Question

In a laundromat, during the spin-dry cycle of a washer, the rotating tub goes from rest to its maximum angular speed of 2.2 rev/s in 8.8 s. You lift the lid of the washer and notice that the tub accelerates and comes to a stop in 20.0 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:

$n_{T} = 31.68\,rev$

Explanation:

The angular acceleration is:

$\ddot n_{1} = \frac{2.2\,\frac{rev}{s} -0\,\frac{rev}{s} }{8.8\,s}$

$\ddot n_{1} = 0.25\,\frac{rev}{s^{2}}$

And the angular deceleration is:

$\ddot n_{2} = \frac{0\,\frac{rev}{s}-2.2\,\frac{rev}{s} }{20\,s}$

$\ddot n_{2} = -0.11\,\frac{rev}{s^{2}}$

The total number of revolutions is:

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

$n_{T} = \frac{\left(2.2\,\frac{rev}{s} \right)^{2}-\left(0\,\frac{rev}{s} \right)^{2}}{2\cdot \left(0.25\,\frac{rev}{s^{2}} \right)} + \frac{\left(0\,\frac{rev}{s} \right)^{2}-\left(2.2\,\frac{rev}{s} \right)^{2}}{2\cdot \left(-0.11\,\frac{rev}{s^{2}} \right)}$

$n_{T} = 31.68\,rev$