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

`\omega_2=5.1rad/s`

Step-by-step 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=(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=(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=((I_D+I_(S1))\omega_1)/((I_D+I_(S2)))`

or:

`\omega_2=(((m_Dr_D^2)/(2)+m_Sr_(S1)^2)\omega_1)/(((m_Dr_D^2)/(2)+m_Sr_(S2)^2))`

which for our values is:

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