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Zina [86]
3 years ago
8

A rod of length r and mass m is pivoted at its center, and given an angular velocity, ω1. What would be the angular velocity of

a second rod, which has the same angular momentum as the first, but whose length is 3r and whose mass is 2m?
Physics
1 answer:
amid [387]3 years ago
8 0

Answer:

ω₁ / 18

Explanation:

Angular momentum is the moment of inertia times the angular velocity.

L = Iω

For a rod pivoted at its center, the moment of inertia is:

I = mr² / 12

where m is the mass and r is the length.

For the first rod:

L = (mr² / 12) ω₁

For the second rod:

L = ((2m) (3r)² / 12) ω₂

L = (18mr² / 12) ω₂

They have the same angular momentum, so:

(mr² / 12) ω₁ = (18mr² / 12) ω₂

mr² ω₁ = 18mr² ω₂

ω₁ = 18 ω₂

ω₂ = ω₁ / 18

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