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A merry-go-round rotates at the rate of 0.17 rev/s with an 79 kg man standing at a point 1.6 m from the axis of rotation.
What is the new angular speed when the man walks to a point 0 m from the center? Consider the merry-go-round is a solid 34 kg cylinder of radius of 1.6 m.
Answer in units of rad/s.

Sagot :

leena

Hi there!

We can use the conservation of angular momentum to solve.

[tex]L_i = L_f\\\\I\omega_i = I\omega_f[/tex]

I = moment of inertia (kgm²)

ω = angular velocity (rad/sec)

Recall the following equations for the moment of inertia.

[tex]\text{Solid cylinder:} I = \frac{1}{2}MR^2\\\\\text{Object around center:} = MR^2[/tex]

Begin by converting rev/sec to rad sec:


[tex]\frac{0.17rev}{s} * \frac{2\pi rad}{1 rev} = 1.068 \frac{rad}{s}[/tex]

According to the above and the given information, we can write an equation and solve for ωf.

[tex]1.068(\frac{1}{2}(34)(1.6)^2 + (79)(1.6)^2) = \omega_f(\frac{1}{2}(34)(1.6^2) + 79(0^2))\\\\\omega_f = \boxed{6.03 \frac{rad}{sec}}[/tex]

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