# Is it possible to whirl a point mass (attacted to a string) around in a horizontal circular motion *above* my hand?

I'm studying circular motion and centripetal force in college currently and there is a very simple question but confuses me (our teacher doesn't know how to explain either :/), so I hope we can sort it out here >< So I draw two pictures to show what I was thinking on it.

In pic 1 there is a hand rotating a ball attached to a piece of string in a circular motion, by free body diagram we can easily see that the net force produced by tension and gravity is centripetal force, and it towards to the center of the circle.

But in pic 2 as shown below

When the hand is below the ball, the net force is actually towards downwards, not to the center of the circle. How would that circular motion happen if this free body diagram doesn't make sense? Or is there any other force acting on it?

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Possible Duplicate .See this question.physics.stackexchange.com/questions/62887/… –  ABC May 16 '13 at 14:58
I've seen the second scenario happen, in a toy airplane flying around in such a circle. but it only works because the airplane is generating lift from its wings (also it needs a propellor to counteract the horizontal drag force). back where I'm from this toy is quite common, and shopkeepers will put models out on display, but I've not been able to find a youtube video of them anywhere!! :( ah well. the point is that in order for the second scenario to happen you need an additional vertical force. so unless the ball is moving really really fast and generating some lift of its own it wont happen –  nervxxx May 16 '13 at 15:14
The system in the second diagram will always have a net force downwards as there is no upward component to any of the forces shown. So the ball will fall (accelerate by $F=ma$) until the system resembles the first diagram. The system could only work if the ball was generating some lift (like from the wings of @nervxxx 's plane or a helium balloon) or if the string is rigid (a stick) and held at that angle. –  ejrb May 16 '13 at 15:39

I think the answer is that the second diagram you drew won't happen. I just picked up a string and tried this. What happened is that the first diagram is easy. For the second, I have to twirl the string faster, and I can't quite get it to stay above my hand. The best I can do is to get the mass to swing in a plane almost even with my hand.

Note: it's a little difficult to do this fairly with your hand. When I tried it, I had a tendency to slightly adjust the plane of motion of the mass, so that it oscillated slightly. This was particularly bad when trying the second situation. If I didn't do that, the sting would hit one of my knuckles every time it passed. That's another indication that the plane of revolution is actually below my hand.

Your force diagrams are qualitatively correct. Gravity points down towards the floor, and the tension points along the string at some angle. It's easiest to break the tension into a vertical component (which will either add to or subtract from gravity), and a radial component, which lies in the plane of the ball's orbit and provides the centripetal force.

To be concrete, take $\theta$ to be the angle between the string and the vertical. The string hanging under just gravity means $\theta = 0$, and the ball orbiting in the horizontal plane is $\theta = 90^\circ$. You want the vertical component to cancel out gravity, so $w = T\cos{\theta}$. As you increase $\theta$, $\cos{\theta}$ decreases, so you have to increase $T$ to keep the vertical component balanced. The centripetal force is $F_c = T\sin{\theta}$, which increases both because you increased $\theta$ and because you increased $T$. The ball will then need to move faster to account for the increased centripetal force. I'll let you work out the actual details, but you should get that the ball will need to move infinitely fast to get to $\theta = 90^\circ$.

An illustration of the wobble effect I noticed is when cowboys attempt to lasso animals. You can watch this video starting at 1:20 to see this.

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So let's see from where does the centripetal force comes from.

Imagine we have a body that moves in 2 dimensions. Let's then describe the system using the polar coordinates, such that: $x=r\cos(\theta)$ and $y=r\sin(\theta)$. Let's define two vectors, $\vec{r}$ and $\vec{\theta}$, such that: $$\begin{cases} \vec{r}=x\vec{e}_x+y\vec{e}_y \\ \vec{\theta}=-y\vec{e}_x+x\vec{e}_y\end{cases},$$

where the vector $\vec{r}$ is such that is magnitude is the distance from the origin to the particle's position and $\vec{\theta}$ the vector whose magnitude is the angle between the positive half of the $xx$ axis and the vector $\vec{r}$.

Now, for commodity, let's define the vectors $\vec{e}_r$ and $\vec{e}_\theta$ the unit vectors that have the same direction of the vectors $\vec{r}$ and $\vec{\theta}$, respectively. Using this vectors I can write

$$\begin{cases} \vec{r}=r~\vec{e}_r \\ \vec{\theta}=\theta~ \vec{e}_\theta\end{cases}.$$

Comparing the above equation with the previous definition in terms of the Cartesian coordinates we have that $$\begin{cases}\vec{e}_r=\cos(\theta)\vec{e}_x+\sin(\theta)\vec{e}_y \\ \vec{e}_\theta=-\sin(\theta)\vec{e}_x+\cos(\theta)\vec{e}_y \end{cases}.$$

Because we'll need it a little further let's take the derivative with respect to time of $\vec{e}_r$ and $\vec{e}_\theta$. This is quite straight forward and we find: $$\begin{cases}\frac{d\vec{e}_r}{dt}=\frac{d\theta}{dt}\vec{e}_\theta \\ \frac{d\vec{e}_\theta}{dt}=-\frac{d\theta}{dt}\vec{e}_r \end{cases}.$$

Finnaly we're in position to answer your question. Let's compute the expression for the velocity of the particle: $$\vec{v}=\frac{d\vec{r}}{dt}=\frac{dr}{dt}\vec{e}_r+r\frac{d\theta}{dt}\vec{e}_\theta.$$

Let's now compute the acceleration: $$\vec{a}=\frac{d^2\vec{r}}{dt^2}=\left[ \frac{d^2r}{dt^2}-r\left(\frac{d\theta}{dt}\right)^2 \right]\vec{e}_r+\left[ 2 \frac{dr}{dt}\frac{d\theta}{dt}+r\frac{d^2\theta}{dt^2} \right]\vec{e}_\theta.$$

In your specific problem the radius of the orbit doesn't change so the derivatives of $r$ are zero. We can further assume an idealized situation where the angular velocity of the particle doesn't change. So we're left with:

$$\vec{a}=-r\left(\frac{d\theta}{dt}\right)^2~\vec{e}_r.$$

And this is what we call the centripetal acceleration and, as you see, it always has the direction of $\vec{e}_r$ (in your case, horizontal).

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This is a good and correct development of centripetal acceleration but it doesn't really answer the OP question. That is, why the free body diagram of the second situation doesn't make sense. –  OSE May 16 '13 at 14:30
@OSE Indeed. Well, it answers the fact that in vacuum, with just gravity and angular velocity there's no way a body's movement can be the one from the second picture. It can't even go horizontal, since, even if the angular velocity goes to infinity (maybe the string is made of carbon nanotubes), there's nothing that cancels the vertical gravitational force... –  PML May 16 '13 at 14:38

The first diagram you drew was absolutely right and the free diagram was also right .

The second rotation would only be possible if the angular velocity is enough to displace the molecules of the air above the circle of rotation such that the molecules below the circle of rotation would push the bob with a force that will be more than the weight of the bob i.e your thought of presence of another force was absolutely right which is in the opposite direction of the weight and more than it.

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