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A simple pendulum of length l and mass m free to oscillate in vertical plane. A nail is located at a distance d=l-a vertically below the point of suspension of a simple pendulum. The pendulum bob is released from the position where the string makes an angle of 90^@ from vertical. Discuss the motion of the bob if (a) l=2a and (b) l=2.5a.

Answer» <html><body><p></p>Solution :a. `l=2a`, velocity at lowest point from <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> conservation <br/> `0+mg2a=1/2mv^2` <br/> `v=sqrt(4ga)` <br/> Here the radius of <a href="https://interviewquestions.tuteehub.com/tag/circle-916533" style="font-weight:bold;" target="_blank" title="Click to know more about CIRCLE">CIRCLE</a> is a about nail and velocity at lowest point is not sufficient to <a href="https://interviewquestions.tuteehub.com/tag/complete-423576" style="font-weight:bold;" target="_blank" title="Click to know more about COMPLETE">COMPLETE</a> the loop. Therefore, <a href="https://interviewquestions.tuteehub.com/tag/motion-1104108" style="font-weight:bold;" target="_blank" title="Click to know more about MOTION">MOTION</a> of bob is the combination of circular and projectile motion. Because the velocity at lowest point <a href="https://interviewquestions.tuteehub.com/tag/lies-1073086" style="font-weight:bold;" target="_blank" title="Click to know more about LIES">LIES</a> between `sqrt(3ga)` and `sqrt(5ga)`. <br/> b. `l=2.5a`, Velocity at lowest point from energy conservation <br/> `0+mg(2.5a)=1/2mv^2impliesv=sqrt(5ga)` <br/> here radius of circle is a about nail and velocity at lowest point is just sufficient to complete the loop so that here looping the loop about nail.</body></html>


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