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aksik [14]
3 years ago
11

A carousel has a radius of 1.70 m and a moment of inertia of 110 kg · m2. A girl of mass 44.0 kg is standing at the edge of the

carousel, which is rotating with an angular speed of 3.40 rad/s. Now the girl walks toward the center of the carousel and stops at a certain distance from the center d. The angular speed of the carousel is now 5.4 rad/s. How far from the center, in meters, did the girl stop?
Physics
1 answer:
bonufazy [111]3 years ago
8 0

Answer:

Explanation:

Initial moment of inertia of the carousel + girl

I₁ = 110 + 44 x 1.7²

= 110 + 127.16

= 237.16 kgm².

final moment of inertia of carousel + girl

I₂ = 110 + 44 x d²

applying law of conservation of angular momentum

I₁ ω₁ = I₂ω₂

ω₁ and ω₂ are angular velocities of the carousel before and after .

237.16 x 3.4 = (110 + 44 x d²)x 5.4

806.34 = 594 + 237.6 d²

237.6 d² = 212.34

d²= .8936

d = .9453 m

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Answer:

The speed of the block is 8.2 m/s

Explanation:

Given;

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height above the top of the spring, h = 5.5 m

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¹/₂K(0.25)² = 2.1 x 9.8(5.5 +0.25)

0.03125K = 118.335

K = 118.335 / 0.03125

K = 3786.72 N/m

Total energy stored in the block at rest is only potential energy given as:

E = U = mgh

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This is equal to elastic potential energy stored in the spring,

Then, kinetic energy of the spring is given as:

K.E = E - W

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K.E = 70.59 J

To determine the speed of the block due to this energy:

KE =  ¹/₂mv²

70.59 =  ¹/₂ x 2.1 x v²

70.59 = 1.05v²

v² = 70.59 / 1.05

v² = 67.229

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