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

Question

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?

Answer 1

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.

Answer 2

It is not possible to whirl it above your hand because the tension in the string will always pull on the object and the sum of the vertical component of the tension and the weight will always accelerate the object downward.

As for the second diagram, it is the horizontal component of the tension that provides centripetal force. The vertical component simply cancels out the weight of the object and prevents the object from accelerating up or down.

Answer 3

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.

Answer 4

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).