Question

A woman with mass 50 kg is standing on the rim of a large disk that is rotating at 0.80 rev/s about an axis through its center. The disk has mass 270 kg and radius 4.0 m.

Calculate the magnitude of the total angular momentum of the woman–disk system. (Assume that you can treat the woman as a point.)

Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer:

angular momentum, $L=1.4\times 10^4\ kg-m^2/s$

Explanation:

Given that,

Mass of the woman, m = 50 kg

Angular velocity of the disk, $\omega=0.8\ rev/s=5.02\ rad/s$

Mass of the disk, m' = 2670 kg

Radius of the disk, R = 4 m

We need to find the magnitude of the total angular momentum of the woman–disk system. The moment of inertia of the system is equal to the sum of moment of inertia of women and the moment off inertia of the disk.

$I=mR^2+\dfrac{1}{2}m'R^2\\\\I=R^2(m+\dfrac{1}{2}m')\\\\I=(4)^2\times (50+\dfrac{1}{2}\times 270)\\\\I=2960\ kg-m^2$

The angular momentum is given by :

$L=I\omega\\\\L=2960\times 5.02\\\\L=14859.2\ kg-m^2/s$

or

$L=1.4\times 10^4\ kg-m^2/s$

So, the magnitude of the total angular momentum of the woman–disk system is $1.4\times 10^4\ kg-m^2/s$. Hence, this is the required solution.

Answer 2

The magnitude of the total angular momentum of the woman–disk system is 14,879.9 kgm²/s.

The given parameters;

• mass of the woman, m₁ = 50 kg
• angular speed of the disk, ω = 0.8 rev/s
• mass of the disk, m₂ = 270 kg
• radius of the disk, R = 4 m

The moment of inertia of the disk and the woman is calculated as follows;

$I = M_2R^2 \ + \ \frac{1}{2} M_1R^2\\\\I = R^2 (M_2+ 0.5M_1)\\\\I = 4^2(50 \ + \ 0.5(270))\\\\I = 2,960\ kgm^2$

The angular speed in rad/s is calculated as;

$\omega = 0.8 \ \frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} = 5.027 \ rad/s$

The magnitude of the total angular momentum of the woman–disk system is calculated as follows;

$L = I \omega \\\\L = 2,960 \times 5.027\\\\L = 14,879.9 \ kg m^2/s$

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