About

<h2>Horizontal Circular Motion of Mass on a Table</h2>

<p>A particle with mass m is moving with constant speed v along a circular orbit (radius r ). The centripetal force \( F=\frac{mv^2}{r} \) is provided by gravitation force from another mass \(M=\frac{F}{g} \). A string is connected from mass m to the origin then connected to mass M . Because the force is always in the r direction, so the angular momentum \( \widehat{L} = m\widehat{r} \widehat{v} \)  is conserved. i.e. \(L=mr^2\omega \)  is a constant. For particle with mass m:</p>

<p> \(m \frac{d^2r}{dt^2}=m\frac{dv}{dt}=mv^2r−Mg=\frac{L^2}{mr^3}−Mg \)</p>

<p> \( \omega = Lmr^2 \) </p>

<h2>Controls</h2>

<p>You can change the hang mass M or the on the table 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. </p>

 

Translations

Code	Language		Translator	Run
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Credits

Fu-Kwun Hwang - Dept. of Physics, National Taiwan Normal Univ. and lookang; lookang; tinatan

http://iwant2study.org/lookangejss/02_newtonianmechanics_6circle/ejss_model_circular3dfr02wee/circular3dfr02wee_Simulation.xhtml

Learning Content

Motion in a Circle Content taken from http://www.seab.gov.sg/content/syllabus/alevel/2017Syllabus/9749_2017.pdf

1. Kinematics of uniform circular motion
2. Centripetal acceleration
3. Centripetal force

 

Learning Outcomes

Candidates should be able to:

1. express angular displacement in radians
2. show an understanding of and use the concept of angular velocity to solve problems
3. recall and use v = rω to solve problems
4. describe qualitatively motion in a curved path due to a perpendicular force, and understand the centripetal acceleration in the case of uniform motion in a circle
5. recall and use centripetal acceleration a = rω ² , and \( a = \frac{v^2 }{r} \) to solve problems
6. recall and use centripetal force F = mrω ² , and \( F = \frac{mv^2}{r} \) to solve problems.

 

For Teachers

A particle with mass \(m\) is moving with constant speed \(v\) along a circular orbit (radius \(r\)). The centripetal force \(F=m\frac{v^2}{r}\) 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 \(\vec{L}=m\,\vec{r}\times \vec{v}\) is conserved. i.e. \(L=mr^2\omega\) is a constant.

For particle with mass m:

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

The following is a simulation of the above model.

When mass m or radius r is changed with sliders, equilibrium condition is recalulated for constant circular motion.
However, if mass M is changed, the equilibrium condition will be broken, and the system will oscilliate up and down. 

Research

[text]

Video

 Ejs Open Source Horizontal Circular Motion java applet by lookang lawrence wee

 Version:

1. http://weelookang.blogspot.sg/2016/03/horizontal-3d-webgl-circular-motion-of.html HTML5 JavaScript WebGL version by Loo Kang Wee and Tina Tan
2. http://weelookang.blogspot.sg/2010/07/lesson-on-circular-motion-with-acjc.html 09 July 2010 Computer Lab hands on learning session on  Ejs Open Source Vertical Circular Motion of mass m attached to a rod java applet side view of the same 3D view with teacher explaining the physical setup of the mass m and mass M attached by a string through a table with a fricitionless hole in the middle of table for string to go through and student working on their own desktop
3. http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=1883.0 remixed Java applet by Loo Kang Wee
4. http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=1454.0 original Java applet by Fu-Kwun Hwang

Other Resources

https://www.geogebra.org/m/rnkrmcx8 by Tan Seng Kwang

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