A playground merry-go-round of radius 2.00 m has a moment of inertia I = 275 kg · m2 and is rotating about a frictionless vertical axle. As a child of mass 25.0 kg stands at a distance of 1.00 m from the axle, the system (merry-go-round and child) rotates at a rate of 14.0 rev/min. The child then proceeds to walk toward the edge of the merry-go-round. What is the angular speed of the system when the child teaches the edge?

Respuesta :

Answer:

0.2932 rad/s

Explanation:

r = Radius = 2 m

[tex]I_i[/tex] = Initial angular momentum = [tex]275\ kgm^2[/tex]

[tex]\omega_i[/tex] = Initial angular velocity = 14 rev/min

[tex]I_f[/tex] = Final angular momentum

[tex]\omega_f[/tex] = Final angular velocity

Here the angular momentum of the system is conserved

[tex]I_i\omega_i=I_f\omega_f\\\Rightarrow 275\times 14\times \dfrac{2\pi}{60}=(275+275(2)^2)\omega_f\\\Rightarrow \omega_f=\dfrac{275\times 14\times \dfrac{2\pi}{60}}{275+275(2)^2}\\\Rightarrow \omega_f=0.2932\ rad/s[/tex]

The final angular velocity is 0.2932 rad/s

The angular speed of the system when the child reaches the edge is 1.37 rad/s.

The given parameters;

  • radius of the merry-go-round, r = 2.0 m
  • moment of inertia, I = 275 kg.m²
  • mass of the car, m = 25 kg
  • position of the child, r₁ = 1 m
  • initial angular speed, ω₁ = 14 rev/s

The angular speed in rad/s is calculated as follows;

[tex]\omega = 14 \ \frac{rev}{\min} \times \frac{1\min}{60 \ s} \times \frac{2\pi \ rad}{rev} = 1.47 \ rad/s[/tex]

Apply the principle of conservation of angular momentum to determine the final angular speed of the child;

[tex]I_i \omega _i = I_f \omega_f \\\\\omega_i (I_1 + I_2) = \omega_f (I_1 + I_2)\\\\\omega_i (I_1 + I_2) = \omega_f (I_1 + I_2)\\\\1.47(275 \ + 25\times 1^2)= \omega_f (275 \ + \ 25\times 2)\\\\ 441 = \omega (325)\\\\\omega = \frac{441}{325} \\\\\omega = 1.37 \ rad/s[/tex]

Thus, the angular speed of the system when the child reaches the edge is 1.37 rad/s.

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