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Vika [28.1K]
4 years ago
12

In a playground there is a small merry-go-round of radius 1.25 m and mass 175 kg. Assume the merry-go-round to be a uniform disc

. A child of mass 45 kg runs at a speed of 3.0 m/s tangent to the rim of the merry-go-round (initially at rest) and jumps on. If we neglect function. what is the angular speed of the many-go- round after the child has jumped on and is standing at its outer rim?
Physics
1 answer:
Margaret [11]4 years ago
3 0

Answer:

ωf = 0.82 rad/s

Explanation:

By conservation of the angular momentum:

I_c * \omega_o = (I_c+I_m)*\omega_f

Where I_c is the inertia of the child, I_m is the inertia of the merry-go-round.

Solving for ωf:

\omega_f = (m_c*R^2)*(V_o/R)/(m_c*R^2+M_m*R^2/2)

\omega_f=0.82rad/s

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