Question

6. The tub of a washer goes into its spin cycle, starting from rest and gaining angular speed steadily for 8.00 s, at which time it is turning at 5.00 rev/s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub smoothly slows to rest in 12.0. Through how many revolution does the tub turn while it is in motion?

Answer 1

When an object traces a circular path about an axis of rotation, its motion is considered to be rotatory motion.

We categorize the problem to be rotational kinematics under constant angular acceleration.

It is given that

Angular velocity

$ω= 5.0 \;rev/s \\ = 5(2 \pi \;rad/s) \\ = 10 \pi \;rad/s$

Time of constant angular velocity t = 8 s

Time of angular retardation t’= 12 s

Equations to use

(i) angular acceleration

$α = \frac{ω-ω_o}{t}$

(ii) angular displacement

$θ = ω_ot +(\frac{1}{2})αt^2$

Substitution of the values

When the tub is speeding up, the angular acceleration of the tub is

$α_1 = \frac{ω-0}{t} \\ = \frac{10 \pi \;rad/s -0}{8 \;s} \\ = 1.25 \pi \;rad/s^2$

The angular displacement

$θ_1 = ω_ot +(\frac{1}{2})α_1t^2 \\ = 0 + (\frac{1}{2})(1.25 \pi \;rad/s^2)(8 \;s)^2 \\ = 40 \pi \;rad$

However, one revolution

$=2 \pi \;rad$

Therefore, the number of revolutions made by the tub in 8s when it is accelerating is

$n_1 =\frac{40 \pi \;rad}{2 \pi \;rad/rev} \\ = 20 \;rev$

When the tub is slowing down from

$10 \pi \;rad/s$

to zero in 12 s, then the angular acceleration of the tub is

$α_2 = \frac{0 -ω}{t} \\ = \frac{0-10 \pi \;rad/s}{12 \;s} \\ = -0.833 \pi \;rad/s^2$

The angular displacement

$θ_2 =ω_ot + (\frac{1}{2})α_2t^2 \\ = (10 \;rad/s)(12 \;s) + (\frac{1}{2})(-0.833 \pi \;rad/s^2)(12 \;s)^2 \\ = 120 \pi \;rad -60 \pi \;rad \\ = 60 \pi \;rad$

Therefore, the number of revolutions made by the tub in 12s when it is decelerating is

$n_2= \frac{60 \pi \;rad}{2 \pi \;rad/rev} \\ = 30 \;rev$

The total number of revolutions the tub makes when it is in motion

= 20 rev + 30 rev

= 50 rev