for angles that are not very small). – 2017-01-03
  • 0
    @shost71 Don't worry, thank you for writing. But now here's the catch! How come I get a perfectly easy and smooth solution (and equations as well) using the energy, but when I use coordinates I don't get an exact solution? Surely I should get the same solution, i.e. $v^2 = 2gl (1-\cos\theta_0)$, which I don't! – 2017-01-03
  • 0
    One of the answers below has touched on this - from the nonlinear equation you can solve for the velocity integrating the contribution from $\theta = \theta_0$ to the bottom position – 2017-01-03
  • 0
    I believe you should get to the same result - just that it will be a more complicated procedure because of the ugly integral. – 2017-01-03
  • 0
    @shost71 okay I will try to first makes sense of that question and then do the integration. I thought that using coordinates would have gave be the solution in a more elegant way but I was clearly wrong – 2017-01-03
  • 0
    @shost71 so I get $w_b =\dot{\theta}(\theta_0) =\frac{g}{l}\sin{\theta_0}$ – 2017-01-03
  • 0
    @shost71 which is not what I got above! – 2017-01-03
  • 1
    In terms of $\dot \theta = \omega$, the energy argument gives ${\dot \theta}^2 = \omega^2 = 2 \frac g l (1 - cos \theta_0)$. The differential equation for $\theta$ is $\ddot \theta = - \frac g l sin \theta$. Now we want from this equation to get the value of $\omega$ at the bottom position - we know $\omega(\theta = \theta_0) = 0$, so intuitively we have to sum the contribution of force the angle gets at every change of the angle. We have $\omega = \dot \theta$, so in our equation we get $\dot \omega = - \frac g l sin \theta$. – 2017-01-03
  • 1
    Now here is the catch (and the reason we were getting the a different answer): we want to integrate along time (to get rid of the dot in $\omega$), but our equation relates $\omega$ to $\theta$, not $t$ explicitly. Using the chain rule $\dot \omega = \frac{dw}{dt} = \frac{dw}{d\theta} \frac{d\theta}{dt}$. But $\frac{d\theta}{dt} = \dot \theta = \omega$, so we have $\frac\omega {d\omega}{d\theta} = - \frac g l sin \theta$, which is now a differential equation in terms only of $\omega$ and $\theta$. – 2017-01-03
  • 1
    "Multiplying" by $d\theta$ we get $\omega d\omega = - \frac g l sin \theta d\theta$. We integrate along the angles that interest us: from $\theta = \theta_0$ to $\theta = 0$ (bottom position). We know that at $\theta = \theta_0 \rightarrow \omega = 0$ (initial rest), and we want $\omega_f$ at $\theta = 0$. So the integral looks like $\int_{\omega = 0}^{\omega = \omega_f} \omega d\omega = \int_{\theta = \theta_0}^{\theta = 0} - \frac g l sin \theta d\theta$. Integrating we get $\frac{\omega_f^2}{2} = \frac g l (1 - cos \theta_0)$ which matches the energy argument. – 2017-01-03
  • 0
    Basically the answer that was originally posted, I hope this helps to see the intuition behind it at least – 2017-01-03
  • 0
    @shost71 thank you for the lengthy explaination! Ive understood most of it but I'm unsure about the chain rule step. We have $\dot{w} = \frac{dw}{d\theta}\frac{d\theta}{dt} = \frac{dw}{d\theta}w$ but how do we get $\dot{w} = \frac{w}{d w}d\theta$ ? Cause taking your expression and substituting $w$ back again, for a double check, we get $\frac{d\theta}{dw}\frac{d\theta}{dt}$ which is different from what we had – 2017-01-03
  • 0
    I made a typo, it should be $\dot \omega = \omega \frac{d\omega}{d\theta} $ – 2017-01-03
  • 0
    @shost71 okay I think now it makes sense! Thank you! – 2017-01-03

2 Answers 2

3

Consider the following angular position of the pendulum at which it has an angular velocity $\omega$

enter image description here

The angular acceleration of the pendulum is given by

$$\alpha=\frac{g}{l}\sin\theta$$

The moment of inertia of the pendulum is given by $I=ml^2$, and the torque acting in the pendulum bob is $$\tau=mgl\sin\theta$$

And we know that torque is given by $\tau=I\alpha$. Hence, we get

$$\tau=mgl\sin\theta\implies ml^2\cdot\alpha=mgl\sin\theta\implies \alpha=\dfrac{g}{l}\sin\theta$$

