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1 year ago

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]1 year 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

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(A) is correct option.

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