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Angular Momentum of a sliding disc about a point on the floor 
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#1
Feb2013, 06:21 PM

P: 20

Hi everybody,
A seemingly straightforward example from lecture is causing me some confusion. The example was about calculating the angular momentum of a sliding disk (not rolling) about a point on the floor. The result given in lecture says the distance to the point on the floor is unrelated to the angular momentum: [tex]\vec{L}=\vec{r} \times \vec{p}=\sum{\vec{r}_i \times \vec{p}_i}=\sum{\vec{r}_i \times m_i\vec{v}_i}=MRv [/tex] where M is the total mass of the disc, R is the radius of the disc, and v is the translational velocity of the sliding disc. Now my confusion comes in at the last equal sign. I think it should read: [tex]\sum{\vec{r}_i \times m_i \vec{v}_i}=\sum{m_i r_i v_i \sin{\theta_i}}[/tex] where [itex]\theta_i[/itex] is the angle between each particles radius vector and the constant velocity vector. I don't see how this sum ends up as [itex]MRv[/itex] as was claimed in lecture. I tried to write [itex] \theta_i [/itex] as a function of each ri and integrate over the disc, but didn't make progress. I know I can take the mi and vi out of the sum because they are the same for each i, but I still can't deal with the [tex]\sum{r_i\sin{\theta_i}}[/tex] Any help would be greatly appreciated! Maybe I'm just missing something really obvious, I don't know. Thanks, Sean 


#2
Feb2113, 08:35 AM

Sci Advisor
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Thanks
P: 26,148

Hi Sean!
∑ r_{i} x m_{i} v_{i} = (∑ m_{i}r_{i}) x v since v_{i} = a constant, v and then use ∑ m_{i}(r_{i}  r_{o}) = 0 by definition, where r_{o} is the centre of mass 


#3
Feb2113, 12:50 PM

P: 20

Ok, I see. That makes sense. In my situation it just so happens that the perpendicular distance is equal to the radius of the disc. In a more general scenario this would be different. Thank you!



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