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As seen from above, a playground carousel is
rotatingcounterclockwise about its center on frictionless bearings.
Aperson standing still on the ground grabs onto one of the bars
onthe carousel very close to its outer edge and climbs aboard.
Thus,this person begins with an angular speed of zero and ends up
with anonzero angular speed, which means that he underwent
acounterclockwise angular acceleration.

The carousel has a radius of 1.50 m,an initial angular speed of
3.14 rad/s, and a moment of inertia of125 kgm2.
Themass of the person is 48 kg. Find thefinal angular speed of the
carousel after the person climbsaboard.

Respuesta :

Answer:

The final angular speed of the carousel is 1.68 rad/s.

Explanation:

This is simply a conservation of angular momentum problem.

[tex]\vec{L}_0 = \vec{L}_1\\L_0 = I_0\omega_0 = 125*3.14 = 392.5\\L_1 = I_1\omega_1 = (mR^2 + I_0)\omega_1 = (48*1.5^2 + 125)\omega_1 = 233\omega_1[/tex]

Combining the two equations yields the angular momentum of the carousel after the man climbs.

[tex]392.5 = 233\omega_1\\\omega_1 = 1.68~rad/s[/tex]

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