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What is the angular momentum [tex]L_2[/tex] of a 0.500 kg tetherball when it whirls around the central pole at 70 rpm and at a radius of 115 cm?

Answer :

Final answer:

The angular momentum L2 of a 0.500-kg tetherball, rotating at a radius of 1.15 m and a speed of 70 rpm, is approximately 4.863 kg·m²/s.

Explanation:

To find the angular momentum (L2) of a tetherball, we use the formula L = Iω, where I is the moment of inertia and ω is the angular velocity. For a point mass, the moment of inertia is I = mr², where m is the mass and r is the radius of rotation. The angular velocity ω can be converted from revolutions per minute (rpm) to radians per second by using the conversion factor: 1 rpm = (2π radians)/(60 seconds).

The tetherball has a mass of 0.500 kg and moves at 70 rpm. The radius of the tetherball's circular path is 115 cm, which we convert to meters (1.15 m). First, we convert the angular velocity to radians per second:

ω = 70 rpm × (2π radians) / (60 seconds) = 7.33 radians/s

Now we calculate the moment of inertia for the tetherball:

I = (0.500 kg) × (1.15 m)² = 0.66375 kg·m²

Finally, we find the angular momentum:

L2 = I ω = (0.66375 kg·m²) × (7.33 radians/s) = 4.863 kg·m²/s

Therefore, the angular momentum L2 of the tetherball is approximately 4.863 kg·m²/s.

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