Question

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.

Answer 1

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