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Ludmilka [50]
2 years ago
10

A merry-go-round has a small box placed on it at a distance R' from its axis of

Physics
1 answer:
Lady_Fox [76]2 years ago
7 0

The time after it starts spinning, that the box will slip off is 0.1 s.

<h3>Apply the principle of conservation of angular momentum</h3>

I₁ω₁ - I₂ω₂ = 0

I₁ω₁  =  I₂ω₂

where;

  • I₁ is initial moment of inertia
  • I₂ is final moment of inertia
  • ω₁ is initial angular speed
  • ω₂ is final angular speed

The final angular speed when the box slides off;

ω₂ = I₁ω₁ / I₂

ω₂  = [0.5(m₁ + m₂)(R + r)²] / (0.5MR²)

ω₂  = [0.5(30 + 0.3)(3.1 + 1.4)²] / (0.5 x 30 x 3.1²)

ω₂  = 2.13 rad/s

Time taken for the for box to slide off;

τ = Iα

τ = I(ω/t)

τ = (Iω)/t

t =  (Iω)/τ

t = (0.5 x 0.3 x 1.4² x 2.13)/6

t = 0.1 s

Thus, the time after it starts spinning, that the box will slip off is 0.1 s.

Learn more about moment of inertia here: brainly.com/question/3406242

#SPJ1

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Explanation:

The package in relation to the point where it was released falls a certain distance that is calculated by applying the horizontal motion formulas , as the horizontal speed of the plane and the height above the ground are known, the time that It takes the package to reach its destination and then the horizontal distance (x) is calculated from where it was dropped, as follows:    

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