","comment":{"@type":"Comment","text":" The moment of inertia about the own axis must be determined, and the moment of inertia about the other axis is $m{{d}^{2}}$, where $d$ is the distance from the axis. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$2m{{R}^{2}}+2m{{b}^{2}}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{m{{R}^{2}}}{4}+m{{b}^{2}}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$2m{{R}^{2}}+\\dfrac{m{{b}^{2}}}{2}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$m{{R}^{2}}+2m{{b}^{2}}$$","position":2,"answerExplanation":{"@type":"Comment","text":" $$ \\\\ \\text{Moment of Inertia of disc of radius }R\\,\\text{is }=\\dfrac{m{{R}^{2}}}{4} \\\\ \\text{By parallel axis theorem,} \\\\ \\text{Moment of Inertia of disc at }C\\,\\text{is} \\\\ \\Rightarrow{{I}_{C}}=\\dfrac{m{{R}^{2}}}{4}+m{{\\left( BC \\right)}^{2}} \\\\ \\Rightarrow{{I}_{C}}=\\dfrac{m{{R}^{2}}}{4}+m{{b}^{2}} \\\\ \\text{Total moment of Inertia }{{\\text{I}}_{net}}, \\\\ \\Rightarrow{{\\text{I}}_{net}}=\\left( \\dfrac{m{{R}^{4}}}{4} \\right)+\\left( \\dfrac{m{{R}^{4}}}{4} \\right)+\\left( \\dfrac{m{{R}^{4}}}{4}+m{{b}^{2}} \\right)+\\left( \\dfrac{m{{R}^{4}}}{4}+m{{b}^{2}} \\right) \\\\ \\therefore \\text{Total moment of Inertia }{{\\text{I}}_{net}}=m{{R}^{2}}+2m{{b}^{2}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Rotational motion Quiz 1","text":"A constant force acts in the line represented by $$y=4x-8$$ in $$3-D$$ space. If the magnitude of the force is $$5N$$, what would be the resultant torque due to this force about the origin. ","comment":{"@type":"Comment","text":"Calculate shortest distance of the curve from the origin as $\\dfrac{c}{\\sqrt{{{a}^{2}}+{{n}^{2}}}}$, then calculate $\\text{torque}=F\\times {{r}_{\\bot }}.$ "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{40}{\\sqrt{41}}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{64}{\\sqrt{82}}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{64}{\\sqrt{41}}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{40}{\\sqrt{82}}$$","position":2,"answerExplanation":{"@type":"Comment","text":" $$\\text{For a line }y=mx+C, \\\\ \\text{The shortest distance from origin }'d'\\text{ is } \\\\d=\\dfrac{C}{\\sqrt{{{m}^{2}}+1}} \\\\ \\text{For line }y=9x-8 \\\\ \\Rightarrow d=\\dfrac{8}{\\sqrt{{{9}^{2}}+1}} \\\\ \\Rightarrow d=\\dfrac{8}{\\sqrt{82}} \\\\ \\text{Thus, torque }Z\\text{ about origin will be} \\\\ \\Rightarrow Z=\\text{Shortest distance}\\times \\text{Force} \\\\ \\Rightarrow Z=\\dfrac{8}{\\sqrt{82}}\\times 5 \\\\ \\therefore Z=\\dfrac{40}{\\sqrt{82}}\\,Nm$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Rotational motion Quiz 1","text":"A massless charge $$Q$$ is added to a solid sphere of radius $$R$$ and mass $$M$$ and connected to the end of a massless rod of length $$\\ell $$. The opposite end of the rod is set at $$O$$ and serves as a pivot. The figure depicts the setup of a uniform electric field $$E$$. When the electric field is turned on, calculate the instantaneous angular acceleration. $$ \\\\ $$
","comment":{"@type":"Comment","text":"The rod is mass less, So $I$ is only due to the solid sphere $\\tau =I\\alpha $. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{QER}{M\\left( \\dfrac{7}{5}{{R}^{2}}+{{\\ell }^{2}} \\right)}$$","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{QE\\ell }{M\\left( {{R}^{2}}+{{\\ell }^{2}} \\right)}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{QE\\left( R+\\ell \\right)}{M\\left( \\dfrac{2}{5}{{R}^{2}}+{{\\ell }^{2}}+2R\\ell \\right)}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{QE\\left( R+\\ell \\right)}{M\\left( \\dfrac{7}{5}{{R}^{2}}+{{\\ell }^{2}}+2R\\ell \\right)}$$","position":2,"answerExplanation":{"@type":"Comment","text":"
