# Maximum tangential speed

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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Doc Al
Mentor
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. thats 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..

Doc Al
Mentor
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?

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?

Doc Al
Mentor
No, not correct.
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?

Doc Al
Mentor
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.)

okay thanks a lot!! :):)