Question

A washing machine drum has an internal diameter of 47.0 cm. It spins at 16.0 revolutions per second. As it spins, a sock is pressed against the side of the drum. The sock is stationary with respect to the drum. a) What is the speed of the sock? b) After 3.25 revolutions, what is the average velocity of the sock?

Answer 1

Answer:

a = 23.6

b = 1.64

Explanation:

A)

$16 R/s$ is the frequency, and $0.47 m$ is the diameter. Hence:

$16R/s = f = \frac{1}{T}$,

and

$d = 0.47m = 2r$, $r = \frac{d}{2} = 0.235m$

So, we can solve for the period (T).

$16Rs^{-1} = f = \frac{1}{T}$

$\frac{1}{16} =T$

Then, we have to remember that, in a uniform circular motion, the centripetal acceleration is given by $a=\frac{v^2}{r} =\frac{4\pi ^2r}{T^2}$ .

Now we have to substitute.

$a=\frac{4\pi^{2}(0.235) }{(1/16)^{2}}$

$a=2,375.02 ms^{-2}$

Then, we can use the equation $a=\frac{v^2}{r}$ to solve for $v$.

$2,375.02=\frac{v^2}{r}\\2,375.02=\frac{v^2}{0.235} \\(0.235)(2,375.02)=v^2\\\sqrt{(0.235)(2,375.02)} =v\\23.6=v$

B)

3.25 revolutions means that the displacement will be 0.25 or 1/4 of the circumference.

Hence, the displacement becomes the hypotenuse in a right triangle is where we can use the Pythagorean theorem.

$a=b=r\\a=b=0.235$

$a^{2} +b^{2} =c^2\\(0.235)^2 + (0.235)^2=c^2\\\sqrt{(0.235)^2 + (0.235)^2} =c\\\frac{0.47}{\sqrt{2} } =c$

The total time for 3.25 revolutions has to be equal to

$3.25T=(3.25)(1/16)$

So, since the average velocity is $\frac{displacement}{time}$, we can input the values we have and solve for the result.

$\frac{displacement}{time} = v_a\\\frac{0.47/\sqrt{2}}{(3.25)(1/16)} = v_{a} \\1.64ms^{-1} = v_{a}$