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

AI Thread Summary
The discussion centers on calculating the speed of a dot on a disk after it accelerates and coasts. The disk, with a diameter of 8.0 cm, accelerates at 600 rad/s² for 0.5 seconds, resulting in an angular velocity of 300 rad/s. The speed of the dot is calculated using the formula V = rw, yielding a result of 24 m/s, which is confirmed as reasonable. For the total number of revolutions, the calculations involve using angular displacement formulas for both the acceleration and coasting phases. The final results indicate that the dot completes a significant number of revolutions during the given time frame.
guru
Messages
38
Reaction score
0
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:
Physics news on Phys.org
For constant angular acceleration:

\omega = \omega_0 + \alpha t
\theta = \theta_0 + \omega_0 t + (1/2)\alpha t^2
 
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.
 
Thread 'Question about pressure of a liquid'
I am looking at pressure in liquids and I am testing my idea. The vertical tube is 100m, the contraption is filled with water. The vertical tube is very thin(maybe 1mm^2 cross section). The area of the base is ~100m^2. Will he top half be launched in the air if suddenly it cracked?- assuming its light enough. I want to test my idea that if I had a thin long ruber tube that I lifted up, then the pressure at "red lines" will be high and that the $force = pressure * area$ would be massive...
I feel it should be solvable we just need to find a perfect pattern, and there will be a general pattern since the forces acting are based on a single function, so..... you can't actually say it is unsolvable right? Cause imaging 3 bodies actually existed somwhere in this universe then nature isn't gonna wait till we predict it! And yea I have checked in many places that tiny changes cause large changes so it becomes chaos........ but still I just can't accept that it is impossible to solve...
Back
Top