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saveliy_v [14]
3 years ago
8

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 5.0 kg and diameter 0.36 m . The potter then throws a 3.1-kg chunk 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? Ignore friction.
Physics
1 answer:
weeeeeb [17]3 years ago
8 0

Answer:

ωf =1.78 rev/s

Explanation:

it is a non elastic collision, so conservation of angular momentum is used.  and there id no friction

I₁ωi₁ + I₂ωi₂ = (I₁ + I₂)ωf

ωi₁ = 2.0 rev/s*(2π) rad/rev = 4.0π rad/s

assume ωi₂ = 0

Sol the equation to wf

ωf =  (½m₁r₁²)ωi₁ /  (½m₁r₁² + ½m₂r₂²)

ωf = (m₁r₁²)*ωi₁  /  (m₁r₁² + m₂r₂²)

ωf = (5.0kg*(0.18m)²)4.0 π rad/s / ((5.0kg*(0.18m)²) + (3.1*(0.08m)²))

ωf =11.19 rad/s/ 2π

ωf =1.78 rev/s

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The pressure outside

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             P_interior = P_exterior

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            P₀ = 10⁵ (40 + h ’) / 10 = 4 10⁵ + h’ 10⁴

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             (4 10⁵ + h’ 10⁴) + ρ g h’ = Pₐ + ρ g h

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            h' (1.98 10⁴) = -3 105 + 3.92 10⁵

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