Angular Acceleration of a centrifuge

Click For Summary

Homework Help Overview

The problem involves a centrifuge in a medical laboratory that rotates at a specified angular speed and decelerates to a stop after a certain number of revolutions. The subject area includes angular kinematics and the application of rotational motion equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss methods for calculating angular acceleration, exploring the use of angular kinematic equations and the relationship between angular and linear motion. Questions arise about the validity of time calculations and the conversion of units from revolutions to radians.

Discussion Status

The discussion is ongoing, with participants providing guidance on the appropriate equations to use and questioning assumptions about time and unit conversions. There is an acknowledgment of the need to relate angular motion to linear kinematics, and some participants are correcting previous calculations.

Contextual Notes

Participants are navigating the challenge of converting between different units of measurement and ensuring that all variables are correctly accounted for in their calculations. There is a focus on understanding the implications of negative acceleration as the centrifuge comes to a stop.

vipertongn
Messages
97
Reaction score
0
A centrifuge in a medical laboratory rotates at an angular speed of 3400 rev/min. When switched off, it rotates 48.0 times before coming to rest. Find the constant angular acceleration of the centrifuge.

relevant equations: ωf = ωi + αt

My work:
I found out the time by (48 rev)(1min/3400 rev)(60 sec/min) = 0.847 seconds

I then converted the rev/min to rad/sec (3400 rev/min)(2pi/1rev)(1min/60sec)= 356 rad/sec

So I set the problem up like 0=356 rad/sec + α(0.847 seconds)
-356/0.847=α
α = -420 rad/s^2
 
Physics news on Phys.org
Last edited:
yes but couldn't you find time using the equation i did earlier? I'm trying to solve for a tho...

However, if I were to calculate what you put in...

0=3400 rev/min+2a(48rev)
-3400/96=a
a=-35 min^-1
 
Last edited:
vipertongn said:
yes but couldn't you find time using the equation i did earlier? I'm trying to solve for a tho...

Yes. You can use that equation for time, once you've found the α based on the θ given and the initial ω.
 
i edited my last post, is that correctly calculated?
 
vipertongn said:
i edited my last post, is that correctly calculated?

Well I left an exponent out of my earlier post and I corrected that apparently after you used it.

Instead of blindly using formulas, I encourage you to review the link and think about how the rotational kinematic equations relate to the regular one dimensional kinematics that I think you already know.

But that aside (and the exponent of the initial ω corrected) your calculation hasn't accounted for the conversion from min to sec. (Unless you don't need your answer in sec.)
 
if i put it in seconds it ends up being 0.583 s^-1 but that' can't be the acceleration...shouldn't it be to the -2?
 
vipertongn said:
if i put it in seconds it ends up being 0.583 s^-1 but that' can't be the acceleration...shouldn't it be to the -2?

Look you need to understand the equations.

ωf2 = ωi2 + 2*α*θ

But they give you revolutions. And there are 2π radians in a revolution and θ is in radians.
 
ohhhh i see now, thanks lowly ^^
 
  • #10
Remember the acceleration α they want is (-) because you are slowing down to a stop.
 

Similar threads

Replies
10
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
14K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K