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A merry-go-round has a small box placed on it at a distance R' from its axis of
rotation. Initially (at time = 0), the merry-go-round is stationary. At time t = 0.0, a
constant torque of 6 Nm is applied to it, causing it to start rotating. As the merry-
go-round spins faster, a point comes when the small box slides off of it. Given the
following data, calculate the time after it starts spinning, that the box will slip off
the disk. Assume that the merry-go-round is a solid disc. Write your answer in
seconds.
Mass of the merry-go-round (disk) = 30 kg.
Radius of the merry-go-round = 3.1 m
Distance from the axis where the box is placed = R = 1.4 m
Coefficient of friction between merry-go-round and box: Static = 0.5, kinetic

A merrygoround has a small box placed on it at a distance R from its axis of rotation Initially at time 0 the merrygoround is stationary At time t 00 a constant class=

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onyebuchinnaji

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

Apply the principle of conservation of angular momentum

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: https://brainly.com/question/3406242

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