Calculate the magnitude of the total angular momentum of the woman

[Black_Hole???]

Question:

A woman with a mass of 55.0 is standing on the rim of a large disk that is rotating at an angular velocity of 0.550 about an axis through its center. The disk has a mass of 106 and a radius of 4.10 .


a.)Calculate the magnitude of the total angular momentum of the woman-plus-disk system. (Assume that you can treat the woman as a point.)


Here is my attempt.

L = I(omega)
= [(1/2)mr^2]omega
= [(1/2)(106 kg + 55 kg)(4.10m)^2]*3.456 rad/s
= 5143.8 kg m^2/s

=4676.7 which is wrong anyone got any incite?

 

[rl.bhat]

Calculate the moment of inertia of disc and woman separately and then add.

 

[Black_Hole???]

so i am going to use the formula

I = R^2 * M

and do

I = 4.1^2*55 + 4.1^2*106?
i got my answer as 5371.41

is that correct?


i get one chance to get it right so just making sure

 

[rl.bhat]

What is I for disc?

 

[Black_Hole???]

It is:

I = (m*r^2)/2

so it would then be

I= (55*4.1^2)/2 + (106*41.^2)/2
=1353.21

Is that right?

 

[rl.bhat]

No. What is I of woman?

 

[Black_Hole???]

I = R^2 * M?

so then add those two together for

I=4.1^2* 55 +4.1^2*106*(1/2)
=1815.48?

 

[rl.bhat]

That is right.

 

[Black_Hole???]

well the answer was wrong.

It was 6270

however i would like to know how this answer came to be

 

[rl.bhat]

In your problem angular velocity is 0,550 ?. What is the unit?= [(1/2)(106 kg + 55 kg)(4.10m)^2]*3.456 rad/s
= 5143.8 kg m^2/s

=4676.7 which is wrong anyone got any incite?
From where did you get 3.456?
Why the two answer are different?

 

[Dr.D]

How many women are you going to incite?