blackstreet23
  • blackstreet23
A particle of mass m moves in a circle of radius R at a constant speed v, as shown below. The motion begins at point Q at time t = 0. Determine the angular momentum of the particle about the axis perpendicular to the page through point P as a function of time. (Use any variable or symbol stated above along with the following as necessary: t.)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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blackstreet23
  • blackstreet23
AihberKhan
  • AihberKhan
The time enters because you need the cross product between the vector from point P and the velocity vector. The distance between these two points and the angle between these two vectors changes with time. A little trigonometry will provide the x and y components of the vector V and the vector to m from P. then use the matrix method of calculating the cross product. You should find that the angular momentum varies as 1+ cos omega t) where omega is the angular velocity about the center and omega = v/R
AihberKhan
  • AihberKhan
Hope this helped! Have a great day! Also a medal would be much appreciated! Just click best response next to my answer. Thank You! @blackstreet23

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blackstreet23
  • blackstreet23
@Agl202 @Mehek14 @Michele_Laino @IrishBoy123 @Hero @pooja195
Michele_Laino
  • Michele_Laino
the angolar momentum is given, by definition, by this subsequent formula: \[\huge {\mathbf{L}} = {\mathbf{OP}} \times m{\mathbf{v}}\] |dw:1449355411005:dw|
blackstreet23
  • blackstreet23
if i did not read wrong it says that i need the variable 't'
Michele_Laino
  • Michele_Laino
using cartesian coordinates, we can write this: \[{\mathbf{L}} = {\mathbf{OP}} \times m{\mathbf{v}} = \left| {\begin{array}{*{20}{c}} {{\mathbf{\hat x}}}&{{\mathbf{\hat y}}}&{{\mathbf{\hat z}}} \\ {R\cos \left( {\omega t} \right)}&{R\sin \left( {\omega t} \right)}&0 \\ { - \omega R\cos \left( {\omega t} \right)}&{\omega R\sin \left( {\omega t} \right)}&0 \end{array}} \right| = ...?\]
Michele_Laino
  • Michele_Laino
please look at this drawing: |dw:1449355732005:dw|
Michele_Laino
  • Michele_Laino
oops...I forgotten the mass \(m\) of the particle: \[{\mathbf{L}} = {\mathbf{OP}} \times m{\mathbf{v}} = m\left| {\begin{array}{*{20}{c}} {{\mathbf{\hat x}}}&{{\mathbf{\hat y}}}&{{\mathbf{\hat z}}} \\ {R\cos \left( {\omega t} \right)}&{R\sin \left( {\omega t} \right)}&0 \\ { - \omega R\cos \left( {\omega t} \right)}&{\omega R\sin \left( {\omega t} \right)}&0 \end{array}} \right| = ...?\]
blackstreet23
  • blackstreet23
wait how did you get that table again to find the determinant?
Michele_Laino
  • Michele_Laino
I wrote the involved vectors, component by component, like below:
blackstreet23
  • blackstreet23
and those cartesian or polar coordinates?
Michele_Laino
  • Michele_Laino
\[\begin{gathered} {\mathbf{OP}} = \left( {R\cos \left( {\omega t} \right),R\sin \left( {\omega t} \right),0} \right) \hfill \\ \hfill \\ m{\mathbf{v}} = m\left( { - \omega R\cos \left( {\omega t} \right),\omega R\sin \left( {\omega t} \right),0} \right) \hfill \\ \end{gathered} \] I have used cartesian coordinates: |dw:1449355973134:dw|
blackstreet23
  • blackstreet23
ohh i think i am understanding it a little bit more now
Michele_Laino
  • Michele_Laino
please keep in mind that the magnitude of the velocity, is \(\omega R\)
blackstreet23
  • blackstreet23
btw why the omega of the second vector is negative?
Michele_Laino
  • Michele_Laino
more precisely, we have this: |dw:1449356057607:dw|
blackstreet23
  • blackstreet23
ohh i see
Michele_Laino
  • Michele_Laino
in my reference system I can write the vector \(\omega\) like below: \[\huge {\mathbf{\omega }} = \left( {0,0,\omega } \right)\]
Michele_Laino
  • Michele_Laino
since the particle is rotating counterclockwise, in my drawing
Michele_Laino
  • Michele_Laino
|dw:1449356274321:dw|
blackstreet23
  • blackstreet23
umm so which would be my equation ?
