Question

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

Answer 1

`N = 42 rev`

Step-by-step 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 = (\omega_f + \omega_i)/(2) t`

`\theta_1 = (2\pi f_2 + 2\pi f_1)/(2)(t)`

`\theta_1 = (2\pi * 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 = (\omega_f + \omega_i)/(2) t`

`\theta_2 = (2\pi f_2 + 2\pi f_1)/(2)(t)`

`\theta_2 = (2\pi * 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 = (\theta)/(2\pi)`

`N = 42 rev`