What is the Angular Velocity of the Runner and Turntable System?

[anubis01]

Homework Statement

A runner of mass 51.0 kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. The runner's velocity relative to the Earth has magnitude 3.60 m/s. The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.160 rad/s relative to the earth. The radius of the turntable is 3.20m , and its moment of inertia about the axis of rotation is 79.0 kg*m^2.

Homework Equations

w=v/r
I₁w₁=I₂w₂

The Attempt at a Solution

okay first we are given the veocity of the runner, to determine his angular velocity its just
w₁=v_runner/r_table=3.6/3.2=1.125 rad/s

I₁(runner)=mr^2=51.0*3.2^2=522.24 kg*m^2

now using the conversation of angular momentum

I₁w₁+I₂w₂=(I₁+I₂)w₂'

the w and I provided in the problem statement can be used as w₂ & I₂ respectivly and we can now solve for w₂' which is

I₁w₁+I₂w₂/(I₁+I₂)=w₂'

subbing in all the values

(522.24*1.125)+(79.0*0.160)/(522.24+79.0)=600.16/601.24=0.9982=0.988(sig figs)

Now the problem I'm having with this is that I still receive an error with this answer and I've been through my work half a dozen times and I don't believe I made any rounding answers so if anyone could tell me what I am doing wrong I would be very thankful.

 

[alphysicist]

Hi anubis01,

Homework Statement

A runner of mass 51.0 kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. The runner's velocity relative to the Earth has magnitude 3.60 m/s. The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.160 rad/s relative to the earth. The radius of the turntable is 3.20m , and its moment of inertia about the axis of rotation is 79.0 kg*m^2.

Homework Equations

w=v/r
I₁w₁=I₂w₂

The Attempt at a Solution

okay first we are given the veocity of the runner, to determine his angular velocity its just
w₁=v_runner/r_table=3.6/3.2=1.125 rad/s

I₁(runner)=mr^2=51.0*3.2^2=522.24 kg*m^2

now using the conversation of angular momentum

I₁w₁+I₂w₂=(I₁+I₂)w₂'

the w and I provided in the problem statement can be used as w₂ & I₂ respectivly and we can now solve for w₂' which is

I₁w₁+I₂w₂/(I₁+I₂)=w₂'

subbing in all the values

(522.24*1.125)+(79.0*0.160)/(522.24+79.0)=600.16/601.24=0.9982=0.988(sig figs)

I have not checked all of your numbers, but remember that the runner and platform are rotating in opposite direction. What has to be changed here?

 

[anubis01]

Oh so then since the table is rotating in the opposite direction of the runner w₂=-0.160 rad/s

I₁w₁+I₂w₂/(I₁+I₂)=w₂'

subbing in the values

(522.24*1.125)+(79*-0.160)/(522.24+79)=574.88/601.24
w₂'=0.95615=0.956(sig figs)

Thanks for the help, I would have never have caught that error by myself.