Friday, October 22, 2010

Irodov Problem 4.28


First let us consider by understanding all the forces acting on the rod. The rod experiences five forces - i) the two normal reactions N1 and N2 from each of the two rotating wheels ii) the forces of friction kN1 and kN2 acting in a direction so as to oppose the relative motion between the rotatig wheel and the rod and iii) the force of gravity mg the pulls the rod down.



Considering the forces on the rod in the vertical direction we have,



Considering forces on the rod in the horizontal direction we have,



In (2), m is the mass of the rod and x the location of the CG of the rod. The location of the CG at the equilibrium position at the center of the wheels is considered to be the origin.

Now let us consider the torques acting on the rod due to the five forces about the CG of the rod. Since the forces of gravity and those due to friction pass through the CG of the rod, they impose no orque on the rod. The two normal reactions from the wheel however do generate torques on the rod. Let l be the distance between the wheels' centers. When the rod's CG is located at x, the distance between the point of contact of the left wheel and the rod and the rod's CG is given by l/2+x. The distance between the point of contact of the right wheel and the rod and the rod's CG is given by, l/2 - x. Thus, the net torque is given by,




Since the rod does not rotate and has no angular acceleration, then net torque on the rod must be zero i.e. T=0.








From (2) and (4) we have,



From, (5) the time period can be computed as,




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