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Title: Rolling with or without slippingPost by: ahmedelshfie on April 23, 2010, 01:38:03 am
This applet created by Prof Hwang
Modified by Ahmed Original project Rolling with or without slipping (http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=706.msg2558#msg2558) You can drag both end points of the plane to change the slope (angle) If the friction coefficient is large enough, the circular object will rolling (without slipping) along the plane. (energy is conserved) However, if the coefficient is too small, the object will slip along the plane and energy will be loss. Plotting panel show traces for different energy: Kinetic energy: (1/2) m v2 rotational energy: (1/2) I w2 potential energy: mgh Can you identify all the traces? -*- You can drag cyan circle inside the rotating object, the trace for this small circle will be shown and the velocity vector(relative to object center ot ground) will be shown,too. Title: Re: Rolling with or without slippingPost by: ahmedelshfie on April 23, 2010, 01:43:26 am
Assume the angle of the slope is $\theta$, mass of the disk is m, radius is R. The
momentum of inertia is $I=\frac{1}{2}mR^2.$ Let analysis this problem from the contact point between the disk and the slope. The normal force between the disk and the slope is $mg \cos\theta$, and the force along the slope is $mg \sin\theta$, Assume the friction between the disk and the slope is f. The net force is $mg \sin\theta + f = m a$, where a is the acceleration along the slope. (the friction force from the slope to the disk is in the same direction as acceleration a) The condition for rolling without slipping is $a=R\alpha$ The torque is $\tau= R mg\sin\theta = I \alpha =\frac{1}{2}mR^2 \alpha=\frac{1}{2}m R R\alpha=\frac{1}{2}m Ra.$ So $mg\sin\theta=\frac{1}{2}m a, i.e. a =2g \sin\theta and f=ma- mg\sin\theta= mg\sin\theta$ Since the maximum static friction force f_{max}=mg\cos\theta\mu \ge mg\sin\theta, it imply that$ \mu\ge \tan\theta$ for the disk to rolling without slipping. If it is a ball instead of a disk, then $I=\frac{2}{5}mR^2$. $\tau= R mg\sin\theta = I \alpha =\frac{2}{5}mR^2\alpha=\frac{2}{5}m R a$ So$ mg\sin\theta=\frac{2}{5}ma, or a=\frac{5}{2}g\sin\theta.$ $f=ma -mg\sin\theta=\frac{3}{5}ma=\frac{3}{2}g\sin\theta$. So the condition for rolling without slipping becomes$ \mu \ge \frac{2}{3}\tan\theta$. |