# A disk 8.0 cm in diameter is initially at rest. A small dot is painted

1. Nov 14, 2004

### guru

A disk 8.0 cm in diameter is initially at rest. A small dot is painted on the edge of the disk. The disk accelerates at $$600rad/s^2$$ for .5s, then coasts at a steady angular velocity for another .5s.
What is the speed of the dot at t = 1 s?
Through how many revolutions has it turned?

I approached the first question by drawing a graph of the angular acceleration Vs. time. I then obtained the angular velocity by finding the area under the graph.
I then used the formula V=rw to get the speed.
I obtained 24m/s as my final result.

Is this reasonable?

Last edited: Nov 14, 2004
2. Nov 15, 2004

### James R

For constant angular acceleration:

$$\omega = \omega_0 + \alpha t$$
$$\theta = \theta_0 + \omega_0 t + (1/2)\alpha t^2$$

3. Nov 15, 2004

### HallsofIvy

Staff Emeritus
Your graph of the angular acceleration should be a horizontal straight line so the area under it is just the area of a rectangle: the length of the segment on the "t" axis is 0.5 seconds and the height is 600 rad/s2. That's exactly the same as multiplying (0.5 s)(600 rad/s2)= 300 rad/s. "Graphing" works, but seems unecessary to me.
To get the speed of the dot, note that at 300 rad/s it completes 300/(2pi) revolutions per second and each revolution has length 2pi(8) cm: the dot completes 300(8)= 2400 cm per second, exactly what you did.

To get the number of revolutions completed you will need the second formula James R gave for the first 0.5 second, then add 150/2pi revolutions for the second 0.5 second.