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Vikki [24]
3 years ago
10

At the outer edge of a rotating space habitat, 130 m from the center, the rotational acceleration is g. What is the rotational a

cceleration at a distance of 65 m from the center of the habitat? Be sure to show all work and steps to support your answer.
Physics
1 answer:
enyata [817]3 years ago
6 0

Answer:

Explanation:

Given:

R1 = 130 m

R2 = 65 m

w^2R = g

Assume, g = 9.81 m/s^2

w^2 = 9.81/130

w = 0.275 rad/s

At R2 = 65 m

g = w^2R

= (0.275^2) × 65

= 4.905 m/s^2

In conclusion,

g × R = k

g1/R1 = g2/R2

g2 = (g1 × 65)/130

= g1 ×1/2

= g1/2

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How fast, in rpm, would a 5.6 kg, 25-cm-diameter bowling ball have to spin to have an angular momentum of 0.26 kgm2/s
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