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So, I solved a question about acceleration in polar coordinates, but most people in my class (Classical Physics, first year at university studying Physics) disagree with my answer. So the question is about a playground roundabout with radius 3m rotating at a speed of 10m/s. A person on the roundabout throws a snowball towards the centre of the roundabout at a speed of 20m/s. What acceleration will the snowball experience that will cause it to miss the centre?

So, I started with the formula for acceleration in polar coordinates:

$$\vec{a} = \hat{r}(\ddot{r} - r\dot{\theta}^2) + \hat{\theta}(r\ddot{\theta} + 2\dot{r}\dot{\theta}).$$

I let the centre of the roundabout be the origin, and interpreted the problem like this: $r$ is the distance from the snowball to the centre of the roundabout, $\dot{r}$ is then the one dimensional velocity of the snowball, and $\ddot{r}$ would be the rate of change of $\dot{r}$. $\dot{\theta}$ is the angular velocity of the roundabout, and $\ddot{\theta}$ the angular acceleration.

So, from the problem statement, I picked out the following values:

$\ddot{r} = 0, \dot{r} = -20 \text{ ms}^{-1}, r = (3 - 20t)m, \ddot{\theta} = 0, \dot{\theta} = 3.33 \text{ rad s}^{-1}$

And then I just plugged these values into the formula above. Then I get an answer where the component of $\vec{a}$ in the $\hat{r}$ direction depends on $t$, and there is a constant component in the $\hat{\theta}$ direction.

However, most people in my class didn't use this formula, and are arguing that there should only be a corilios acceleration, e.g. the theta component in this case, as $\ddot{\theta} = 0.$ That doesn't make sense to me, as I was under the impression that the acceleration formula should always work in polar coordinates. It certainly looks like that to me when the first line in the derivation is as general as just $\vec{r} = r\hat{r}$, and then we take first and second derivatives of $\vec{r}$

So, what I want to know is who is right? Am I right that we can simply apply this formula, or are my classmates right, and there is only a corilios acceleration? Also, are the values I picked out for $\ddot{r}, \dot{r}, r, \ddot{\theta}, \dot{\theta}$ correct?

Any help would be much appreciated!

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  • $\begingroup$ You think that $\dot{r}$ is the velocity of the snowball and $\dot{\theta}$ is the angular velocity of the roundabout. Why do you think that, in a formula for acceleration, $r$ and $\theta$ would be coordinates of different objects? $\endgroup$
    – G. Smith
    Oct 18, 2019 at 16:25
  • $\begingroup$ I figured that since the snowball is the thing that will experience an acceleration, we should say that $\dot{r}$ is the velocity of the snowball, and $\dot{\theta}$ is the angular velocity of the snowball. Sorry, I should have made it more clear that I was assuming the snowball and the roundabout would have the same angular velocity as the snowball is thrown by a person on the roundabout. Is this assumption not correct? $\endgroup$ Oct 18, 2019 at 16:34
  • $\begingroup$ If they had the same angular velocity, why would the snowball miss the center? $\endgroup$
    – G. Smith
    Oct 18, 2019 at 16:37
  • $\begingroup$ Oh right, yeah I guess it wouldn't. That was stupid of me. Sorry, my intution isnt the greatest. Now im thinking the angular velocity of the snowball should be 0, but looking at the formula that doesnt seem right, since every term then goes to 0. Can you give me a pointer on how to work out the angular velocity of the snowball? $\endgroup$ Oct 18, 2019 at 16:41
  • $\begingroup$ You can not suppose that $r=r_0-\dot{r}t$. I understand what you were thinking but this would only be true for an intertial frame of refrence. What you have to do, is you have to solve the differential equation with the correct starting conditions. $\endgroup$
    – tomtom1-4
    Oct 18, 2019 at 17:37

4 Answers 4

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Ignoring z motion in the following.

Reference frame:"lab"-- the one where roundabout is rotating. Right handed, origin at roundabout center.

The trajectory is a straight line. There is no acceleration. The reason the ball misses the center is because of its initial conditions being such-there was always an initial tangential($\hat{\theta}$) velocity.

Reference frame:"rotating"-- the one where roundabout is at rest. Coincides with lab at $t=0$

At $\mathbf{t=0}$
The object has only radial velocity($-\hat{r}$). In theory it should hit the center. The only reason it won't is if something accelerated it tangentially. This come from the pseudo-forces. The object does experience acceleration:

  1. Coriolis: $\propto -\vec{\omega} \times \vec{v} $. Here, since $\hat{v}=-\hat{r}$, the acceleration is exactly what we want: along $\hat{\theta}$.
  2. Centrifugal: $\propto -\vec{\omega} \times (\vec{\omega} \times \vec{r}) $. Here, since $\hat{v}=-\hat{r}$, the acceleration is along $\hat{r}$. Won't affect hitting the center.

