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Savatey [412]
3 years ago
14

A horizontal circular platform (m = 119.1 kg, r = 3.23m) rotates about a frictionless vertical axle. A student (m = 54.3kg) walk

s slowly from the rim of the platform toward the center. The angular velocity, omega, of the system is 3.1 rad/s when the student is at the rim. Find omega when the student is 1.39m from the center.
Physics
1 answer:
Murrr4er [49]3 years ago
6 0

Answer:

\omega_2=5.1rad/s

Explanation:

Since there is no friction angular momentum is conserved. The formula for angular momentum thet will be useful in this case is L=I\omega. If we call 1 the situation when the student is at the rim and 2 the situation when the student is at r_2=1.39m from the center, then we have:

L_1=L_2

Or:

I_1\omega_1=I_2\omega_2

And we want to calculate:

\omega_2=\frac{I_1\omega_1}{I_2}

The total moment of inertia will be the sum of the moment of intertia of the disk of mass m_D=119.1 kg and radius r_D=3.23m, which is I_D=\frac{m_Dr_D^2}{2}, and the moment of intertia of the student of mass m_S=54.3kg at position r (which will be r_1=r=3.23m or r_2=1.39m) will be I_{S}=m_Sr_S^2, so we will have:

\omega_2=\frac{(I_D+I_{S1})\omega_1}{(I_D+I_{S2})}

or:

\omega_2=\frac{(\frac{m_Dr_D^2}{2}+m_Sr_{S1}^2)\omega_1}{(\frac{m_Dr_D^2}{2}+m_Sr_{S2}^2)}

which for our values is:

\omega_2=\frac{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(3.23m)^2)(3.1rad/s)}{(\frac{(119.1kg)(3.23m)^2}{2}+(54.3kg)(1.39m)^2)}=5.1rad/s

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Answer:

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