And if you are not familiar with the notion of moment of inertia, then just consider the relation b/w the length of arc and the angle subtended by that arc, which is

$$\theta=\dfrac{L}{l}, \text{ where $L$ is the length of the circular arc}$$

Differentiate the equation twice and you get $$\alpha=\dfrac{a_t}{l}\text{, where $a_t$ is the tangential acceleration}$$

Now we do know that angular acceleration in terms of angular velocity and angular position can be written as $\alpha=\omega\dfrac{d\omega}{d\theta}$.Hence, we get

$$\omega\dfrac{d\omega}{d\theta}=\dfrac{g}{l}\sin\theta\implies \int{\omega\cdot d\omega}=\int{\left(\dfrac{g}{l}\sin\theta \right)(-d\theta)}$$

Setting proper limits for the above integration gives you a relation b/w $\theta$ and $\omega$ from which you can easily find the relation b/w time $t$ and angular position $\theta$

The proper limits for finding the velocity of the object at the lowest point is as follows

$$\int_{0}^{\omega_l}{\omega\cdot d\omega}=-\int_{\theta_0}^{0}{\dfrac{g}{l}\sin\theta\cdot d\theta}\\ \implies \dfrac{{\omega_{l}}^2}{2}=\dfrac{g}{l}(1-\cos\theta_0)\\ \implies v^2=\sqrt{2gl\left(1-\cos\theta_0\right)}$$

  • 0
    I might be pedantic, but how did you get the angular acceleration? From $mg\sin\theta$ ? Can you expand it a little please? – 2017-01-03
  • 0
    @Euler_Salter - The $\alpha=\omega\dfrac{d\omega}{d\theta}$ part or the $\alpha=\dfrac{g}{l}\sin\theta$, I just edited the second part though.. – 2017-01-03
  • 0
    the $\alpha = g\sin\theta$ part at the beginning. I'm still stuck there, how did you get it? – 2017-01-03
  • 0
    oh okay it was $\alpha = g\sin\theta$. However I still don't see where you got that from to be honest – 2017-01-03
  • 0
    @Euler_Salter - Made a mistake it should be $\alpha=\dfrac{g}{l}\sin\theta$ instead of $g\sin\theta$, made the edit. – 2017-01-03
  • 0
    okay thank you, I am trying to make sense of it! – 2017-01-03
  • 0
    okay so how do I even find the result of $\int w dw$ ? – 2017-01-03
  • 1
    @Euler_Salter - I made some edits to answer your concern regarding $\int{\omega\cdot d\omega}$ – 2017-01-03
1

(Sorry, can't add a comment yet)

$Hint$: Try to find a relation between $\dot{\theta}$ and $u$. Then multiply the equation you get from Newton's second law by $\dot{\theta}$, use appropriate differentiation rules and integrate.

  • 0
    I hope you can add a comment here. Anyway, the point of using the Polar Coordinates (apart from more simplicity, I hope, wrt the cartesian) was to actually find a relation involvin $\dot{\theta}$. However I don't seem to find any in cartesian coordinates!! – 2017-01-03
  • 1
    You could write down your position vector in polar coordinates. Then differentiate it. You get $u$ on the one side, but on the other... – 2017-01-03
  • 0
    yes the usual as I did, but I get $\underline{r} = l\underline{e_r}$ and hence $\dot{\underline{r}} = \dot{l}\underline{e_r}+l\dot{\theta}\underline{e_\theta}$ where I have two unknowns anyway – 2017-01-03
  • 1
    Notice $l$ is constant in your problem! – 2017-01-03
  • 0
    okay, but I still have the unknowns, as I have $\dot{\underline{r}} = l\dot{\theta}\underline{e_\theta}$ where both the LHS and $\dot{\theta}$ are unknowns – 2017-01-03
  • 0
    Taking norms, what you write is $u=l \dot{\theta}$ – 2017-01-03
  • 0
    I still have the unknown $u$ though. Recall that I am trying to find $u$ at the lowest point – 2017-01-03
  • 0
    Now, armed with this relation, multiply the equation you get from Newton's second law by \dot{\theta}, and then integrate. – 2017-01-03
  • 0
    so you mean $\frac{u}{l}m\ddot{\theta} = -\frac{u}{l}mg\cos(\theta)$? – 2017-01-03