What Is the Final Angular Speed of a Spinning Disk with a Running Person?

[teddygrams]

A flat uniform circular disk (radius = 4.69 m, mass = 280 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a frictionless axis perpendicular to the center of the disk. A 55.0-kg person, standing 1.59 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 1.70 m/s relative to the ground. Find the resulting angular speed (in rad/s) of the disk.

i know that i have to use :

0=I(person)w(final person) + I(disk)w(final disk)

and i know I=mr^2

but i just don't know how to rearrange everything...correctly..

 

[Nuclear on the Rocks]

post this in the homework section

 

[m2003]

Using the equation 0=I(person)w(final person) + I(disk)w(final disk), we can rearrange it to solve for the final angular speed of the disk:

w(final disk) = -(I(person)w(final person)) / I(disk)

Since we know that I(person) = mr^2 and I(disk) = 1/2mr^2 (for a thin disk), we can substitute these values into the equation:

w(final disk) = -((mr^2)(w(final person))) / (1/2mr^2)

Simplifying, we get:

w(final disk) = -2w(final person)

Now, we can substitute in the values given in the problem to solve for the final angular speed of the disk:

w(final disk) = -2(1.70 m/s) = -3.40 rad/s

Therefore, the resulting angular speed of the disk is -3.40 rad/s. The negative sign indicates that the disk will be rotating in the opposite direction of the person running on it.