Question

a small cylindrical satellite is spinning at 40.0 rpm (revolutions per minute) to stabilize it. in order to start its mission the satellite must be de-spun to 2.00 rpm. the satellites cylindrical radius is 0.700 m with a mass of 400 kg. the de-spin will happen with two 3.00 kg masses on very light wires. (a) what length of wire, r, must the masses go out to in order to de-spin the satellite to 2.00 rpm? (b) what will be the initial moment of inertia? (c) what will be the final moment of inertia? (d) check your answer to (a) by using (b), (e) and conservation of angular momentum to calculate w.

Answer 1

Given,

_o = 40.0 rpm

= 2.00 rpm

r = 0.70 m

M = 400 kg'

m = 3 kg

a)

By applying the conservation of angular momentum

final angular momentum = initial angular momentum

I * = I_o * o

(0.5*M*R² + 2*m*r²)*w = (0.5*M*R² + 2*m*R²)*_o

(0.5*M*R² + 2*m*r²) = (0.5*M*R² + 2*m*R²⁾*_o /

2*m*r² = (0.5*M*R² + 2*m*R²)*_o / - 0.5*M*R²

r² = ((0.5*M*R² + 2*m*R²)* _o / - 0.5*M*R²)/(2*m)

r = √ (((0.5*M*R2 + 2*m*R^2)*wo/w - 0.5*M*R^2)/(2*m)

= sqrt(( (0.5*400*0.7² + 2*3*0.72)*40/2 - 0.5*400*0.7²)/(2 * 3) )

= 17.9 m

b)

I_o = 0.5*M*R² + 2*m*R²

= 0.5*400*0.7² + 2*3*0.7²

= 101 kg.m²

c) I = 0.5*M*R² + 2*m*r²

= 0.5*400*0.7² + 2*3*17.9²

= 2020 kg.m²

d) I * = I_o * _o

w = I_o * _o / I

= 101 * 40 / 2020

= 2.00 rpm