Question

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.)

Answer 1

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

Answer 2

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

Answer 3

@Agl202
@Mehek14
@Michele_Laino
@IrishBoy123
@Hero
@pooja195

Answer 4

the angolar momentum is given, by definition, by this subsequent formula:
\[\huge {\mathbf{L}} = {\mathbf{OP}} \times m{\mathbf{v}}\]

Answer 5

if i did not read wrong it says that i need the variable 't'

Answer 6

using cartesian coordinates, we can write this:
\[{\mathbf{L}} = {\mathbf{OP}} \times m{\mathbf{v}} = \left| {\begin{array}{*{20}{c}}
{{\mathbf{\hat x}}}&{{\mathbf{\hat y}}}&{{\mathbf{\hat z}}} \\
{R\cos \left( {\omega t} \right)}&{R\sin \left( {\omega t} \right)}&0 \\
{ - \omega R\cos \left( {\omega t} \right)}&{\omega R\sin \left( {\omega t} \right)}&0
\end{array}} \right| = ...?\]

Answer 7

please look at this drawing:

Answer 8

oops...I forgotten the mass \(m\) of the particle:
\[{\mathbf{L}} = {\mathbf{OP}} \times m{\mathbf{v}} = m\left| {\begin{array}{*{20}{c}}
{{\mathbf{\hat x}}}&{{\mathbf{\hat y}}}&{{\mathbf{\hat z}}} \\
{R\cos \left( {\omega t} \right)}&{R\sin \left( {\omega t} \right)}&0 \\
{ - \omega R\cos \left( {\omega t} \right)}&{\omega R\sin \left( {\omega t} \right)}&0
\end{array}} \right| = ...?\]

Answer 9

wait how did you get that table again to find the determinant?

Answer 10

I wrote the involved vectors, component by component, like below:

Answer 11

and those cartesian or polar coordinates?

Answer 12

\[\begin{gathered}
{\mathbf{OP}} = \left( {R\cos \left( {\omega t} \right),R\sin \left( {\omega t} \right),0} \right) \hfill \\
\hfill \\
m{\mathbf{v}} = m\left( { - \omega R\cos \left( {\omega t} \right),\omega R\sin \left( {\omega t} \right),0} \right) \hfill \\
\end{gathered} \]
I have used cartesian coordinates:

Answer 13

ohh i think i am understanding it a little bit more now

Answer 14

please keep in mind that the magnitude of the velocity, is \(\omega R\)

Answer 15

btw why the omega of the second vector is negative?

Answer 16

more precisely, we have this: