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