Question

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

Answer 1

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:

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