Question

A space station consists of three modules, connected to form an equilateral triangle of side length 82.0 m. Suppose 100 people, with an average mass of 75.0 kg each, live in each capsule and the mass of the modules is negligible compared to the mass of the people. At the current rotational rate the effective acceleration of gravity is g/2. (a) What angular momentum of the system

Answer 1

Answer:

Angular momentum of the system is 16221465.4617 kgm²/s

Explanation:

Given that;

length of the side of the triangle L =  82 M

m = 75.0 kg × 100 = 7500 kg

distance of each vertex from center R = L/√3 = 82/√3 = 47.34 m

effective acceleration a = 9.8 / 2 = 4.9 m/s²

we know that; effective acceleration is being provided by centripetal acceleration.

so

a = R × w²

rate of rotation w = √( a / R) = √( 4.9 / 47.34)  = 0.3217 rad/seconds

Moment of Inertia I = 3mR²

we substitute

I = 3 × 7500 × (47.34)²

Also, Angular momentum L is expressed as;

L = I × w

so

L = 3 × 7500 × (47.34)² × 0.3217

L =  16221465.4617 kgm²/s

Therefore, Angular momentum of the system is 16221465.4617 kgm²/s

Answer 2

The Angular momentum of the system is 16221465.4617 kgm²/s

• The calculation is as follows:

Given that;

Length of the side of the triangle L =  82 M

$m = 75.0 kg \times 100 = 7500 kg$

Now

distance of each vertex from center R = L/√3

= 82/√3

= 47.34 m

Now

effective acceleration $a = 9.8 \div 2$ = 4.9 m/s²

So,

a = R × w²

Now

rate of rotation $w = \sqrt( a \div R) = \sqrt( 4.9 \div 47.34)$  = 0.3217 rad/seconds

Moment of Inertia I = 3mR²

Now

$I = 3 \times 7500 \times (47.34)^2$

Also, Angular momentum L is expressed as;

L = I × w

so

L = 3 × 7500 × (47.34)² × 0.3217

=  16221465.4617 kgm²/s

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