blackstreet23
  • blackstreet23
I inserted OP * MV but it did not work. I am doing webassign.net
blackstreet23
  • blackstreet23
i think there is not cross symbol 'X'
Michele_Laino
  • Michele_Laino
after a simple computation of the value of the determinant, we get: \[\large \begin{gathered} {\mathbf{L}} = {\mathbf{OP}} \times m{\mathbf{v}} = m\left| {\begin{array}{*{20}{c}} {{\mathbf{\hat x}}}&{{\mathbf{\hat y}}}&{{\mathbf{\hat z}}} \\ {R\cos \left( {\omega t} \right)}&{R\sin \left( {\omega t} \right)}&0 \\ { - \omega R\sin \left( {\omega t} \right)}&{\omega R\cos \left( {\omega t} \right)}&0 \end{array}} \right| = \hfill \\ \hfill \\ = m\omega {R^2}{\mathbf{\hat z}} \hfill \\ \end{gathered} \] oops... I made an error in my determinant above, now I fixed my error above
Michele_Laino
  • Michele_Laino
or, if we want to introduce the speed \(v\), then we can write this: \[\large \begin{gathered} {\mathbf{L}} = {\mathbf{OP}} \times m{\mathbf{v}} = m\left| {\begin{array}{*{20}{c}} {{\mathbf{\hat x}}}&{{\mathbf{\hat y}}}&{{\mathbf{\hat z}}} \\ {R\cos \left( {\omega t} \right)}&{R\sin \left( {\omega t} \right)}&0 \\ { - \omega R\sin \left( {\omega t} \right)}&{\omega R\cos \left( {\omega t} \right)}&0 \end{array}} \right| = \hfill \\ \hfill \\ = m\omega {R^2}{\mathbf{\hat z}} = mvR{\mathbf{\hat z}} \hfill \\ \end{gathered} \]
blackstreet23
  • blackstreet23
umm is it there as a function of time?
Michele_Laino
  • Michele_Laino
since the speed \(v\) is constant, also the magnitude of the vector angular momentum is constant
Michele_Laino
  • Michele_Laino
we can write this: at time \(t_0\) wherein the particle passes at point P, the traveled space \(s\) is: \[\huge s = \pi R\] and such time \(t_0\), is: \[\huge {t_0} = \frac{{\pi R}}{v}\]
Michele_Laino
  • Michele_Laino
so we can write: \[\huge R = \frac{{v{t_0}}}{\pi }\] and after a simple substitution, we get: \[\huge {\mathbf{L}}\left( {{t_0}} \right) = mv\frac{{v{t_0}}}{\pi }{\mathbf{\hat z}} = \frac{{m{v^2}{t_0}}}{\pi }{\mathbf{\hat z}}\]
blackstreet23
  • blackstreet23
so that final one is the equation as a function of time?
blackstreet23
  • blackstreet23
and s = pi * R becuase?
Michele_Laino
  • Michele_Laino
yes! Nevertheless the magnitude of the vector angular momentum is constant, as I said before
Michele_Laino
  • Michele_Laino
since it is half circumference
blackstreet23
  • blackstreet23
ohh we do not care about the other half in this case?
Michele_Laino
  • Michele_Laino
no, no, since your problem asks for the angular momentum when the particle passes at point P for first time
blackstreet23
  • blackstreet23
the answer was wrong
Michele_Laino
  • Michele_Laino
I'm very sorry, I misunderstood your question. We have to compute the angular momentum with respect to point P. So we have the subsequent result: \[\large \begin{gathered} {\mathbf{L}} = {\mathbf{PG}} \times m{\mathbf{v}} = \left| {\begin{array}{*{20}{c}} {{\mathbf{\hat x}}}&{{\mathbf{\hat y}}}&{{\mathbf{\hat z}}} \\ {R + R\cos \left( {\omega t} \right)}&{R\sin \left( {\omega t} \right)}&0 \\ { - \omega R\sin \left( {\omega t} \right)}&{\omega R\cos \left( {\omega t} \right)}&0 \end{array}} \right| = \hfill \\ \hfill \\ = \omega {R^2}\left( {1 + \cos \left( {\omega t} \right)} \right){\mathbf{\hat z}} = vR\left( {1 + \cos \left( {\omega t} \right)} \right){\mathbf{\hat z}} = \hfill \\ \hfill \\ = 2vR{\left\{ {\cos \left( {\frac{{\omega t}}{2}} \right)} \right\}^2}{\mathbf{\hat z}} \hfill \\ \end{gathered} \] |dw:1449410482453:dw| so, I was wrong, and @AihberKhan gave the right answer
Michele_Laino
  • Michele_Laino
Thanks!! :) @AihberKhan for your reply
Michele_Laino
  • Michele_Laino
oops.. I have forgotten the mass \(m\) of the particle. Here are the right formulas: \[\Large \begin{gathered} {\mathbf{L}} = mvR\left\{ {1 + \cos \left( {\omega t} \right)} \right\}{\mathbf{\hat z}} \hfill \\ \hfill \\ {\mathbf{L}} = 2mvR{\left\{ {\cos \left( {\frac{{\omega t}}{2}} \right)} \right\}^2}{\mathbf{\hat z}} \hfill \\ \end{gathered} \]

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