# Homework Help: Circular motion prac

1. Apr 24, 2007

### ellese

I recently completed an experiment about centripetal motion in which we attached a string to a weighed mass and passed the string through a glass tube, tying the free end to another weighed mass. We then moved the tube so the upper mass traced out circular motion and recorded the period of rotation. We then varied independently each of the masses and the radius of rotation.

On plotting the graphs, however, I wasn't able to clearly work out the relationship between the variables.

My best calculations are that

T squared is proporional to R
T squared is inversely proportional to the mass used to provide the centripetal force
T is proportional to the moving mass

I'm not sure of the equations I should be using to determine these relationships. Can anyone help?

2. Apr 24, 2007

### chaoseverlasting

I dont really understand the experiment youre carrying out. The forces that are acting seem to be the centrifugal force: mw^2r, gravity:mg, and the tension in the string. Draw a fbd and try to work out a general expression for the forces on your system. The expression you get, you can try verifying it with your experimental results.

3. Apr 25, 2007

### ellese

Sorry, let me rephrase the question.

Basically what I am trying to work out is how the radius is related to the period of an object undergoing circular motion with a constant centripetal force.

My experimental results don't seem to show any discernable relationship.

4. Apr 25, 2007

### chaoseverlasting

The force is constant. So then v=rw. $$T=\frac{2\pi r}{rw}$$. As you see, the radius cancels out, so the time period is independent of radius. This is only valid since the angular speed is constant.

5. Apr 25, 2007

### andrevdh

The centripetal force that the moving mass, $$m_v$$, is experiencing is given by

$$F_c = m_c g$$

and it is also represented by

$$F_c = \frac{m_v v^2}{R}$$

while the speeed of the moving mass is

$$v = \frac{2 \pi R}{T}$$

this gives the relation

$$T^2 = \frac{4 \pi ^2 m_v R}{g m_c}$$

The square of the period is therefore directly proportional to $$m_v$$ and $$R$$, and it is inversely proportional to $$m_c$$. So your approach is correct. As a further test you could calculate and plot the square of the period as a function of the values

$$\frac{m_v R}{m_c}$$

this graph should then be a directly proportional graph with a gradient of

$$\frac{4 \pi ^2}{g}$$

Just be sure to use values in your calculations that represent a single set of measurements.

Last edited: Apr 25, 2007