Question

A woman with mass 50 kg is standing on the rim of a large horizontal disk that is rotating at 0.80 rev/s about an axis through its center. The disk has mass 110 kg and radius 4.3 m.
Calculate the magnitude of the total angular momentum of the woman–disk system. (Assume that you can treat the woman as a point.)

Answer 1

Answer:

Angular momentum of disc + woman is given as

$L = 9759 kg m^2/s$

Explanation:

As we know that the angular momentum of disc + woman is given as

$L = (I_{disk} + I_{woman}) \omega$

here we know for disk moment of inertia is given as

$I_{disk} = \frac{1}{2}mR^2$

$I_{disk} = \frac{1}{2}(110)(4.3)^2 = 1017 kg m^2$

similarly for woman we will have

$I_{woman} = MR^2$

$I_{woman} = (50)(4.3^2)$

$I_{woman} = 924.5 kg m^2$

so we will have

$\omega = 2\pi (0.80) rad/s$

so angular momentum is given as

$L = (1017 + 924.5)(2\pi 0.80)$

$L = 9759 kg m^2/s$

Answer 2

The magnitude of the total angular momentum of the woman–disk system is 9,759.7 kgm²/s.

The given parameters;

• mass of the woman, M = 50 kg
• angular speed of the disk, ω = 0.8 rev/s
• mass of the disk, m = 110 kg
• radius, r = 4.3 m

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

$I_t = I_{disk}\omega + I_{woman}\omega\\\\I_t = ( I_{disk} + I_{woman})\omega\\\\I_t = (\frac{1}{2} mr^2 \ + \ Mr^2) \omega \\\\I_t = (\frac{1}{2}\times 110 \times 4.3^2 \ \ + \ \ 50\times 4.3^2)\times 0.8 \times \frac{rev}{s} \times \frac{2\pi \ rad}{rev} \\\\I_t = (1941.45 \ kgm^2 ) \times 5.027 \ rad/s \\\\I_t = 9,759.7 \ kgm^2/s$

Thus, the magnitude of the total angular momentum of the woman–disk system is 9,759.7 kgm²/s.

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