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Old Mar 30th 2009, 08:51 AM   #1
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speed of falling rod

A thin uniform rod is initially positioned in the vertical direction, with its lower end attached to a frictionless axis that is mounted on the floor. The rod has a length of 2.00 m and is allowed to fall, starting from rest. Find the tangential speed of the free end of the rod, just before the rod hits the floor after rotating through 90 degrees.

a. 3.59 m/s
b. 4.26 m/s
c. 5.89 m/s
d. 7.67 m/s
e. 6.83 m/s

Thanks for your help.
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Old Mar 30th 2009, 10:14 AM   #2
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We have to use law of conservation of enwrgy in this case.
When the rod is vertical, its centre of mass is at a hieght of 1 m from the ground ( centre of mass of the rod is at its mid point ). Hence the gravitational potential energy of the rod wrt floor is mgh = Mg*1 = Mg.

When the rod has turned through 90 degree, Let at that moment its angular velocity be w. Hence its rotational kinetic energy at that moment will be (1/2)Iw^2 = (1/2)(ML^2/3)w^2.

The loss of potential energy must be equal to the this kinenetic enegy. Hence-
(1/2)(ML^2/3)w^2 = Mg. Put l = 2 m, g = 9.8 and get w = 3.834

Now v=rw = 2*3.834 = 7.67 m/s
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