At $\mathbf{t\gt0}$

  1. The object is starting to move tangentially. At the same time its radial velocity is being decreased by the centrifugal force. Also Coriolis force from tangential motion is also along centrifugal. All in all the object moves as if it was going forth while curving in the direction of rotation.(see fig. 1 below.)
  2. Eventually the object outright turns back and seems to be escaping the roundabout(see fig. 2 below).
  3. now the direction of Coriolis force switches....

All in all, the object moves in an ever increasing spiral. Note that the acceleration keeps changing in time.

Conclusion

So whose frame should we consider?.Depends on the observer-if its the person on the roundabout, its the rotating frame. The final acceleration must, of course include gravity. The stated values for $\ddot{r}, \dot{r}, r, \ddot{\theta}, \dot{\theta}$ seem correct for the lab frame.

In rotating frame
$\ddot{r} \ne 0, \dot{r} = -20 \text{ ms}^{-1}, r \ne(3 - 20t)m, \ddot{\theta} \ne 0, \dot{\theta} = 0 \text{ rad s}^{-1}$


fig 1: trajectory at some initial time
fig 2: Trajectory initially. (The blue curve is the trajectory seen in rotating frame. Orange is where the person would be in the lab frame.X and y axes are x and y postions in meters)

fig 2: trajectory at some later time
fig 2: Trajectory after some time. (The blue curve is the trajectory seen in rotating frame. Orange is where the person would be in the lab frame.)


As far as the formula you have stated, its applicable to inertial frames only. In particular, for rotating frames, use ( $'$ denotes rotating frame)
$$ m\vec{a}'=\vec{F}' -m\frac{d\vec{\omega}}{dt} \times \vec{r}' -2 m \vec{\omega} \times m \vec{v}' -m\vec{\omega} \times (\vec{\omega} \times \vec{r}') $$ with $$ \vec{F}'= \hat{r}’(\ddot{r}’ - r'\dot{\theta}’^2) + \hat{\theta}’(r'\ddot{\theta}’ + 2\dot{r}’\dot{\theta}’)\\ \vec{v}'=\dot{r}'\hat{r}'+r'\dot{\theta}'\hat{\theta}' $$
At $t=0$, with no applied force
$$ \vec{\omega}=\omega\hat{z}'\\ \vec{r}'=R \vec{\hat{r}}'\\ \vec{v}=-v \vec{\hat{r}}'\\ \vec{F}'=0\\\ $$ we get $$ \vec{a}'=2 m \omega v \hat{\theta}'+m \omega^{2}R\hat{r}' $$

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Perhaps this defeats the point of the exercise, but it seems to me that if you want to know what the snowball does then why not just calculate the motion of the snowball, which has nothing at all to do with the motion of the roundabout. It just flies in a parabola, whose horizontal part is a straight line relative to the ground. If the initial velocity is towards the centre of the roundabout then it will hit the centre of the roundabout.

Even if you are intended to carry out a calculation in some other frame, it surely helps to know what the answer is when calculated the easiest way.

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  • $\begingroup$ I think the problem means the initial vector velocity is not toward the center; it is the component of velocity toward the center from the throw, plus the component of velocity due to rotation of the merry-go-round. $\endgroup$
    – John Darby
    Jan 17, 2023 at 2:19
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The initial radial velocity is the cause of veer-off in a rotating frame.

In circumferential dynamic equilibrium Coreolis acceleration balances the angular acceleration

$$ \alpha_r = -2 \omega \dot {r}_{radial} $$ and is the de-tracking component.

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Updated answer

If gravity is present, it adds displacement in the vertical direction but does not affect displacement in the plane of the rotating round-about, assuming the ball does not hit the surface of the round-about due to gravity; therefore, I will ignore the effect of gravity.

enter image description here

For your problem, the velocity is constant in the inertial, non-rotating frame, and the problem can be evaluated without solving the general equation of motion. However, to address the effects of the fictitious forces in the rotating frame, I will also evaluate the equation of motion in that frame. For your problem, consider the motion in the plane of the round-about from three viewpoints: (1) Cartesian coordinates in the inertial, non-rotating frame, (2) polar coordinates in the inertial, non-rotating frame, and (3) polar coordinates in the non-inertial frame rotating with the round-about. See Figure 1. $t$ is time. For a point P in space, we can express the coordinates of $P(t)$ as $(x, y)$ in the inertial frame using Cartesian coordinates, $(x^*, y^*)$ in the rotating frame using Cartesian coordinates, $(r, \theta)$ in the inertial frame using polar coordinates, and $(r^*, \beta)$ in the rotating frame using polar coordinates. $r$ and $r^*$ are the same for your problem, since there is no translational motion of the origin of the rotating frame with respect to the origin of the inertial frame. Hereafter, $r$ will denote the radial distance in both frames.

