Lab 8: Circular Motion

Here is a video to show you what we are doing:

In this lab, you are going to play with a toy. Here is a diagram of this toy.

Basically, this is a mass on a string attached to a rubber stopper. The string passes through a glass tube. There is a piece of tape to help you keep the length of the string outside the glass tube constant. The way you use this toy is to spin the rubber stopper around in a circle. Here is an example:

Circular Motion Lab from Rhett Allain on Vimeo.

So, what are you going to do? First, the physics. If I were to draw free body diagram for the stopper at the above instant, it would look like this:

There are only two forces on the stopper, the tension from the string and the gravitational force. The net force is NOT zero. This is because the stopper is accelerating. It is moving at a constant speed, but changing direction. When an object moves in a circle, it has a magnitude of acceleration:

a_\text{circular} = \frac{v^2}{r}

The direction of the acceleration is towards the center of the circle. If you want to know where this formula comes from, check this out (which is also in my book). So, for the above instant, the acceleration would be in the negative x-direction. If I write down the force equation for the x- and y-direcitons, I get (note that hte acceleration in the y-direction is zero m/s2).

F_\text{net-x} = -T\cos \theta = - \frac{m_2 v^2}{r}
F_\text{net-y} = T\sin \theta - m_2 g = 0

(note – this mass is the mass of the stopper) What can you measure and what can you solve for? First, there is a relationship between r and L:

r = L \cos \theta

You can measure:

  • The tension ( T ). This is just the mass times g (the mass of the hanging mass).
  • The length ( L ). Remember, this is the distance from the point of rotation to the center of the stopper.
  • The period of the rotation – how long it takes to complete one revolution. This will be easiest to measure by timing say 10 rotations and dividing by 10.

From the period, you can find the velocity:
\text{Period} = \frac{2\pi r}{v} = \frac{2\pi L \cos \theta}{v}
v = \frac{2\pi L \cos \theta}{\text{Period}}

What about theta? Is there anyway to get the angle from the diagram? Yes. Just look at the y-direction and solve for theta:

T\sin \theta = m_2 g
\theta = \sin^{-1} \left(\frac{m_2 g}{T} \right)
\theta = \sin^{-1} \left(\frac{m_2 g}{M_1 g} \right)
\theta = \sin^{-1} \left(\frac{m_2}{M_1} \right)

I am not going to put all this together for you. But, let me say this. If you keep the mass on the bottom of the string (M2) constant, then there is a relationship between the length of the string and the period – I am not saying it is a linear relationship. So, here is what you are going to do:

– Pick a mass to hang on the end of the string. If you pick something really high, you are going to have to swing that stopper really fast and bad things may happen.
– For that mass, pick a length of string to be hanging out of the string holder. You can keep track of what this length is by putting a piece of tape at the bottom of the string holder (glass rod). Make sure while you are swinging the stopper that the tape stays stationary but does not touch the glass.
– Get the thing swinging.
– The swinging person should only concentrate on making sure the tape stays constant
– Someone needs to count 10 or so periods and time it. Repeat this process 5 times to get an uncertainty.
– Change the length of the string and do it again.
– Do this for 5 different lengths of string.
– Make an a graph that shows a linear function (it will involve the period and the length). Find the slope of this linear graph and figure out what it means.

Extra stuff:
– Calculate what your angle is. If you want to check it, you can take a picture of the stopper swinging (I can help you with this).


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