","comment":{"@type":"Comment","text":" For the part $QR$, the translational kinetic energy $K.E.=\\dfrac{1}{2}m{{v}^{2}}$ will be converted into potential energy."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 1","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" 0.75","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" 0.5","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 0.6","position":2,"answerExplanation":{"@type":"Comment","text":" For pure rolling part on $$PQ$$, the mechanical energy is conserved and hence, $$ \\\\ \\dfrac{1}{2}m{{v}^{2}}\\left[ 1+\\dfrac{{{k}^{2}}}{{{R}^{2}}} \\right]=mgH \\\\ \\Rightarrow \\dfrac{5}{6}m{{v}^{2}}=mgH \\\\ $$ For the part $$QR$$, the translational kinetic energy $$K.E.=\\dfrac{1}{2}m{{v}^{2}}$$ will be converted into potential energy hence, $$ \\\\ \\dfrac{1}{2}m{{v}^{2}}=mgh \\\\ \\therefore h=\\dfrac{3}{5}H$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Kinematics of rotation Quiz 1","text":" Garima cuts a semi-circular disc of $$5\\,g$$ with radius $$2\\,cm$$ and rotates it by keeping the largest chord as the axis of rotation by applying a force of $$N$$ on an arbitrary point of the arc of the semicircle. If the region is said to be free from gravity, what will be the value of angular acceleration? ","comment":{"@type":"Comment","text":" Find the value of torque and moment of inertia to find angular acceleration given as, $ \\\\ $ Angular acceleration $=\\dfrac{T}{I}$. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{5}{2}\\,rad/{{s}^{2}}$$ ","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{7}{5}\\,rad/{{s}^{2}}$$ ","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{2}{5}\\,rad/{{s}^{2}}$$ ","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$5\\,rad/{{s}^{2}}$$ ","position":1,"answerExplanation":{"@type":"Comment","text":" For a semi-circular disc it is known that $$ \\\\ I=\\dfrac{M{{R}^{2}}}{4}=\\dfrac{5\\times {{2}^{2}}}{4}=5\\,kg-{{m}^{2}} \\\\ $$ We know that torque is given as $$ \\\\ torque=rF\\sin \\theta \\\\ torque=2\\times 1\\times \\dfrac{1}{2}=1N-m \\\\ $$ Angular acceleration is given as $$ \\\\ $$ Angular acceleration $$=\\dfrac{T}{I}$$ $$ \\\\ $$ $$\\therefore$$ Angular acceleration $$=\\dfrac{5}{1}\\,rad/{{s}^{2}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}