Part A. Motion using Cartesian coordinates in the inertial, non-rotating frame

In the inertial frame, there is no force hence no acceleration. The ball of mass $m$ has constant velocity $\vec v = 10 \hat i - 20 \hat j$, where $\hat i$ and $\hat j$ are unit vectors in the $x$ and $y$ directions, respectively, in the inertial frame. The motion is $x(t) = 10 t$ and $y(t) = 3 – 20 t$ and is shown in Figure 2 for time from 0 until the ball crosses the $x$ axis at time 0.15. The motion is a straight line.

y(x) in Inertial Frame

Figure 2. y(x) in Inertial Frame

Part B. Motion using polar coordinates in the inertial, non-rotating frame

Your relationship for the acceleration $\vec a$ in polar coordinates is correct. [Symon, Mechanics] But, in the inertial frame $\vec a = 0$ since there is no net force in the inertial frame. Therefore, using polar coordinates, the equation of motion in the inertial frame is: $$ m(\ddot r - r\dot \theta^2)\hat n + m(r\ddot\theta + 2\dot r\dot \theta)\hat l = 0 \enspace (1) $$ where $\hat n$ and $\hat l$ are unit vectors in the positive $r$ and $\theta$ directions, respectively. $\theta$ is considered positive in the counterclockwise direction. We could solve (1) for the motion, but since the velocity is constant, we can easily find the motion in polar coordinates as follows.

In general, $r = \sqrt{x^2 + y^2}$ and $\theta = tan^{-1}({y \over x})$. So for your problem $ r = \sqrt{(10t)^2 + (3 - 20t)^2} $ and $\theta = tan^{-1}({(3 - 20 t) \over 10t})$. Figure 3 shows $r(t)$ and $\theta (t)$, and Figure 4 shows $r(\theta)$, for time from 0 until the ball crosses the $x$ axis at time 0.15.

Figure 3

Figure 3. $r(t)$ and $\theta(t)$ in Inertial Frame

Fig 4

Figure 4. $r(\theta)$ in Inertial Frame

In Figure 4, $\theta$ goes from ${\pi \over 2}$ to $0$, and $r$ goes from $3$ to $1.5$.

Figure 5 shows $\theta(t)$ for longer times; as time increases $\theta(t)$ approaches about -63 degrees.

Fig 5

Figure 5. $\theta(t)$ In Inertial Frame, Longer Time

Part C. Motion using polar coordinates in the non-inertial, rotating frame

First, I will describe the motion without evaluating the equation of motion, then I will evaluate the equation of motion.

Part C1. Motion in Rotating Frame without Evaluating Equation of Motion

We can consider the rotating system relative to the inertial system as a passive rotation. [Goldstein, Classical Mechanics]. See Figure 1. Positive angles are measured counterclockwise, $\beta = \theta - \omega t$. Given Cartesian coordinates $(x, y)$ for a point in the inertial frame, the polar coordinates are $(r, \theta)$, where $x = r cos(\theta)$ and $y = r sin(\theta)$. In the rotating frame the polar coordinates are $(r, \theta - \omega t)$, and the Cartesian coordinates are $(x^*, y^*)$. $$x^* = xcos(\theta) + ysin(\theta) \enspace(7)$$ $$y^* = -xsin(\theta) + ycos(\theta) \enspace(8)$$ $$\beta = tan^{-1}({y^* \over x^*}) \enspace (9)$$ In the rotating frame, $r = \sqrt{x^{*2} + y^{*2}}$. This the same value as in the inertial frame, $r = \sqrt{x^2 + y^2}$, as expected since $\vec r$ is the same in both frames.

Figure 6 shows $r(\beta)$ and Figure 7 shows both $\beta(t)$ and $\theta(t)$, for time from 0 until the ball crosses the $x$ axis at time 0.15. At time 0.15, $\theta$ is zero degrees; $\beta$ should be $\theta - \omega t = 0 – {-10 \over 3} 0.15$, or 28.65 degrees in agreement with Figure 7.

