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Part 1: A baseball has a mass of 145g. The radius of gyration is 3cm from the center. What is the moment of inertia of a baseball?

Part 2: A curveball rotates at 330rev/min. What is the Angular momentum of a curveball?

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Answer:

Explanation:

Part 1: Moment of Inertia of a Baseball

A baseball has a mass of 145 grams

The radius of gyration is 3 centimeters from the center

To find the moment of inertia, we need to use a special formula:

Moment of Inertia = Mass x (Radius of Gyration)^2

So, the moment of inertia of a baseball would be:

145 grams x (3 centimeters)^2 = 1,305 gram-centimeter^2

Part 2: Angular Momentum of a Curveball

A curveball rotates at 330 revolutions per minute (that's really fast!)

Angular momentum is a measure of how much a spinning object is twisting and turning

To find the angular momentum, we need to use another formula:

Angular Momentum = Moment of Inertia x Angular Velocity

The angular velocity is the same as the rotation speed, which is 330 revolutions per minute

So, the angular momentum of a curveball would be:

1,305 gram-centimeter^2 x (330 revolutions per minute) = 430,650 gram-centimeter^2 per minute

MathPhys

Answer:

I = 1.31×10⁻⁴ kg m²

L = 4.51×10⁻³ kg m²/s

Explanation:

The moment of inertia of an object is equal to its mass times the square of its radius of gyration. The angular momentum of an object is equal to its moment of inertia times its angular velocity.

Part 1:

I = mk²

where I is moment of inertia, m is mass, and k is radius of gyration.

Given:

m = 145 g × (1 kg / 1000 g) = 0.145 kg

k = 3 cm × (1 m / 100 cm) = 0.03 m

Find I:

I = (0.145 kg) (0.03 m)²

I = 1.31×10⁻⁴ kg m²

Part 2:

L = Iω

where L is angular momentum, I is moment of inertia, and ω is angular velocity.

Given:

I = 1.31×10⁻⁴ kg m²

ω = 330 rev/min × (2π rad/rev) × (1 min / 60 s) = 34.6 rad/s

Find: L

L = (1.31×10⁻⁴ kg m²) (34.6 rad/s)

L = 4.51×10⁻³ kg m²/s

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