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Tema [17]
3 years ago
6

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?
Physics
1 answer:
denis23 [38]3 years ago
0 0

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

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Answer:

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two billiard balls moving along the same line hit each other head-on. each has a mass of 0.220 kg; one has an initial velocity o
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Hi there!

Since the collision is elastic, we must also satisfy the following condition:

Ei = Ef, or:

KEi = KEf

Begin by writing an expression for momentum. (p = mv) Remember that one ball's direction is negative; in this instance, we can let the second ball be moving LEFT.

mv1 + mv2 = mvf1 + mvf2

0.220(1.84) + 0.220(-.530) = 0.220(vf1 + vf2)

0.2882/0.220 = vf1 + vf2

1.31 = vf1 + vf2

Now, we can express this as a conservation of energy:

1/2mv1² + 1/2mv2² = 1/2mvf1² + 1/2mvf2²

Plug in values and simplify:

0.403315 = 1/2m(vf1² + vf2²)

Simplify further:

3.6665 = vf1² + vf2²

Use the equation derived from momentum above and solve for one variable:

vf2 = 1.31 - vf1

Plug in this expression for vf2:

3.6665 = vf1² + (1.31 - vf1)²

Expand:

3.6665 = vf1² + 1.7161 - 2.62vf1 + vf1²

Simplify:

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Solve for vf1 using a graphing calculator:

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