Maximum Tangential Speed of Laundry in Washing Machine Spin Cycles

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The discussion focuses on calculating the maximum tangential speed and radial acceleration of laundry in a washing machine during spin cycles at two different angular speeds: 448 rev/min and 619 rev/min, with a drum diameter of 0.670 m. Users have successfully calculated the ratios of maximum radial force and tangential speed for the two speeds, yielding values of 1.91 and 1.38, respectively. The correct approach to finding maximum tangential speed involves converting angular speed from revolutions per minute to radians per second before applying the formula v = r * omega. For maximum radial acceleration, the formula a = v^2/r is confirmed to be appropriate, as it calculates centripetal acceleration without tangential acceleration. The discussion emphasizes the importance of unit conversion and correct formula application in solving the problem.
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Homework Statement

The spin cycles of a washing machine have two angular speeds, 448rev/min and 619rev/min . The internal diameter of the drum is 0.670m.

a. What is the ratio of the maximum radial force on the laundry for the higher angular speed to that for the lower speed?

b. What is the ratio of the maximum tangential speed of the laundry for the higher angular speed to that for the lower speed?

c.Find the laundry's maximum tangential speed .


d.Find the laundry's maximum radial acceleration, in terms of g.


The attempt at a solution

I have found the answer ro part a and b, which are 1.91 and 1.38 respectively.
But how do i find part c?


can someone help me pls? thanks!
 
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How are angular and tangential speed related?
 
v = r * omega
 
yah. i have found linear speed of the higher omega to be 3.456 m/s and that of the lower omega to be 2.5013 m/s. that's how i get part a and b answer..

but for part c, the ans is not 3.456 m/s.. Then i duno wad to do le..
 
janettaywx said:
yah. i have found linear speed of the higher omega to be 3.456 m/s and that of the lower omega to be 2.5013 m/s.
Show how you got those answers. What did you get for omega?
 
Doc Al said:
Show how you got those answers. What did you get for omega?

omega is as given, 619 rev/min and 448rev/min?
so v = (0.67/2)(619) = 207.365 rev/min = 3.46m/s
and v = (0.67/2)(448) = 150.08 rev/min = 2.5013m/s
 
hmm.. correct?
 
No, not correct.
janettaywx said:
v = r * omega
To use this equation, omega must be in radians/second.
 
oh! yah! okay, i got it :) thanks!

but for the last part, how to find the radial accel?

if i use a = v^2/r, I'm finding tangential accel too rite?
 
  • #10
janettaywx said:
if i use a = v^2/r, I'm finding tangential accel too rite?
That formula gives you the radial (centripetal) acceleration. (There's no tangential acceleration in this problem.)
 
  • #11
okay thanks a lot! :):)
 
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