Title: Circular motion: acceleration always perpendicular to velocity Post by: FuKwun Hwang on June 10, 2008, 10:26:58 pm Acceleration is the rate of the velocity change, i.e. $\vec{a}\equiv\frac{d\vec{v}}{dt}$.
Because velocity is a vector, there are many ways to change velocity: 1. change the magnitude of the velocity (change speed) without change it's direction. 2. change direction of the velocity without change it's magnitude: this is the circular motion. 3. change both magnitude and directtion of the velocity. If the acceleration is in the direction of velocity, the acceleration will change the magnitude the of velocity(change speed). However, if the acceleration is always perpendicular to the velocity, the velocity will change direction without change it's magnitude. This is the circular motion. So, for circular motion, the force(acceleration) always perpendicular to it's velocity. You can drag the velocity to change its direction and magnitude, the magnitude of the force is changed by slider bar. Title: Re: Circular motion: acceleration always perpendicular to velocity Post by: lookang on December 09, 2008, 08:29:41 am under evolution
dvx/dt = ( fx=calf(0,vx,vy)/m) dvy/dt = ( fy=calf(1,vx,vy)/m) custom: public double calf(int xid,double vx,double vy){ cta=Math.atan2(vy,vx); if(xid==0)return f*Math.cos(cta+pi/2); else return f*Math.sin(cta+pi/2); } Question: why is it so complicated, need int xid ? is it correct to just let under evolution dvx/dt = f/m*Math.cos(cta+pi/2) dvy/dt = f/m*Math.sin(cta+pi/2) Title: Re: Circular motion: acceleration always perpendicular to velocity Post by: FuKwun Hwang on December 09, 2008, 11:03:23 pm If you use the following at evolution page:
dvx/dt = f/m*Math.cos(cta+pi/2) dvy/dt = f/m*Math.sin(cta+pi/2) But the above equations need "cta" value during the evolution page. If you use cta at constraint page, the "cta" will be at tdt; You will find much larger calculation error. It has something to do with how the evolution were done numerically. The program need to calculate several values between t to t+dt to know how to changed variables to next time step with high preission (You can check out RungeKutta's 4th order method if you are interested in how it was done). If you do not like xid; You can also define two functions: public double calfx(double vx,double vy){ cta=Math.atan2(vy,vx); return f*Math.cos(cta+pi/2); else return f*Math.sin(cta+pi/2); } public double calfy(double vx,double vy){ cta=Math.atan2(vy,vx); return f*Math.sin(cta+pi/2); } combined with dvx/dt = (fx=calfx(vx,vy)/m); dvy/dt = ( fy=calfy(vx,vy)/m); Title: Re: Circular motion: acceleration always perpendicular to velocity Post by: juliaamit on May 22, 2009, 07:38:20 pm Dear sir,
In this simulation, u assumed that the body is rotating at constant angular speed. So there will not be any force along the tangential direction ( along v). The only force is along the radial directed towards centre. The force direction is independent even if the body is rotating clockwise or anti clockwise. Now as the rotating body is subjected to a force directed towards centre, it should move towards centre according to Newton's 2nd law. Will u please explain to me. More over where can i get the answer of Halliday Resnik physics book. if i ask some questions from this book, will u kindly answer me? Regards, Amit Title: Re: Circular motion: acceleration always perpendicular to velocity Post by: FuKwun Hwang on May 22, 2009, 07:52:36 pm Quote Now as the rotating body is subjected to a force directed towards centre, it should move towards centre according to Newton's 2nd law. Newton's Law only indicated that acceleration is in the same direction of the net force. F=m a Acceleration is the rate of velocity change. So it is in the direction of difference between two velocity. By the way, circular motion or means that it is moving in circle. Title: Re: Circular motion: acceleration always perpendicular to velocity Post by: sean1016 on February 16, 2010, 10:04:37 pm This is very interesting. I got the files for offline use but it doesn't include the *.java files (after extracting the jar file).
How did you find the perpedicular force in your simulation? Because if you simulate every one second the direction/distance moved is not your true heading. Rather, because the path you take is curved the heading is tangent to the point you are at. Would you mind posting some of the code you used to show finding the perpendicular acceleration? Thanks. Title: Re: Circular motion: acceleration always perpendicular to velocity Post by: FuKwun Hwang on February 17, 2010, 01:57:25 pm The above simulation was created with EJS (all the simulations under category: Easy Java Simulations (2001 ) (http://www.phy.ntnu.edu.tw/ntnujava/index.php#3) are the same)
The ejs source code is an xml file. Assume you have install Easy java simulation (http://fem.um.es/EjsWiki/index.php): You can open the source with EJS by Right click inside the simulation region, and select open EJS model . The perpendicular force is the force required for circular motion $\vec{F}=\frac{v^2}{r}\hat{r}$ Title: Re: Circular motion: acceleration always perpendicular to velocity Post by: freddygon on October 21, 2015, 01:32:10 pm It was the great information to me.
