### Author Topic: Conservation of Angular momentum and 3D circular motion  (Read 20568 times)

#### Fu-Kwun Hwang

• Hero Member
• Posts: 3062
##### Conservation of Angular momentum and 3D circular motion
« on: March 13, 2010, 12:06:08 am »
A particle with mass \$m\$ is moving with constant speed \$v\$ along a circular orbit (radius \$r\$).
The centripetal force \$F=mfrac{v^2}{r}\$ is provided by gravitation force from another mass \$M=F/g\$.
A string is connected from mass m to the origin then connected to mass \$M\$.
Because the force is always in the \$hat{r}\$ direction, so the angular momentum \$vec{L}=m,vec{r}   imes vec{v}\$ is conserved. i.e. \$L=mr^2omega\$ is a constant.

For particle with mass m:

\$ m frac{d^2r}{dt^2}=mfrac{dv}{dt}= m frac{v^2}{r}-Mg=frac{L^2}{mr^3}- Mg \$
\$ omega=frac{L}{mr^2}\$

The following is a simulation of the above model.

You can change the mass M or the radius r with sliders.
The mass M also changed to keep the mass m in circular motion when you change r.
However, if you change mass M , the equilibrium condition will be broken.

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#### macfamous

• Newbie
• Posts: -5
##### Re: Conservation of Angular momentum and 3D circular motion
« Reply #1 on: March 19, 2010, 11:53:15 am »
there is more simple formula for this simulation

#### Fu-Kwun Hwang

• Hero Member
• Posts: 3062
##### Re: Conservation of Angular momentum and 3D circular motion
« Reply #2 on: March 19, 2010, 02:22:45 pm »
Quote
there is more simple formula for this simulation

Could you explain what do you mean in detail?

#### lookang

• Hero Member
• Posts: 1772
• http://weelookang.blogspot.com
##### Re: Conservation of Angular momentum and 3D circular motion
« Reply #3 on: July 07, 2010, 01:32:23 pm »
question:
how is the tension in the string calculated?

My understanding:
T = m*v*v/r; // added lookang for tension only work for circular motion

my hypothesis:
T1 = m*v*v/r + m*dvr/dt ? // radial acceleration due to tangential change in velocity and radial acceleration due to radial change in velocity ?

where dvr/dt = (cst*cst/(m*r*r*r)-M*g)/m

My equation:
T1 = m*v*v/r +(cst*cst/(m*r*r*r)-M*g) ;
can help take a look if my equation of Tension is correct ?
did i get the generalized equation for tension in any elliptical motion?
the logic seems correct and i implement the equation looks correct. is it correct conceptually in the absence of frictional forces
« Last Edit: July 07, 2010, 01:54:54 pm by lookang »

#### Fu-Kwun Hwang

• Hero Member
• Posts: 3062
##### Re: Conservation of Angular momentum and 3D circular motion
« Reply #4 on: July 07, 2010, 07:05:49 pm »
If the mass is in circular motion, then the tension \$T=mfrac{v^2}{r}\$.

What if we increase the tension so that \$T>mfrac{v^2}{r}\$.
Will the mass moving inward(\$frac{dv_r}{dt}<0\$) or moving outward(\$frac{dv_r}{dt}>0\$)?

#### lookang

• Hero Member
• Posts: 1772
• http://weelookang.blogspot.com
##### Re: Conservation of Angular momentum and 3D circular motion
« Reply #5 on: July 07, 2010, 07:16:08 pm »
If the mass is in circular motion, then the tension \$T=mfrac{v^2}{r}\$.

What if we increase the tension so that \$T>mfrac{v^2}{r}\$.
Will the mass moving inward(\$frac{dv_r}{dt}<0\$) or moving outward(\$frac{dv_r}{dt}>0\$)?

I see I got the sign convention wrong!
I forgot the convention for r unit vector is positive outward.
Will change it to
T1 = m*v*v/r -(cst*cst/(m*r*r*r)-M*g) ;
thanks !!!

your masterful questioning allows to think and figure out the answer myself.