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Explain the method to find the centre of gravity of irregularly shaped lamina.

Answer» <html><body><p></p>Solution :(i) The center of gravity of a uniform lamina of even an irregular <a href="https://interviewquestions.tuteehub.com/tag/shape-1204673" style="font-weight:bold;" target="_blank" title="Click to know more about SHAPE">SHAPE</a> is determined by pivoting it at various <a href="https://interviewquestions.tuteehub.com/tag/points-1157347" style="font-weight:bold;" target="_blank" title="Click to know more about POINTS">POINTS</a> by trial and error. <br/> (ii) The lamina remains horizontal when pivoted at the point where the net gravitational <a href="https://interviewquestions.tuteehub.com/tag/force-22342" style="font-weight:bold;" target="_blank" title="Click to know more about FORCE">FORCE</a> acts, which is the center of gravity. <br/> (<a href="https://interviewquestions.tuteehub.com/tag/iii-497983" style="font-weight:bold;" target="_blank" title="Click to know more about III">III</a>) When a body is supported at the center of gravity, the sum of the torques acting on all the point masses of the rigid body becomes <a href="https://interviewquestions.tuteehub.com/tag/zero-751093" style="font-weight:bold;" target="_blank" title="Click to know more about ZERO">ZERO</a>. Moreever the weight is compensated by the normal reaction force exerted by the pivot. <br/> (iv) The body is in static equilibrium and hence it remains horizontal. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/SUR_PHY_XI_V01_C05_E03_002_S01.png" width="80%"/> <br/> (v) There is also another way to determine the center of gravity of an irregular lamina. If we suspend the lamina from different points like P, Q, R. The vertical lines PP', QQ', RR' all pass through the center of gravity. <br/> (vi) Here, reaction force acting at the point of suspension and the gravitational force acting at the center of gravity cancel each other and the torques caused by them also cancel each other.</body></html>


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