I\'ve been working on this problem for a while, but don\'t knowif my method is v

Question

I've been working on this problem for a while, but don't knowif my method is valid. Maybe someone can give me a fewpointers? The problem is: a long rectancular bar swings freely from apivot point at one end. When at rest, the bar hangs freely, barelycoming into contact with a solid barrier. Except, there is acircular rod extruding a small distance x from the barrier,free to move along the x-axis, so when the bar falls from an angletheta, it first strikes the bar, imparting momentum to it. See theimage. The bar and rod have similar, but different massesmb and mr. Assume no friction, and elasticcollisions. The question is, given an angular velocity ofi of the rectangular bar, what is the velocityvf of the bar the instant theta = 0 (i.e. the bar comesinto contact with the barrier). My approach has been to take 1) the conservation of kinetic energy (rotationalfor the bar and translational for the rod) and 2) consider the bar as a point mass and use theconservation of angular momentum. I get an answer of vf = i(2 Ir)/(I + r2 mb), where I is the moment ofinertia of the bar and r is the vertical distance from the pivotpoint to the bar. Does that approach make sense? How would you go about solving this problem? Thanks for any suggestions.

Explanation / Answer

the conservation of kinetic energy gives (1/2)Iwi2 +(1/2)Iwf2 = (1/2)m * v2 I is the moment of inertia of the bar,wi andwf are the angular speeds of the bar,m is the mass ofthe rod and v is the speed of the rod. or I * (wi2 + wf2)= m * v2 or (wi2 +wf2) = (m * v2/I)----------(1) the conservation of angular momentum gives I * (wi + wf) = m * v * x or (wi + wf) = (m * v * x/I) or wi = (m * v * x/I) - wf------------(2) from (1) and (2) we get (((m * v * x/I) - wf)2 +wf2) = (m * v2/I) solving the above equation for wf we get the finalangular speed of the bar wf = (vf/l) or vf = wf * l l is the length of the bar

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