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

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 1/2 MR^2, and the ant is basically a point particle. What is the angular deceleration of the disc as the ant crawling outward (as a function of the ant's 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

let the angular deceleration be a

angular deceleration = torque . I

I = inertia of disc

when ant is at distance r ,

the moment of inertia of the ant = M*R^2

the force exerted by ant = F

considering the ant and disc as a system

initial angular momentum = final angular momentum

M2R^2/2*W1 + 0 = M2R2^2*W2 + M1*r^2*Want

Want = Vant/r = 0.01m/s / r

r in m

W1 = 60 rpm = 60 * 2pi/60 = 6.28 rad/s

M2 = 0.1 kg

M1 = 10*10^-3 g = 10^-5 kg

therefore

putting these values in the angular momentum balance equation

we get

W2 = R2^2/2*(6.28) - M1/M2*(r^2)*(0.01/r)

W2 = 0.25/2*(6.28) - 10^-4(0.01)*r

W2 = (0.785 - 10^-6r ) rad/s

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