The spin cycles of a washing machine have two angular speeds, 416 rev/min and 63

ID: 1371935 • Letter: T

Question

The spin cycles of a washing machine have two angular speeds, 416 rev/min and 632 rev/min . The internal diameter of the drum is 0.560 m

Part A

What is the ratio of the maximum radial force on the laundry for the higher angular speed to that for the lower speed? (F higher/F lower)=?

Part B

What is the ratio of the maximum tangential speed of the laundry for the higher angular speed to that for the lower speed? (V higher/ V lower)=?

Part C

Find the laundry's maximum tangential speed . (m/s)

Part D

Find the laundry's maximum radial acceleration, in terms of g . (a rad=? g)

Given that

The spin cycles of a washing machine have two angular speeds(w1) =416 rev/min =416*2pi/60 =43.541rad/s

angular speed(w2) = 632 rev/min =632*2pi/60 =66.149rad/s

The internal diameter of the drum is(D) = 0.560 m and the radius is (R) =0.280m

a)What is the ratio of the maximum radial force on the laundry for the higher angular speed to that for the lower speed is given by

Force (F) =ma =mw2r

The maximum radial force is =mw2R

Therefore

Fhigher/Flower =(whigher/wlower)2 =(66.149/43.541)2 =(1.5912)2 =2.308

What is the ratio of the maximum tangential speed of the laundry for the higher angular speed to that for the lower speed is given by

Tangential speed(v) =wr

Max tangential speed is v =wR

So, vhigher/vlower =(whigher/wlower) =66.149/43.541 =1.5912

Find the laundry's maximum tangential speed is given by v =whigher*R =(66.149)(0.280) =18.521m/s

Find the laundry's maximum radial acceleration, in terms of g is given by

a =whigher2R =(66.149)2*(0.280) =1225.193m/s2=124.89g

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