A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the center of the disk. The platform has a radius of 3.22 m and a rotational inertia of 275 kg·m2 about the axis of rotation. A 62.3 kg student walks slowly from the rim of the platform toward the center. If the angular speed of the system is 1.33 rad/s when the student starts at the rim, what is the angular speed when she is 0.861 m from the center?

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Poltergeist Poltergeist · Answer 1 · 2020-03-18T08:14:31+01:00

Answer:

The angular speed is [tex]\omega_2= 3.81s^{-1}[/tex].

Explanation:

The law of conservation of angular momentum says that for an isolated system

[tex]I_1\omega_1 = I_2 \omega_2[/tex]

Now, when the student is at the rim of the platform the moment of inertia of the system is

[tex]I_1 = mr_1^2+275kg\cdot m^2[/tex]

[tex]I _1 = (62.3kg)(3.22m)^2+275kg\cdot m^2[/tex]

[tex]I_1 = 920.95kg\cdot m^2[/tex],

and the angular speed is

[tex]\omega_1 = 1.33s^{-1}[/tex].

When the student is [tex]r_2 = 0.861m[/tex] from the center,the moment of inertia of the system becomes

[tex]I_2 =mr_2^2+275kg\cdot m^2[/tex]

[tex]I_2 =(62.3kg)(0.861m)^2+275kg\cdot m^2[/tex]

[tex]I_2= 321.18kg\cdot m^2[/tex]

Thus, from conservation of angular momentum

[tex](920.95kg\cdot m^2)(1.33s^{-1})= (321.18kg\cdot m^2)\omega_2[/tex]

[tex]\omega_2=\dfrac{ (920.95kg\cdot m^2)(1.33s^{-1})}{(321.18kg\cdot m^2)}[/tex]

[tex]\boxed{\omega_2= 3.81s^{-1}}[/tex]

which is the angular speed when the student is 0.861 meters from the center.