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Fu-Kwun Hwang
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Posts: 3085   « 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 ...

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Fu-Kwun Hwang
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Posts: 3085   « 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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Posts: 1 « Embed this message Reply #2 on: August 11, 2014, 08:32:07 pm » posted from:,,Satellite Provider It's really dynamic theory, So nice.-*- Logged
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