Question

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 7.0 rev/s in 7.0 s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 13.0 s. Through how many revolutions does the tub turn during this 20 s interval? Assume constant angular acceleration while it is starting and stopping.

Answer 1

Answer:

70 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

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{7-0}{7}\\\Rightarrow a=1\ rev/s^2$

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}\\\Rightarrow \theta=\frac{7^2-0^2}{2\times 1}\\\Rightarrow \theta=24.5\ rev$

Number of revolutions in the 7 seconds is 24.5

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-7}{13}\\\Rightarrow a=-0.5384\ rev/s^2$

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-7^2}{2\times -0.5384}\\\Rightarrow \theta=45.5\ rev$

Number of revolutions in the 13 seconds is 45.5

Total total number of revolutions in the 20 second interval is 24.5+45.5 = 70 revolutions