Suppose a pendulum bob is suspended from a pivot by a masslessstring of length l

Question

Suppose a pendulum bob is suspended from a pivot by a masslessstring of length l, measured to the center of the bob. a) show that if I draw the pendulum back by an angle ,the center of the bob is elevated by the distanceh=l(1-cos)=2lsin2(/2) b) using the results of part (a), show that the speed of thependulum bob at the bottom of its travel should bev=2(gl)1/2sin(/2) Suppose a pendulum bob is suspended from a pivot by a masslessstring of length l, measured to the center of the bob. a) show that if I draw the pendulum back by an angle ,the center of the bob is elevated by the distanceh=l(1-cos)=2lsin2(/2) b) using the results of part (a), show that the speed of thependulum bob at the bottom of its travel should bev=2(gl)1/2sin(/2)

Explanation / Answer

Problem occured during loading the figure. The potential energy of the pendulum is          PE = mg( L - Lcos )        mgh = mg ( L -Lcos )            h = L( 1- cos ) We know that         cos2 = 1 - 2sin2 so we can write cos = 1 - 2 sin2(/2) Therefore          h = 2 Lsin2(/2) Speed of the pendulum at the bottom is        v = (2*g*h)           = 2(gL)1/2sin(/2)

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