What Determines the Peak Angular Speed of a Rotating Cylinder?

[menglish20]

Homework Statement

In a manufacturing process, a large, cylindrical roller is used to flatten material fed beneath it. The diameter of the roller is 1.00 m, and, while being driven into rotation around a fixed axis, its angular position is expressed as

θ =2.50t² - 0.600t³

where θ is in radians and t is in seconds. (a) Find the maximum angular speed of the roller. (b) What is the maximum tangential speed of a point on the rim of the roller? (c) At what time t should the driving force be removed from the roller so that the roller does not reverse its direction of rotation? (d) Through how many rotations has the roller turned between t=0 and the time found in part (c)?

Homework Equations

I think this has to do with translational and angular quantities. a_c=v²/r=rω² might be useful.

For part b, a_t=rα

The Attempt at a Solution

I took the derivative of the rotational position to get angular speed in terms of t. I know the radius is .5 m. I don't understand how a max speed can be reached, as it would increase indefinitely with time. I don't think I'm grasping the problem. I also don't understand how the roller could reverse its direction. Any help is much appreciated.

 

[I like Serena]

Welcome to PF, menglish20!

In the manufacturing process, the roller would be first accelerated to a max speed and then decelerated to a stop.

You said you calculated the derivative. So what did you get?
You did solve it for being equal to zero?
Note that you're only interested in solutions where t > 0 and the where t is smaller than the time where the angular velocity becomes zero again.

 

[menglish20]

Glad to be here

Right so the derivative would be
ω=5t - 1.8t²
So I solve for that set to zero, and i get t = 0, 2.78.

So I know the time where the velocity peaks is between those times. I'm going to guess that it acts parabolic, so the peak must be the midpoint, so t=1.389 s.
With that, I find v_max=3.47 rad/s

For part c, the driving force should be removed at 2.78s correct? If not I guess I still don't understand what's going on in terms of the manufacturing process.

 

[I like Serena]

Glad to be here

Right so the derivative would be
ω=5t - 1.8t²
So I solve for that set to zero, and i get t = 0, 2.78.

Good!

So I know the time where the velocity peaks is between those times. I'm going to guess that it acts parabolic, so the peak must be the midpoint, so t=1.389 s.
With that, I find v_max=3.47 rad/s

Actually, you calculated the max angular velocity here.
This is not the tangential velocity.

Edit: Do you know the relation between these two?
Btw, you may have found by now, that the relevant equations you mentioned are not needed in this problem.

For part c, the driving force should be removed at 2.78s correct? If not I guess I still don't understand what's going on in terms of the manufacturing process.

Yes!
At this time the angular velocity is zero, so if the angular acceleration is set to zero, it will remain zero, which is intended.

 

[menglish20]

Do you know the relation between these two?
Btw, you may have found by now, that the relevant equations you mentioned are not needed in this problem.

v=rw

So, I'd take .5 * 3.47 = 1.74 m/s.

Then for part d,

θ = 6.43, so the number of rotations would be 1.02.

All these answers match the given solutions, thanks for clarifying this problem!

 

[I like Serena]

You're welcome!

And thanks for taking the time to finish this thread and say thanks.