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A mass m is released from the top of a vertical circular track of radius r with a horizontal speed v_(theta). Calculate the angle theta with respect to the vertical where it leaves contact with the track.

Answer» <html><body><p></p>Solution :The <a href="https://interviewquestions.tuteehub.com/tag/forces-16875" style="font-weight:bold;" target="_blank" title="Click to know more about FORCES">FORCES</a> acting on the body are its weight mg and <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a> <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> as shown in Fig. <br/> So for circular motion of the body at any position `theta` <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_MP2_C06_SLV_031_S01.png" width="80%"/> `(mv^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))/(r)=mgcostheta-NorN=mgcostheta-(mv^(2))/(r)` <br/> The body will have contact where N = 0 <br/> i.e., mg `cos theta = (mv^(2))/(r) = 0` <br/> i.e, `cos theta = (v^(2))/(rg) "" .............(2)` <br/> Now applying conservationof mechanical energy between H and P, we get <br/> `(1)/(2)mv^(2)=(1)/(2)mv_(0)^(2)+<a href="https://interviewquestions.tuteehub.com/tag/mgr-2831250" style="font-weight:bold;" target="_blank" title="Click to know more about MGR">MGR</a>(1-costheta)` <br/> where .v. is the velocity at .P. <be> `[as y = r(1-cos theta)]` <br/> `or v^(2)=v_(0)^(2)+2gr(1-costheta).........(3)` <br/> substituting the value of `v^(2)` in Equation (2) <br/> `cos theta=(v_(0)^(2))/(rg)+2(1-costheta)` <br/> `i.e., cos theta=[(v_(0)^(2))/(3rg)+(2)/(3)]ortheta=cos^(-1)[(v_(0)^(2))/(3rg)+(2)/(3)]`</be></body></html>


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