# Using gradient values of a graph to determine angular acceleration

1. Feb 23, 2013

### vodkasoup

1. The problem statement, all variables and given/known data

I recently conducted an experiment to determine the moment of inertia of a disc using a tachometer attached to a disc marked with reflected strips, a weight, and an oscilloscope. The resulting oscilloscope data was plugged into fitplot to generate a graph of voltage (Y-axis) against time (X-axis). I am asked to convert the gradient of the graph to angular acceleration by using the equation ω=1.4V, where ω is the angular velocity and V is voltage.

2. Relevant equations

ω=1.4V

3. The attempt at a solution

I have no idea how to approach this. I know that the gradient of the graph is the change in voltage over the change in time, dV/dT , but I don't know how to proceed from here.

Many thanks for any help.

2. Feb 23, 2013

### Staff: Mentor

How is angular acceleration defined in terms of angular velocity ω?

3. Feb 23, 2013

### vodkasoup

Angular acceleration is the rate of change of angular velocity, right? But I'm still not sure how that relates to the gradient of my graph of changing voltage over time.

4. Feb 23, 2013

### Staff: Mentor

Take your equation for ω and differentiate it with respect to time.

You do know that rate of change of angular velocity means "rate of change of angular velocity with respect to time," correct?

Chet

5. Feb 23, 2013

### vodkasoup

Thanks Chet. How do I derive voltage with respect to time? If ω=1.4V and I derive with respect to time, I just end up with 0. Though obviously I'm not doing it correctly.

Also, what is the value of 'V' that I am supposed to use? The gradient of my graph?

6. Feb 23, 2013

### Staff: Mentor

$$\frac{d\omega}{dT}=1.4\frac{dV}{dT}$$

The gradient on your graph is dV/dT. So the rate of change of ω with respect to time is 1.4 times the gradient on your graph. Make sure everything is in the correct units.