A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/s. The wheel can be considered a uniform disk of mass 4.9 kg and diameter 0.40m. The potter then throws a 2.7 kg chucnk of clay, approximately shaped as a flat disk of radius 8.0 cm, onto the center of the rotating wheel. What is the frequency of the wheel after the clay sticks to it?

1.84 rev/s

Explanation:

Given

Initial frequency, f1 = 2 rev/s, w1 = 2πf1 = 12.568 rad/s

Mass of the wheel, m(w) = 4.9 kg

Diameter of wheel, d(w) = 0.4 m

Mass of clay, m(c) = 2.7 kg

Radius of clay, r(c) = 8 cm = 0.08 m

Moment of inertia of wheel,

I(w) = 1/2m(w)R(w)²

I(w) = 1/2 * 4.9 * (0.4/2)²

I(w) = 1/2 * 4.9 * 0.04

I(w) = 0.098 kgm²

Moment of inertia of clay

I(c) = 1/2m(c)R(c)²

I(c) = 1/2 * 2.7 * 0.08²

I(c) = 1/2 * 2.7 * 0.0064

I(c) = .00864 kgm²

Since chunk of clay attached to the wheel, then perfectly inelastic collision occurred, and as such, momentum is conserved.

I(i)w(i) = I(f)w(f)

I(i)w(i) = (I(c) + I(w)) * w(f), on substituting, we have

0.098 * 12.568 = (0.00864 + 0.098) * w(f)

1.232 = 0.10664 * w(f)

w(f) = 1.232 / 0.10664

w(f) = 11.55 rad/s

remember, w = 2πf, thus

f = w/2π

f = 11.55/6.284

f = 1.84 rev/s

thus, final frequency of the wheel is 1.84 rev/s

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this

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