Rotational Motion Problem #2
One fourth length of a Uniform Rod of mass \(m\) and length \(x\) is placed on rough horizontal surface and it is held stationary by means of a light thread as shown in the figure. The thread is brunt and the rod starts rotating about the edge, the co-efficient of friction of the surface is \(\mu\). Determine the angle between the rod and the horizontal surface at which the rod is about to slip on the edge.
Help on the FBD (Free Body Diagram) would be great.
Nice problem. It's amazing how something so seemingly simple can be quite so complicated. Call the pivot point A. Notice that if the rod is rotating about A, then there is a force at A. The way to see that is there is a centripetal force from the part to the right/below A; and another centripetal force for the part left/above A. The first of those forces is greater, hence there is a force at A and it points along the direction of the rule. Now, friction must provide that force. The ruler will slip when friction cannot provide the necessary force. So to solve the problem: find an expression for that net centripetal force at A. Find an expression for the maximum frictional force at A. Then find when the former equals (and is about to exceed) the latter.
Thanks!
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