Question

A top-loading washing machine has a cylindrical tub that rotates about a vertical axis. The tub of one such machine starts from rest and reaches an angular speed of 5.0 rev/s in 8.0 s. The lid of the machine is then opened, which turns off the machine, and the tub slows to rest in 12.0 s. Through how many revolutions does the tub turn during this 20 s interval? Assume constant angular acceleration while it is starting and stopping.

____ rev

Answer #1

here,

for acceleration period ,

the initial angular veloicty , w0 = 0 rev/s

the final angular velocity , w1 = 5 rev /s

time taken , t1 = 8 s

using first equation of motion

w1 = w0 + alpha * t1

5 = 0 + alpha * 8

alpha = 0.625 rev/s^2

the numner of revolutions , N1 = w0 * t1 + 0.5 * alpha1 * t1^2

N1 = 0 + 0.5 * 0.625 * 8^2 turns

N1 = 20 turns

for deacceleration period

the initial angular veloicty , w1 = 5 rev/s

the final angular velocity , w2 = 0 rev /s

time taken , t2 = 12 s

using first equation of motion

w2 = w1 + alpha * t2

- 5 = alpha * 12

alpha2 = - 0.4167 rev/s^2

the numner of revolutions , N2 = w1 * t2 + 0.5 * alpha2 * t2^2

N2 = 5 * 12 - 0.5 * 0.4167 * 12^2 turns

N2 = 30 turns

the total number of turns , N = N1 + N2

N = 20 + 30 rev

N = 50 rev

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SHOW ALL WORK ON HOW YOU GOT ANSWER PLEASE

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