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Q18  The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?

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Consider the extreme position (horizontal) :- 

                      The kinetic energy at this position is zero as velocity is zero.

                      Thus total energy is given by :    =\ mgl\ +\ 0 \ =\ mgl

Now consider the mean position (lowermost point) :

 Here the potential energy of Bob is zero. 

              Whereas   kinetic energy is  : 

                                                                         =\ \frac{1}{2}mv^2

Further, it is given that 5 per cent of energy is dissipated due to air resistance while coming down.

Thus energy equation becomes (conservation of energy):- 

                                         \frac{1}{2}mv^2\ =\ \frac{95}{100}\times mgl

or                                           

                                          v\ =\ \sqrt{\frac{2\times 95\times1.5\times 9.8}{100}}\ =\ 5.28\ m/s

Posted by

Devendra Khairwa

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