Maximum Tangential Speed of Laundry in Washing Machine Spin Cycles

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Homework Help Overview

The discussion revolves around the physics of a washing machine's spin cycles, focusing on angular speeds and their effects on laundry. The problem includes calculating ratios of maximum radial force and tangential speed at two different angular speeds, as well as determining maximum tangential speed and radial acceleration.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between angular speed and tangential speed, with some providing calculations for linear speeds based on given angular velocities.

Discussion Status

Some participants have shared their calculations for tangential speeds and ratios, while others are seeking clarification on the correct use of formulas and units. There is an ongoing exploration of how to approach the calculation of radial acceleration.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There is a noted need to convert angular speeds from revolutions per minute to radians per second for accurate calculations.

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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!
 
Last edited:
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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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