How Is Angular Momentum Calculated in Rotational Kinematics?

[Rheegeaux]

[Note: Post moved to homework forum by mentor]

So I stumbled upon a reviewer for my physics exam tomorrow and I was wondering how the equation was formulated. Your help is very much appreciated :) ! Normally I would consult my professor for this but it's Sunday in my country today so I can't.

Question:
A uniform stick with length 3.00 [m] and mass
5.00 [kg] is moving and rotating about its center of mass (CM) as
shown in the figure. If the stick and point O both lie in the same
xy-plane, what is the total angular momentum of the stick at point
O at the instant shown?

answer:
L = (r )(p) +1/12mL^2(w) = -76.7kmm^2/s positive k hat

Picture: http://postimg.org/image/u5eeo77el/96cf1d1b/

 

[Brian T]

The first term $r \times p$ comes from the definition of angular momentum of any particle (or center of mass), while the second term comes from the fact that this object has a finite extent (i.e. moment of inertia). However, technically, the formula is incorrect since the first term is a vector while the second term seems to be a scalar (this doesn't disrupt the answer since you're given the direction of the rotation, just note that the formula isn't technically correct)

 

[haruspex]

the second term seems to be a scalar

ω should be a vector. Are you saying it seems to be a scalar because it is not in bold? The value is shown in the diagram as a vector ($\hat k$).

 

[Rheegeaux]

The first term $r \times p$ comes from the definition of angular momentum of any particle (or center of mass), while the second term comes from the fact that this object has a finite extent (i.e. moment of inertia). However, technically, the formula is incorrect since the first term is a vector while the second term seems to be a scalar (this doesn't disrupt the answer since you're given the direction of the rotation, just note that the formula isn't technically correct)

how did it yield a negative answer? What I got was letter D. my solution is: (5kg)(7.50m/s)(4m)sin(35) + 1/12(5kg)(3m)^2(2.50 rad) = 95.411
Thanks for the reply, I really need to learn this before tomorrow. Cheers :)

 

[Rheegeaux]

how did it yield a negative answer? What I got was letter D. my solution is: (5kg)(7.50m/s)(4m)sin(35) + 1/12(5kg)(3m)^2(2.50 rad) = 95.411
Thanks for the reply, I really need to learn this before tomorrow. Cheers :)

I got it already *ZOINKS* I just needed to use the right hand rule