Title: in which direction will it roll? Post by: FuKwun Hwang on June 21, 2013, 11:40:45 am
Click the following image to show the simulation. [eye][ejsapplet][/eye] Title: Re: in which direction will it roll? Post by: FuKwun 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: $Fxf=m*a$ The Net torque is $R*fr*F=I*\alpha$ Assume the disk motion satisfy "Rolling Without Slipping" condition: i.e. $a=R*\alpha$ $R*fr*F=I*\alpha= I*\frac{a}{R}=\frac{I}{R} \frac{FxF}{m}=\frac{I}{m*R}(Fxf)$ 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 $f=\frac{r+\frac{I}{m*R}\frac{Fx}{F}}{R+\frac{I}{m*R}}*F$ $I*\alpha=R*fr*F=R*\frac{r+\frac{I}{m*R}\frac{Fx}{F}}{R+\frac{I}{m*R}}*Fr*F=\frac{R*r+\frac{I}{m}\frac{Fx}{F}r*R\frac{I}{m}\frac{r}{R}} {R+\frac{I}{m*R}}*F=\frac{R*\frac{Fx}{F}r}{I+m*R^2}*F=\frac{R*\cos\thetar}{I+m*R^2}*F$ $I+m*R^2$ 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.*
