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<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>


<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>

Horizontal Circular Motion of Mass on a Table


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Software Requirements


Android iOS Windows MacOS
with best with Chrome Chrome Chrome Chrome
support full-screen? Yes. Chrome/Opera No. Firefox/ Samsung Internet Not yet Yes Yes
cannot work on some mobile browser that don't understand JavaScript such as.....
cannot work on Internet Explorer 9 and below



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

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Learning Content

Motion in a Circle Content taken from

  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. 




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


  1. HTML5 JavaScript WebGL version by Loo Kang Wee and Tina Tan
  2. 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. remixed Java applet by Loo Kang Wee
  4. original Java applet by Fu-Kwun Hwang

Other Resources

end faq


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Article ID: 364
Article Category ID: 21
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  • Dynamics
  • Circle
  • Junior College
  • EasyJavaScriptSimulation
  • Android/iOS including handphones/Tablets/iPads
  • Windows/MacOSX/Linux including Laptops/Desktops
  • ChromeBook Laptops
  • Science
  • Simulations