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A large horizontal circular platform (M=87.1 kg, r=3.09 m) rotates about a frictionless vertical axle. A student (m=96.3 kg) walks slowly from the rim of the platform toward the center. The angular velocity Ï of the system is 2.50 rad/s when the student is at the rim. A) Find Ï (in rad/s) when the student is 1.75 m from the center.B) Find the moment of inertia of platform through the center with respect to the z-axis. C) Find the moment of inertia of the student about the center axis (while standing at the rim) of the platform. D) Find the moment of inertia of the student about the center axis while the student is standing 1.75 m from the center of the platform.

Respuesta :

Answer:

A)ω₂= 4.42 rad/s

B)I=415.82 kg.m²

C)I₁ = 919.48 kg.m²

D)I₂ = 294.91 kg.m²

Explanation:

Given that

M= 87.1 kg

r= 3.09 m

m = 96.3 kg

ω ₁= 2.5 rad/s (when student at rim )

There is no any external torque that is why the total angular momentum of the system will be conserve.

The moment of inertia of circular platform about center about its center

[tex]I=\dfrac{MR^2}{2}\ kg.m^2[/tex]

[tex]I=\dfrac{87.1\times 3.09^2}{2}\ kg.m^2[/tex]

I=415.82 kg.m²

Moment of inertia of the student

[tex]I_1={mr^2}\ kg.m^2[/tex]

[tex]I_1=96.3\times 3.09^2\ kg.m^2[/tex]

I₁ = 919.48 kg.m²

[tex]I_2={mr^2}\ kg.m^2[/tex]

[tex]I_2=96.3\times 1.75^2\ kg.m^2[/tex]

I₂ = 294.91 kg.m²

[tex](I+I_1)\omega_1=(I+I_2)\omega_2[/tex]

[tex]\omega_2=\dfrac{I+I_1}{I+I_2}\times \omega_1[/tex]

[tex]\omega_2=\dfrac{514.82+919.48}{514.82+294.91}\times 2.5[/tex]

A)ω₂= 4.42 rad/s

B)I=415.82 kg.m²

C)I₁ = 919.48 kg.m²

D)I₂ = 294.91 kg.m²

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