Question

During the spin cycle of a washing machine, the clothes stick to the outer wall of the barrel as it spins at a rate as high as 1800 revolutions per minute. The radius of the barrel is 26cm.
a. Determine the speed (in m/s) of the clothes which are located on the wall of the spin barrel.
b. Determine the acceleration of the clothes.

Answer 1

To answer the two questions, we need to know two important equations involving centripetal movement:

v = ωr (ω represents angular velocity in radians)

a = $\frac{v^{2}}{r}$

Let's apply the first equation to question a:

v = ωr

v = ((1800*2π) / 60) * 0.26

Wait. 2π? 0.26? 60? Let's break down why these numbers are written differently. In order to use the equation v = ωr, it is important that the units of ω is in radians. Since one revolution is equivalent to 2π radians, we can easily do the conversion from revolutions to radians by multiplying it by 2π. As for 0.26, note that the question asks for the units to be m/s. Since we need meters, we simply convert 26 cm, our radius, into meters. The revolutions is also given in revs/min, and we need to convert it into revs/sec so that we can get our final units correct. As a result, we divide the rate by 60 to convert minutes into seconds.

Back to the equation:

v = ((1800*2π)/60) * 0.26

v = (1800*2(3.14)/60) * 0.26

v = (11304/60) * 0.26

v = 188.4 * 0.26

v = 48.984

v = 49 (m/s)

Now that we know the linear velocity, we can find the centripetal acceleration:

a = $\frac{v^{2}}{r}$

a = $\frac{49^{2}}{0.26}$

a = 9234.6 (m/$s^{2}$)

Wow! That's fast!

We now have our answers for a and b:

a. 49 (m/s)

b. 9.2 * $10^{3}$ (m/$s^{2}$)

If you have any questions on how I got to these answers, just ask!

- breezyツ