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| 1. |
Find the angle .theta. with the lower vertical at which the resultant acceleration of the bob is along the horizontal, when the perpendicular is released from horizontal position. |
| Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Since the resultant acceleration is along the horizontal, the <a href="https://interviewquestions.tuteehub.com/tag/vertical-726181" style="font-weight:bold;" target="_blank" title="Click to know more about VERTICAL">VERTICAL</a> acceeleration at .P. should be zero. <br/> ie., `a_(c) sin(90 - theta) = a_(t) sin theta` <br/> `(a_(c))/(a_(1)) = tan theta ..... (1)`<br/> But `a_(c) = (v^(2))/(l)(2glcostheta)/(l)=2gcostheta` <br/> and `a_(1) = g sin theta` <br/> Then `(a_(c))/(a_(t)) = 2 cot theta` <br/> <a href="https://interviewquestions.tuteehub.com/tag/substituting-1231652" style="font-weight:bold;" target="_blank" title="Click to know more about SUBSTITUTING">SUBSTITUTING</a> in <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a> (1) <br/> `2 cot theta = Tan theta` <br/> Hence `theta = tan^(-1) (sqrt(2))` <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_MP2_C06_SLV_030_S01.png" width="80%"/></body></html> | |