[tex]L = 2.4 \; \text{kg} \cdot \text{m}^{2} \cdot \text{s}^{-1}[/tex]
The angular momentum of a rolling body is the product of the body's moment of inertia and its angular velocity.
[tex]L = I \cdot \omega[/tex]
where
What's the moment of inertia of this bowling ball?
Assuming that the ball is a solid sphere. For a solid sphere,
[tex]I = \dfrac{2}{5} \; m \cdot r^2[/tex].
where
[tex]r = 12.0 \; \text{cm} = 0.120 \; \text{m}[/tex] for this sphere. [tex]m = 5.5 \; \text{kg}[/tex].
[tex]I = \dfrac{2}{5} \; m \cdot r^{2}\\\phantom{I} = \dfrac{2}{5} \times 5.5 \times 0.120^{2}\\\phantom{I} = 0.0317 \; \text{kg}\cdot m^{2}[/tex]
What's the angular momentum of this bowling ball?
[tex]\omega = 12 \; \text{rev} \cdot s^{-1} = 12 \times 2\; \pi \; \text{rad} \cdot \text{s}^{-1} = 75.4 \; \text{rad}\cdot \text{s}^{-1}[/tex].
[tex]L = I \cdot \omega = 0.0317 \times 75.4 = 2.4 \; \text{kg}\cdot \text{m}^{2} \cdot \text{s}^{-1}[/tex].
The angular momentum can be determined by obtaining the angular speed and moment of inertia.
The magnitude of angular momentum of the bowling ball is [tex]2.4 \;\rm kg.m^{2}/s[/tex].
The phenomenon by which the momentum of a body under rotational motion is determined is known as angular momentum. It is expressed as the product of the angular velocity of the body and moment of inertia of the body.
Given data:
The mass of the bowling ball is, m = 5.5 kg.
The radius of ball is, r = 12.0 cm = 0.12 m.
The rotation of ball is, N = 12 rev/s.
The mathematical expression for the angular momentum of the bowling ball is,
[tex]L = \omega \times I[/tex]
here,
[tex]\omega[/tex] is the angular speed and its value is,
[tex]\omega = 2 \pi N\\\\\omega = 2\pi \times 12\\\\\omega =75.4 \;\rm rad/s[/tex]
And I is the moment of inertia of bowling ball (considering it a sphere). And its value is,
[tex]I = \dfrac{2}{5} mr^{2}\\\\I =\dfrac{2}{5} \times 5.5 \times (0.12)^{2}\\\\I = 0.0317 \;\rm kg.m^{2}[/tex]
Then, the angular momentum of the bowling ball is calculated as,
[tex]L =\omega \times I\\\\L = 75.4 \times 0.0317\\\\L = 2.4 \;\rm kg.m^{2}/s[/tex]
Thus, we can conclude that the magnitude of angular momentum of the bowling ball is [tex]2.4 \;\rm kg.m^{2}/s[/tex].
Learn more about the angular momentum here:
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