Fig 6

Figure 6. $r(\beta)$ in Rotating Frame

Fig 7

Figure 7. $\beta(t)$ and $\theta(t)$

Part C2. Evaluation of Equation of Motion in Rotating Frame

Now, I will evaluate the equation of motion in the rotating frame. In the non-inertial rotating frame, the ball experiences acceleration due to the fictitious forces acting in that frame; the acceleration is not zero in the rotating frame due to the fictitious forces. $P(t)$ is the time dependent position of a point. Using polar coordinates we express $P(t)$ as $(r(t),\theta(t)$ in the inertial frame and $(r(t), \beta(t))$ in the rotating frame, noting that $r$ is the same in both frames. Using polar coordinates, the general equation of motion in the rotating frame is: [Symon, Mechanics] $$ m(\ddot r - r\dot \beta^2)\hat n^* + m(r\ddot\beta + 2\dot r\dot \beta)\hat l^* = \\ \vec F - m\vec \omega \times(\vec \omega \times r) - 2m\vec \omega \times {d^*\vec r \over dt} - m {d\omega\over dt} \times r \enspace (10)$$ $\vec \omega$ is the angular velocity of the rotating frame with respect to the inertial frame. ${d^*\vec r \over dt}$ is the derivative of $r$ in the rotating frame. $\hat n^*$and $\hat l^*$ are unit vectors in the radial and angular directions, respectively, in the rotating frame; $\hat n^* = \hat n$ for your problem.

In equation (10), $\vec F$ is the total true external force, zero for your problem. The fictitious forces on the right side of equation (10) are as follows: the second term is the centrifugal force, the third term is the Coriolis force, and the third term is sometimes called the Euler force. For your problem the Euler force is zero since $\vec \omega$ is constant. We can express the fictitious forces for your problem using Figure 8.

Fig 8

Positive angles are counterclockwise. $\vec \omega = \omega \hat k$. Evaluating the cross products, the centrifugal force is $$m\omega^2r\hat n' \enspace (11)$$ And the Coriolis force is $$2m\omega r\dot \beta \hat n' – 2m\omega \dot r \hat l' \enspace(12)$$ For your problem, $\omega = -{10 \over 3}$, negative, since the rotation is in the clockwise direction. The acceleration (the equation of motion divided by $m$) using polar coordinates in the non-inertial rotating frame is: $$ \ddot r - r\dot \beta^2 = ({10 \over 3})^2r - 2{10 \over 3} r\dot \beta \enspace (13)$$ in the $\hat n’$ direction, and $$r \ddot \beta +2\dot r\dot \beta = 2{10 \over 3} \dot r \enspace(14)$$ in the $\hat l$ direction. For your problem, part of the Coriolis force, $2m{10 \over 3} \dot r$ pushes the ball in the $=\hat l$ direction in the rotating frame, as indicated in (14). This part of the total Coriolis force causes the ball to miss the center of the round-about as viewed in the non-inertial rotating frame. As viewed by an observer fixed in the rotating frame throwing the ball, the ball moves to the right ($\hat l$ direction) due to this part of the Coriolis force.

See Figure 1. For your problem, $R$ equal $3$ and $\omega = -{10 \over 3}$. In the inertial frame the components of velocity are $v_x = 10$ and $v_y = -20$. $x(t) = 10t$ and $y(t) = (3 – 20)t$. Using (7) and (8) $x^*=(3-20 t) \sin \left(-\frac{10 t}{3}\right)+10 t \cos \left(-\frac{10 t}{3}\right)$ and $y^*=(3-20 t) \cos \left(-\frac{10 t}{3}\right)-10 t \sin \left(-\frac{10 t}{3}\right)$. Using (9) $ \beta = \tan ^{-1}\left(\frac{10 t \sin \left(\frac{10 t}{3}\right)+(3-20 t) \cos \left(\frac{10 t}{3}\right)}{10 t \cos \left(\frac{10 t}{3}\right)-(3-20 t) \sin \left(\frac{10 t}{3}\right)}\right)$. $\dot r = \frac{20 (25 t-3)}{\sqrt{500 t^2-120 t+9}}$ and $\dot \beta = \frac{200 t (25 t-6)}{3 \left(500 t^2-120 t+9\right)}$. (Equivalently, $\dot \beta = {d \over dt} (\theta + {10 \over 3}t)$). The initial conditions for evaluating (13) and (14) are:$r(0) = 3$, $\beta(0) = \pi/2$, $\dot r(0) = -20$, and $\dot \beta(0) = 0$.

Using Mathematica, I solved (13) and (14) for the motion in the rotating system for $r(t)$ and $\beta (t)$. The results are the same as obtained earlier in part C1.

The acceleration in the rotating frame, evaluated using the left sides of (13) and (14), is shown in Figure 9 for time up to 0.15. (The acceleration in the inertial frame is zero.)

Fig 9

Figure 9. Acceleration in $n^*$ and $l^*$ Directions in Rotating Frame

The physical position $P(t)$ in Figure 1 is the same in both the inertial and non-inertial rotating frames, even though there is no acceleration in the inertial frame and acceleration in the rotating frame. Since, $\beta = \theta + {10 \over 3} t$, as t increases, $\beta(t)$ increases above $2 \pi$, and should corrected to be within $(0, 2 \pi)$. Figure 10 shows $\beta$ for up to time 10.

Fig 10

Figure 10. $\beta(t)$ in Rotating Frame Long Time

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