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Angular, linear velocity & centripetal acceleration

  • #1

Homework Statement


A washing machine's two spin cycles are 328 rev/min and 542 rev/min. The diameter of the drum is 0.43 m.
What is the ratio of the centripetal accelerations for the fast and slow spin cycles?

Homework Equations


ac=v2/r
linear velocity: v=rw (w is really omega)
angular velocity= change in theta/change in time

I think that's all I need.


The Attempt at a Solution


First I changed the angular velocities from rev/min to rad/s. So, 328 rev/min=34 rad/s, and 542 rev/min=57 rad/s.
I solved for the slow cycle's acceleration first. Since it is in linear velocity instead of angular velocity, i solved for the linear velocities for the slow and fast cycle.
v=rw -> vs=(.215 m)(34 rad/s) -> 7.31 m/s
v=rw -> vf=(.215 m)(57 rad/s) -> 12.3 m/s
Then I plugged these into the centripetal acceleration formula for the slow and fast cycles.
acs=v2/r -> a=(7.31 m/s)(7.31 m/s)/(.215 m) -> 248.54 m/s2
acf=v2/r -> a=(12.3 m/s)(12.3 m/s)/(.215 m) -> 703.67 m/s2


There may not be anything wrong with it, but those numbers for the acceleration seem soooooo big. It's easy physics, so I don't know what I'm doing wrong. >:(
 
Last edited:

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
Yes they are correct. Both those speeds are quite high, so the centripetal accelerations are high as well.
 
  • #3
Oh.
I did this wrong the first time I did it, and I forgot what the answers were (yay spring break)
Thanks. :P
 

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