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mamaluj [8]
3 years ago
7

The tub of a washing machine goes into its spin cycle, starting from rest and gaining angular speed steadily for 6.00 s, at whic

h time it is turning at 4.00 rev/s. At this point, the lid of the washing machine is opened, and a safety switch turns it off. The tub then smoothly slows to rest in 15.0 s. Through how many revolutions does the tub rotate while it is in motion?_____________ rev
Physics
1 answer:
MatroZZZ [7]3 years ago
3 0

Answer:

N = 42 rev

Explanation:

As we know that initial angular speed of the tub was zero and then it increases uniformly to 4 rev/s in t = 6.00 s

so we will have

\theta_1 = \frac{\omega_f + \omega_i}{2} t

\theta_1 = \frac{2\pi f_2 + 2\pi f_1}{2}(t)

\theta_1 = \frac{2\pi \times 4 + 0}{2}(6)

\theta_1 = 75.4 rad

Now when the tub will comes to rest uniformly after opening the lid in time interval of t = 15 s

then we have

\theta_2 = \frac{\omega_f + \omega_i}{2} t

\theta_2 = \frac{2\pi f_2 + 2\pi f_1}{2}(t)

\theta_2 = \frac{2\pi \times 4 + 0}{2}(15)

\theta_2 = 188.5 rad

Now total angular displacement of the tub is given as

\theta = \theta_1 + \theta_2

\theta = 75.4 + 188.5

\theta = 263.9 rad

so number of revolutions is given as

\N = \frac{\theta}{2\pi}

N = 42 rev

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