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Easy Java Simulations (2001- ) => Dynamics => Topic started by: Fu-Kwun Hwang on June 21, 2013, 11:40:45 am



Title: in which direction will it roll?
Post by: Fu-Kwun Hwang on June 21, 2013, 11:40:45 am

    If the spool is pulled horizontally to the right.
      in which direction will it roll?

      1. to the left?
      2. to the right?
      3. others ...



      4. Click the following image to show the simulation.
        [eye][ejsapplet][/eye]


Title: Re: in which direction will it roll?
Post by: Fu-Kwun Hwang on June 22, 2013, 09:30:12 pm
Here is another more general case: You can drag the Force arrow to change it's direction.


1. Choose the center of the disk as origin.
Assume the friction force is f, The x, y component of the external force F is Fx, Fy

Then from Newton's law:  $Fx-f=m*a$
The Net torque is $R*f-r*F=I*\alpha$
Assume the disk motion satisfy "Rolling Without Slipping" condition:  i.e. $a=R*\alpha$

$R*f-r*F=I*\alpha= I*\frac{a}{R}=\frac{I}{R} \frac{Fx-F}{m}=\frac{I}{m*R}(Fx-f)$
So $(R+\frac{I}{m*R})f=(r*F+\frac{I}{m*R}Fx)=(r+\frac{I}{m*R}\frac{Fx}{F})*F$

We will get $f=\frac{r+\frac{I}{m*R}\frac{Fx}{F}}{R+\frac{I}{m*R}}*F$

$I*\alpha=R*f-r*F=R*\frac{r+\frac{I}{m*R}\frac{Fx}{F}}{R+\frac{I}{m*R}}*F-r*F=\frac{R*r+\frac{I}{m}\frac{Fx}{F}-r*R-\frac{I}{m}\frac{r}{R}} {R+\frac{I}{m*R}}*F=\frac{R*\frac{Fx}{F}-r}{I+m*R^2}*F=\frac{R*\cos\theta-r}{I+m*R^2}*F$

$I+m*R^2$ is the new "Moment of inertia" if we move the origin to the point disk in contact with the surface.


Click the following image to show the simulation.
[eye][ejsapplet][/eye]


Title: Academic proofreading services
Post by: rexujnk on August 11, 2014, 08:32:07 pm
It's really dynamic theory, So nice.-*-