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2 years ago

A long uniform thin rod of length 55 cm and mass 1.2 kg in FIGURE 2 rotates in

Physics

1 answer:

arlik [135]2 years ago

4 0

Answer:

(i) 0.121 kg m²

(ii) 0.363 kg m²/s

(iii) 2 rad/s

Explanation:

(i) The moment of inertia of a rod about its end is:

I = 1/3 mr²

where m is the mass and r is the length.

Given m = 1.2 kg and r = 0.55 m:

I = 1/3 (1.2 kg) (0.55 m)²

I = 0.121 kg m²

(ii) Angular momentum is the moment of inertia times the angular velocity:

L = Iω

L = (0.121 kg m²) (3 rad/s)

L = 0.363 kg m²/s

(iii) This time, the moment of inertia of the putty is included.

I = 1/3 mr² + Mr²

I = 0.121 kg m² + (0.2 kg) (0.55 m)²

I = 0.1815 kg m²

Angular momentum is conserved:

L = Iω

0.363 kg m² = (0.1815 kg m²) ω

ω = 2 rad/s

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