Question

The tub of a washer goes into its spin cycle, starting from rest and gaining angular speed steadily for 7.00 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 8.00 s. Through how many revolutions does the tub turn while it is in motion

Answer 1

Answer:

The tub of the washer does 37.5 revolutions while it is in motion.

Explanation:

At first we assume that the tub accelerates and decelerates uniformly. In this case, it is required to determine the number of revolutions done by the tub by means of the following equations of motion:

Acceleration

$\ddot n_{1} = \frac{\dot n_{1}-\dot n_{1,o}}{t_{1}}$ (Eq. 1)

Where:

$\ddot n_{1}$ - Angular acceleration experimented by the tub, measured in revolutions per square second.

$\dot n_{1,o}$, $\dot n_{1}$ - Initial and final angular velocities of the tub, measured in revolutions per second.

$t_{1}$ - Acceleration time, measured in seconds.

$\Delta n_{1} = \dot n_{1,o}\cdot t_{1}+\frac{1}{2}\cdot \ddot n_{1}\cdot t_{1}^{2}$ (Eq. 2)

Where $\Delta n_{1}$ is the change in angular position of the tub during acceleration, measured in revolutions.

Deceleration

$\ddot n_{2} = \frac{\dot n_{2}-\dot n_{1}}{t_{2}}$ (Eq. 3)

Where:

$\ddot n_{2}$ - Angular acceleration experimented by the tub, measured in revolutions per square second.

$\dot n_{1}$, $\dot n_{2}$ - Initial and final angular velocities of the tub, measured in revolutions per second.

$t_{2}$ - Deceleration time, measured in seconds.

$\Delta n_{2} = \dot n_{1}\cdot t_{2}+\frac{1}{2}\cdot \ddot n_{2}\cdot t_{2}^{2}$ (Eq. 4)

If we know that $\dot n_{1,o} =0\,\frac{rev}{s}$, $\dot n_{1} = 5\,\frac{rev}{s}$, $t_{1} = 7\,s$, $\dot n_{2} = 0\,\frac{rev}{s}$ and $t_{2} = 8\,s$, then the amount of revolutions done by the tub during each phase is, respectively:

(Eq. 1)

$\ddot n_{1} = \frac{5\,\frac{rev}{s}-0\,\frac{rev}{s} }{7\,s}$

$\ddot n_{1} = \frac{5}{7}\,\frac{rev}{s^{2}}$

(Eq. 3)

$\ddot n_{2} = \frac{0\,\frac{rev}{s}-5\,\frac{rev}{s} }{8\,s}$

$\ddot n_{2} = -\frac{5}{8}\,\frac{rev}{s^{2}}$

(Eq. 2)

$\Delta n_{1} = \left(0\,\frac{rev}{s} \right)\cdot (7\,s)+\frac{1}{2}\cdot \left(\frac{5}{7}\,\frac{rev}{s^{2}}\right) \cdot (7\,s)^{2}$

$\Delta n_{1} = \frac{35}{2}\,rev$

(Eq. 4)

$\Delta n_{2} = \left(5\,\frac{rev}{s} \right)\cdot (8\,s)+\frac{1}{2}\cdot \left(-\frac{5}{8}\,\frac{rev}{s^{2}}\right) \cdot (8\,s)^{2}$

$\Delta n_{2} = 20\,rev$

The total amount of revolutions done by the tub while it is in motion is:

$\Delta n_{T} = \Delta n_{1}+\Delta n_{2}$

$\Delta n_{T} = 37.5\,rev$

The tub of the washer does 37.5 revolutions while it is in motion.