A thin rod of length h and mass M is heldvertically with its lower end resting o

Question

A thin rod of length h and mass M is heldvertically with its lower end resting on a frictionless horizontalsurface. The rod is then released to fall freely. (a) Determine the speed of its center of massjust before it hits the horizontal surface. (Use h, M, and g asappropriate in your equation.)
vCM = 1

(b) Now suppose the rod has a fixed pivot at its lower end.Determine the speed of the rod's center of mass just before it hitsthe surface. (Use h, M, and g as appropriate in your equation.)
vCM = 2
(a) Determine the speed of its center of massjust before it hits the horizontal surface. (Use h, M, and g asappropriate in your equation.)
vCM = 1

(b) Now suppose the rod has a fixed pivot at its lower end.Determine the speed of the rod's center of mass just before it hitsthe surface. (Use h, M, and g as appropriate in your equation.)
vCM = 2

Explanation / Answer

(a)    in the given problem we can see that thereare not any horizontal forces acting on the rod so the centre ofmass will not    move horizontally    more over the centre of mass drops straightdownward with the rod rotating about the centre of mass as itfalls    from the conservation of energy we can writethat    KEf + PEgf =KEi + PEgi    (1 / 2) M v2CM + (1 /2) I 2 + 0 = 0 + M g (h / 2)    (1 / 2) M v2CM + (1 /2) [(1 / 12) M h2] (vCM / (h /2))2 =  M g (h / 2)    so the speed of the centre of mass willbe    vCM = (3 g h / 4) (b)    in this part the motion is a pure rotation aboputa fixed pivot point with the centre of mass moving in a circularpath of    radius (h / 2)    from the conservation of energy we can writethat    KEf + PEgf =KEi + PEgi    (1 / 2) I 2 + 0 = 0 + M g(h / 2)    (1 / 2) [(1 / 3) M h2](vCM / (h / 2))2 =  M g (h /2)    so the speed of the centre of mass willbe    vCM = (3 g h / 4)

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