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A solid cylinder of mass M and radius .R. is mounted on a frictionless horizontal axle so that it can freely rotate about this axis. A string of negligible mass is wrapped round the cylinder and a body of mass .m. is hung from the string. The mass is released from rest. Find the tension in the string and the angular speed of cylinder as the mass falls a distance h.

Answer» <html><body><p></p>Solution :The <a href="https://interviewquestions.tuteehub.com/tag/acceleration-13745" style="font-weight:bold;" target="_blank" title="Click to know more about ACCELERATION">ACCELERATION</a> .a. of the falling body is given by <br/>`mg-T=ma ""`……(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)<br/>Torque on the cylinder is `tau = TR = I alpha`<br/>`therefore T=(I alpha)/(R )=((<a href="https://interviewquestions.tuteehub.com/tag/mr-549185" style="font-weight:bold;" target="_blank" title="Click to know more about MR">MR</a>^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))/(2))((a)/(R^(2))) "" [because alpha = (a)/(R )]`<br/>or `T=(Ma)/(2)`.....(2)<br/>from (1) and (2) `T=(mMg)/((M+2m))`<br/>from <a href="https://interviewquestions.tuteehub.com/tag/consevation-424818" style="font-weight:bold;" target="_blank" title="Click to know more about CONSEVATION">CONSEVATION</a> of energy,<br/>we have `mgh = (1)/(2)mv^(2)+(1)/(2)I omega^(2)`<br/>`=(1)/(2)m (Romega)^(2)+(1)/(2)((MR^(2))/(2)).omega^(2)`<br/>on solving `omega = [(4mgh)/((M+2m)R^(2))]^((1)/(2))`<br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_DOC_OBJ_PHY_XI_V01_B_C07_SLV_071_S01.png" width="80%"/></body></html>


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