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 Author Topic: in which direction will it roll?  (Read 9428 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
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 « Embed this message on: June 21, 2013, 11:40:45 am » posted from:,,Satellite Provider

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.

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Fu-Kwun Hwang
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 « Embed this message Reply #1 on: June 22, 2013, 09:30:12 pm » posted from:,,Satellite Provider

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.

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rexujnk
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 « Embed this message Reply #2 on: August 11, 2014, 08:32:07 pm » posted from:,,Satellite Provider

It's really dynamic theory, So nice.-*-
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Life well spent is long. ..."da Vinci (1452-1519, Italian artist, sculptor, painter, architect, engineer and scientist) "