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A disc of mass M and radius R is rolling with angular speed omega on a horizontal plane as shown in figure. The magnitude of angular momentum of the disc about the origin O is

Answer» <html><body><p>`(1/2)<a href="https://interviewquestions.tuteehub.com/tag/mr-549185" style="font-weight:bold;" target="_blank" title="Click to know more about MR">MR</a>^(2)<a href="https://interviewquestions.tuteehub.com/tag/omega-585625" style="font-weight:bold;" target="_blank" title="Click to know more about OMEGA">OMEGA</a>`<br/>`MR^(2)omega`<br/>`(3/2)MR^(2)omega`<br/>`2Mr^(2)omega`</p>Solution :The disc has two types of motion, namely, translation and rotational. Therefore there are two types of angular moment and the total angular momentum is the sum of thes two. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_VOL2_C03_E01_339_S01.png" width="80%"/> <br/> `L=L_(T)+L_(R), L_(t)=` angular momentum due to translational motion <br/> `L_(R)=` angular momentum due to rotational motion about `CM` <br/> `L=MVxxR+I_(CM)omegaI_(CM)` <br/> `=MI` about centre of mass `C` <br/> `=M(Romega)R+1/2MR^(2)omega(V=Romega` in case of <a href="https://interviewquestions.tuteehub.com/tag/rolling-1190754" style="font-weight:bold;" target="_blank" title="Click to know more about ROLLING">ROLLING</a> motion and <a href="https://interviewquestions.tuteehub.com/tag/surface-1235573" style="font-weight:bold;" target="_blank" title="Click to know more about SURFACE">SURFACE</a> at rest) <br/> `=3/2MR^(2)omega`</body></html>


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