Question

Spin cycles of washing machines remove water from clothes by producing a large radial acceleration at the rim of the cylindrical tub that holds the water and clothes. Suppose that the diameter of the tub in a typical home washing machine is 50 cm.

1.What is the rotation rate, in rev/min, of the tub during the spin cycle if the radial acceleration of points on the tub wall is 3g?
2.At this rotation rate, what is the tangential speed in m/s of a point on the tub wall?

Answer 1

Given Information:

Diameter of the cylindrical tub = d = 50 cm = 0.50 m  

Acceleration = α = 3g

Required Information:

1. Rotation rate in rev/min = ω = ?

2. Tangential speed in m/s = v = ?

Answer:

1. ω = 103.5 rev/min

2. v = 2.71 m/s

Explanation:

We know that centripetal acceleration is given by

α = ω²r

Where ω is the angular speed or rotation rate and r is the radius.

The relation between diameter and radius is given by

r = D/2

r = 0.50/2

r = 0.25 m

Since it is given that the acceleration is equal to 3g where g is the gravitational acceleration 9.81 m/s².

α = ω²r

3g = ω²r

ω² = 3g/r

ω = √(3g/r)

ω = √(3*9.81/0.25)

ω = 10.84 rad/s

To convert rad/s into rev/s divide it by 2π

ω = 10.84/2π

ω = 1.752 rev/s

To convert rev/s into rev/min multiple it by 60

ω = 1.752*60

ω = 103.5 rev/min

Therefore, the rotation rate is 103.5 rev/min

2. The tangential speed can be found using

v = ωr

Where ω is the rotation rate in rad/s and r is the radius.

v = 10.84*0.25

v = 2.71 m/s

Therefore, the tangential speed is 2.71 m/s

Answer 2

1. To determine the rotation rate, in rev/min, of the tub during the spin cycle is 103.5 rpm.

2. At this rotation rate, the tangential speed in m/s of a point on the tub wall is 25.88 m/s.

Given the following data:

• Diameter = 50 cm to m = 0.5 m

• Centripetal acceleration = 3g

Radius = $\frac{Diameter}{2} = \frac{0.5}{2} =0.25 \;meter$

Acceleration due to gravity = 9.8 $m/s^2$

1. To determine the rotation rate, in rev/min, of the tub during the spin cycle:

Mathematically, the centripetal acceleration in rotational motion is given by the formula:

$\alpha =r\omega^2\\\\3g = r \omega^2\\\\\omega^2 = \frac{3g}{r}\\\\\omega = \sqrt{\frac{3g}{r}}$

Substituting the given parameters into the formula, we have;

$\omega = \sqrt{\frac{3\times 9.8}{0.25}} \\\\\omega = \sqrt{\frac{29.4}{0.25}} \\\\\omega = \sqrt{117.6} \\\\\omega = 10.84\;rad/s$

In rev/min:

1 rad/s = 9.5493 rpm

10.84 rad/s = $9.5493 \times 10.84 = 103.5\;rpm$

Rotation rate = 103.5 rpm.

2. To find the tangential speed in m/s of a point on the tub wall, at the above rotation rate:

Mathematically, tangential speed is given by the formula:

$V = r\omega\\\\V = 0.25 \times 103.5$

Tangential speed, V = 25.88 m/s

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