JEE Advanced 2021 · Paper 1 · Physics · Rotational Motion
A thin rod of mass MandlengthM and lengthMandlengthaisfreetorotateinhorizontalplaneaboutafixedverticalaxispassingthroughpointis free to rotate in horizontal plane about a fixed vertical axis passing through pointisfreetorotateinhorizontalplaneaboutafixedverticalaxispassingthroughpointO.Athincirculardiscofmass. A thin circular disc of mass.AthincirculardiscofmassMandofradiusand of radiusandofradius\dfrac{a}{4}ispivotedonthisrodwithitscentreatadistanceis pivoted on this rod with its centre at a distanceispivotedonthisrodwithitscentreatadistance\dfrac{a}{4}fromthefreeendsothatitcanrotatefreelyaboutitsverticalaxis,asshowninthefigure.Assumethatboththerodandthedischaveuniformdensityandtheyremainhorizontalduringthemotion.Anoutsidestationaryconserverfindstherodrotatingwithanangularvelocityfrom the free end so that it can rotate freely about its vertical axis, as shown in the figure. Assume that both the rod and the disc have uniform density and they remain horizontal during the motion. An outside stationary conserver finds the rod rotating with an angular velocityfromthefreeendsothatitcanrotatefreelyaboutitsverticalaxis,asshowninthefigure.Assumethatboththerodandthedischaveuniformdensityandtheyremainhorizontalduringthemotion.AnoutsidestationaryconserverfindstherodrotatingwithanangularvelocityΩandthediscrotatingaboutitsverticalaxiswithangularvelocityand the disc rotating about its vertical axis with angular velocityandthediscrotatingaboutitsverticalaxiswithangularvelocity4Ω.Thetotalangularmomentumofthesystemaboutthepoint. The total angular momentum of the system about the point.ThetotalangularmomentumofthesystemaboutthepointOisisis\left[(Ma2Ω48)\left(\dfrac{M{a}^{2}Ω}{48}\right)(48Ma2Ω)n\right].</p><p>Thevalueof.</p><p>The value of.</p><p>Thevalueofnis _________ .
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Angular momentum of the system about OOO,
L→=L→Rod+L→Disc\overset{\to}{L} = \overset{\to}{L}Rod + \overset{\to}{L}DiscL→=L→Rod+L→Disc
⇒L→=L→Rod+[(L→CM)Disc+r→CM×(MDiscV→CM)]⇒ \overset{\to}{L} = \overset{\to}{L}Rod + \left[{\left(\overset{\to}{L}CM\right)}_{Disc} + \overset{\to}{r}CM \times \left({M}_{Disc}\overset{\to}{V}CM\right)\right]⇒L→=L→Rod+[(L→CM)Disc+r→CM×(MDiscV→CM)]
⇒L=Ma2Ω3+Ma28Ω+M(3a4)Ω(3a4)⇒ L = \dfrac{M{a}^{2}Ω}{3} + \dfrac{M{a}^{2}}{8}Ω + M\left(\dfrac{3a}{4}\right)Ω\left(\dfrac{3a}{4}\right)⇒L=3Ma2Ω+8Ma2Ω+M(43a)Ω(43a)
⇒L=Ma23Ω+Ma28Ω+9Ma216Ω⇒ L = \dfrac{M{a}^{2}}{3}Ω + \dfrac{M{a}^{2}}{8}Ω + \dfrac{9M{a}^{2}}{16}Ω⇒L=3Ma2Ω+8Ma2Ω+169Ma2Ω
⇒L=16Ma2Ω+6Ma2Ω+27Ma2Ω48⇒ L = \dfrac{16M{a}^{2}Ω + 6M{a}^{2}Ω + 27M{a}^{2}Ω}{48}⇒L=4816Ma2Ω+6Ma2Ω+27Ma2Ω
⇒L=4948Ma2Ω⇒ L = \dfrac{49}{48}M{a}^{2}Ω⇒L=4849Ma2Ω
Compare it with nMa248Ω\dfrac{nM{a}^{2}}{48}Ω48nMa2Ω, we get n=49n = 49n=49
Hence, n=49n = 49n=49
