Question

The tub of a washer goes into its spin-dry cy-cle, starting from rest and reaching an angularspeed of 3.1 rev/s in 10.7 s.At this point theperson doing the laundry opens the lid, anda safety switch turns off the washer. The tubslows to rest in 11.2 s.During the whole process of starting upand stopping, through how many revolutionsdoes the tub turn? Assume constant angularacceleration while it is starting and stopping.Answer in units of rev.

Answer 1

Answer:

35 revolutions

Explanation:

t = Time taken

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

$\theta$ = Number of rotation

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{3-0}{10.7}\\\Rightarrow \alpha=0.28037\ rev/s^2$

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}\\\Rightarrow \theta=\frac{3.1^2-0^2}{2\times 0.28037}\\\Rightarrow \theta=17.13806\ rev$

Number of revolutions in the 10.7 seconds is 17.13806

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-3.1}{11.2}\\\Rightarrow a=-0.27678\ rev/s^2$

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-3.1^2}{2\times -0.27678}\\\Rightarrow \theta=17.36035\ rev$

Number of revolutions in the 11.2 seconds is 17.36035

Total total number of revolutions in the 21.9 second interval is 17.13806+17.36035 = 34.49841 = 35 revolutions