Angular Momentum - quick question

mybrohshi5

1. The problem statement, all variables and given/known data

A young physics student is standing on an initially motionless turntable with rotational inertia 0.330 kg m^2; the turntable is free to turn about a frictionless axle. He is holding a wheel of rotational inertia 0.210 kg m^2 which is spinning at 135.0 rpm about a vertical axis. When he turns the wheel upside down, student and turntable begin rotating at 65.0 rpm. Neglect the distance between the axes of the turntable and the wheel.

What is the mass of the student? Approximate the student as a cylinder of radius 0.14 m

2. Relevant equations

L_i=L_f

L_w = angular momentum of the wheel
L_pt = angular momentum of the person and turntable

3. The attempt at a solution

Ok so the answer is 55.3 kg and i know how to get the answer but i am a little confused about one part (the part below in bold)

L_i = L_w

L_f = L_pt + L_w

L_i=L_f

so.....

L_w = L_pt - L_w
- why is it L_pt MINUS L_w

2L_w = L_pt

2(.210)(135rev/min) = (.33 + 1/2m(.14²)) * 65 rev/min

m = 55.3 kg

Thanks for any help on explaining the part i am confused about :)

Squeezebox

Angular momentum is a vector. Flipping its direction has the effect of changing its sign in the equation.

mybrohshi5

Oh ok so since the student turned the wheel upside down the sign changes.

Got it.

Thanks squeezebox