Question

In a laundromat, during the spin-dry cycle of a washer, the rotating tub goes from rest to its maximum angular speed of 9.2 rev/s in 7.3 s. You lift the lid of the washer and notice that the tub decelerates and comes to a stop in 12.0 s. Assuming that the tub rotates with constant angular acceleration while it is starting and stopping, determine the total number of revolutions undergone by the tub during this entire time interval.

Answer 1

Answer:

90 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

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{9.2-0}{7.3}\\\Rightarrow a=1.26027\ rev/s^2$

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}\\\Rightarrow \theta=\frac{9.2^2-0^2}{2\times 1.26027}\\\Rightarrow \theta=33.5801\ rev$

Number of revolutions in the 7.3 seconds is 33.5801

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-9.2}{12}\\\Rightarrow a=-0.76\ rev/s^2$

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-9.2^2}{2\times -0.76}\\\Rightarrow \theta=55.68421\ rev$

Number of revolutions in the 12 seconds is 55.68421

Total total number of revolutions is 33.5801+55.68421 = 89.26431 = 90 revolutions