Question

Clothes are flung to the sides of a round tub during the spin-cycle of a washing machine.If the tub's period of rotation is 0.2s and the tub's radius is 35cm,answer the related qestions below: (a)The frequency of roation (b)The circumference of rotation (c)The spinning of speed (d)The centripetal acceleration (e)The centripetal force of rotation if total mass rotated is 30kg (f)The power of washing machine​

Answer 1

Answer:

Explanation:

First I need to tell you that I used .20 s for the period of rotation instead of just .2, and I used 30.0 kg for the mass instead of just 30. The reason being that both those numbers as stated in the problem only have 1 significant digit and that's not generally enough to get the accuracy you're looking for. Adding a 0 to the ends of each of those numbers doesn't change the value of the numbers, only the number of sig fig's in each. Beginning with a:

a. $f=\frac{1}{T}$ so $f=\frac{1}{.2}$ and f = 5.0 Hz

b. The circumference is the distance around the outside of the washer's drum. We need to find that, but before we do, I'm going to state the radius in meters instead of cm. 35 cm = .35 m. Therefore,

C = 2(3.1415)(.35) so

C = d = 2.2 m

c. The speed of the washer is found in d = rt, where r is the rate and our velocity and d is the distance around the outside of the drum (circumference). Therefore,

2.2 = v(.20) so

v = 11 m/s

d. The centripetal acceleration has an equation

$a_c=\frac{v^2}{r}$ so

$a_c=\frac{(11)^2}{.35}$ and

$a_c=\frac{121}{.35}$ so

$a_c=350\frac{m}{s^2}$

e. The centripetal force has an equation

$F_c=\frac{mv^2}{r}$ and

$F_c=\frac{(30.0)(11)^2}{.35}$ and

$F_c=$ 1.0 × 10⁴ N

f. The equation for Power is

$P=\frac{W}{t}$ where W is work and W = FΔx (force times displacement). Therefore,

$P=\frac{(1.0*10^4)(2.2)}{.20}$ so

P = 1.1 × 10⁵ Watts