The diagram shows a thin rod of uniform mass distribution pivoted about one end

Question

The diagram shows a thin rod of uniform mass distribution pivoted about one end by a pin passing through that point. The mass of the rod is 0.600 kg and its length is1.90 m. When the rod is released from its horizontal position, it swings down to the vertical position as shown.

The diagram shows a thin rod of uniform mass distribution pivoted about one end by a pin passing through that point. The mass of the rod is 0.600 kg and its length is1.90 m. When the rod is released from its horizontal position, it swings down to the vertical position as shown. Determine the speed of its center of gravity at its lowest position. When the rod reaches the vertical position, calculate the tangential speed of the free end of the rod.

Explanation / Answer

By energy conservation;

mgL/2 = Iw^2/2

I=MI of rod =Ml^2/12

on solving we get

w=7.86 rad/sec

a)Vc=w*l/2 =7.47 m/s

b)Vf=w*l=14.95 m/s

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