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1 month ago

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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?

1 answer:

Ostrovityanka [3.2K]1 month ago

0 0

Provided Details:

Cylinder tub diameter = d = 50 cm = 0.50 m

Acceleration = α = 3g

Needed Information:

1. Rotation speed in rev/min = ω =?

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

Response:

1. ω = 103.5 rev/min

2. v = 2.71 m/s

Explanation:

Centripetal acceleration is defined by

α = ω²r

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

The relationship between diameter and radius is expressed as

r = D/2

r = 0.50/2

r = 0.25 m

Given that the acceleration equals 3g, where g is the gravitational acceleration of 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 to rev/s, divide by 2π

ω = 10.84/2π

ω = 1.752 rev/s

To convert rev/s to rev/min, multiply by 60

ω = 1.752*60

ω = 103.5 rev/min

Hence, the rotation rate is determined to be 103.5 rev/min

2. The tangential velocity can be computed 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

Thus, the tangential speed is 2.71 m/s.

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