$$ \\\\ \\text{Moment of Inertia of a solid sphere about it }\\!\\!'\\!\\!\\text{ s center =}\\dfrac{2}{5}M{{R}^{2}} \\\\{{I}_{o}}\\text{ about point }O\\text{ (by parallel axis theorem )}=\\dfrac{2}{5}M{{R}^{2}}+M{{\\left( R+\\ell \\right)}^{2}} \\\\ \\Rightarrow {{I}_{o}}=M\\left( \\dfrac{7}{5}{{R}^{2}}+{{\\ell }^{2}}+2R\\ell \\right) \\\\ \\text{This is total }I\\text{ as rod is massless}\\text{.} \\\\ \\text{Torque on the sphere }F=QE \\\\ \\Rightarrow \\text{ Torque about point }O\\text{ }{{\\tau }_{o}}=\\vec{r}\\times \\vec{F} \\\\ \\Rightarrow {{\\tau }_{o}}=\\left( R+\\ell \\right)\\times F\\times \\sin 90{}^\\circ \\\\ \\Rightarrow {{\\tau }_{o}}=\\left( R+\\ell \\right)QE \\\\ \\text{As }{{\\tau }_{o}}={{I}_{o}}\\alpha \\\\ \\left( R+\\ell \\right)QE={{I}_{o}}\\alpha \\\\ \\therefore\\alpha =\\dfrac{QE\\times \\left( R+\\ell \\right)}{M\\left( \\dfrac{7}{5}{{R}^{2}}+{{\\ell }^{2}}+2R\\ell \\right)}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Rotational motion Quiz 1","text":"If a disc with mass $$\\left( m \\right)$$ is allowed to fall down the plane as shown in the figure, and friction is sufficient to prevent slipping, then the time taken by the disc to reach the bottom is $$ \\\\ $$ ","comment":{"@type":"Comment","text":" Use the formula $t=\\dfrac{1}{\\sin \\theta }{{\\left( \\dfrac{2h}{g} \\right)}^{\\tfrac{1}{2}}}{{\\left( 1+\\dfrac{I}{M{{R}^{2}}} \\right)}^{\\tfrac{1}{2}}}$"},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$2\\sqrt{\\dfrac{6h}{g}}$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{\\sqrt{3}}{2}\\times {{\\left( \\dfrac{3h}{2} \\right)}^{\\dfrac{1}{2}}}$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\dfrac{2}{\\sqrt{3}}\\times {{\\left( \\dfrac{3h}{2} \\right)}^{\\dfrac{1}{2}}}$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$2{{\\left( \\dfrac{3h}{g} \\right)}^{\\dfrac{1}{2}}}$$","position":0,"answerExplanation":{"@type":"Comment","text":" $$\\text{For motion of disc on wedge,} \\\\ \\text{Moment of inertia of disc about it }\\!\\!'\\!\\!\\text{ s centre is } \\\\I=\\dfrac{M{{R}^{2}}}{2} \\\\ \\text{Using the formula} \\\\ \\Rightarrow t=\\dfrac{1}{\\sin \\theta }{{\\left( \\dfrac{2h}{g} \\right)}^{\\dfrac{1}{2}}}{{\\left( 1+\\dfrac{I}{M{{R}^{2}}} \\right)}^{\\dfrac{1}{2}}} \\\\ \\text{Given sin}\\theta \\text{ =}\\sin 30{}^\\circ =\\dfrac{1}{2} \\\\ \\Rightarrow t=2{{\\left( \\dfrac{2H}{g} \\right)}^{\\dfrac{1}{2}}}\\left( 1+\\dfrac{M{{R}^{2}}}{2{{\\left( MR \\right)}^{2}}} \\right) \\\\ \\Rightarrow t=2{{\\left( \\dfrac{2H}{g} \\right)}^{\\dfrac{1}{2}}}\\left( 1+\\dfrac{1}{2} \\right) \\\\ \\therefore t=2{{\\left( \\dfrac{3H}{g} \\right)}^{\\dfrac{1}{2}}}$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Rotational motion Quiz 1","text":"Calculate the moment of Inertia about an axis perpendicularly passing to the plane of the plate through centre $$O$$.
","comment":{"@type":"Comment","text":"The total moment of inertia is a sum of $I$ of the disc that is curved out of the remaining position of the plate. "},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\rho {{a}^{4}}\\left( \\dfrac{8}{3}-\\dfrac{\\pi }{2} \\right)$$","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\rho {{a}^{4}}\\left( \\dfrac{128}{3}-\\dfrac{\\pi }{2} \\right)$$","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" $$\\rho {{a}^{4}}\\left( \\dfrac{8}{3}-\\dfrac{3\\pi }{2} \\right)$$","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" $$\\rho {{a}^{4}}\\left( \\dfrac{128}{3}-\\dfrac{3}{2}\\pi \\right)$$","position":0,"answerExplanation":{"@type":"Comment","text":" Moment of inertia of a complete solid square plate $$=\\dfrac{M{{\\left( 4a \\right)}^{2}}}{6} \\\\M=total\\,mass=density\\times area=\\rho \\times {{\\left( 4a \\right)}^{2}} \\\\ \\Rightarrow {{I}_{t}}=\\dfrac{\\rho {{\\left( 4a \\right)}^{4}}}{6} \\\\ $$ Moment of Inertia of circular disc curved out, about its edge is $$ \\\\ \\Rightarrow {{I}_{D}}=\\dfrac{3}{2}m{{a}^{2}} \\\\ \\Rightarrow m=\\text{mass of disc}=\\rho \\times \\pi {{a}^{2}} \\\\ \\Rightarrow {{I}_{D}}=\\dfrac{3}{2}\\rho \\pi {{a}^{4}} \\\\ $$ If I' is the moment of Inertia of the final arrangement, then $$ \\\\ \\Rightarrow {{I}_{t}}={{I}_{D}}+I' \\\\ \\Rightarrow \\dfrac{\\rho {{\\left( 4a \\right)}^{4}}}{6}=\\dfrac{3}{2}\\rho \\pi {{a}^{4}}+I' \\\\ \\therefore I'=\\rho {{a}^{4}}\\left( \\dfrac{128}{3}-\\dfrac{3}{2}\\pi \\right)$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}