A giant ant (10 mg) is crawling out of a freely-rotating disc (100g) at a speed

Question

A giant ant (10 mg) is crawling out of a freely-rotating disc (100g) at a speed of 1 cm/s. The disc

has a diameter of 100 cm and is rotating at an initial angular speed of 60 RPM (with the ant when

it was just crawling out from the small hole). The moment of inertia of the disc can be

approximated as ½ MR2, and the ant is basically a point particle. What is the angular

deceleration of the disc as the ant is crawling outward (as a function of the ants position)? [Hint:

What is the angular speed of the disc when the ant is at a distance r from the center? And how

does this speed change when r changes?

A giant ant (10 mg) is crawling out of a freely-rotating disc (100g) at a speed of 1 cm/s. The disc has a diameter of 100 cm and is rotating at an initial angular speed of 60 RPM (with the ant when it was just crawling out from the small hole). The moment of inertia of the disc can be approximated as Â1/2 MR2, and the ant is basically a point particle. What is the angular deceleration of the disc as the ant is crawling outward (as a function of the ants position)? [Hint: What is the angular speed of the disc when the ant is at a distance r from the center? And how does this speed change when r changes?

Explanation / Answer

initial angular momentum = I1w1 = 0.5*0.1*0.5^2*60*2*3.142/60 = 0.07855

I1w1 = I(r)w(r)

I(t) = 0.5*0.1*0.5^2 + 10^-5*(r)^2

0.07855 = (0.0125 + 10^-5*r^2)*w(r)

w(r) = 0.07855/(0.0125 + 10^-5*r^2)

angular acceleration = dw/dt = 0.07855*-1*10^-5*2*r*dr/dt/(0.0125 + 10^-5*r^2)^2

= -1.571*10^-6*r*0.01/(0.0125 + 10^-5*r^2)^2 = -1.571*10^-8*r/(0.0125 + 10^-5*r^2)^2

angular deceleration = -(acceleration) = 1.571*10^-8*r/(0.0125 + 10^-5*r^2)^2

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