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Question-31292

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Question Number 31292 by ajfour last updated on 05/Mar/18
Commented by ajfour last updated on 05/Mar/18
Length of tape is L.
$${Length}\:{of}\:{tape}\:{is}\:{L}. \\ $$
Commented by mrW2 last updated on 07/Mar/18
any solution given?
$${any}\:{solution}\:{given}? \\ $$
Commented by ajfour last updated on 07/Mar/18
Answer given.
$${Answer}\:{given}. \\ $$
Answered by ajfour last updated on 07/Mar/18
Commented by ajfour last updated on 07/Mar/18
mgrsin θ=((3/2)mr^2 )((1/r))(((vdv)/dx)) ⇒ ∫_0 ^( v) vdv = ((2gsin θ)/3)∫_0 ^( x) dx ⇒ (v^2 /2)=((2gxsin θ)/3) ⇒ v=(√((4gsin θ)/3)) (√x) ∫_0 ^( L) (dx/( (√x))) = (√((4gsin θ)/3))∫_0 ^( t) dt 2(√L) = ((√((4gsin θ)/3))) t or t=(√((3L)/(gsin θ))) .
$${mgr}\mathrm{sin}\:\theta=\left(\frac{\mathrm{3}}{\mathrm{2}}{mr}^{\mathrm{2}} \right)\left(\frac{\mathrm{1}}{{r}}\right)\left(\frac{{vdv}}{{dx}}\right) \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\:\:{v}} {vdv}\:=\:\frac{\mathrm{2}{g}\mathrm{sin}\:\theta}{\mathrm{3}}\int_{\mathrm{0}} ^{\:\:{x}} {dx} \\ $$$$\Rightarrow\:\frac{{v}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{2}{gx}\mathrm{sin}\:\theta}{\mathrm{3}} \\ $$$$\Rightarrow\:\:{v}=\sqrt{\frac{\mathrm{4}{g}\mathrm{sin}\:\theta}{\mathrm{3}}}\:\sqrt{{x}} \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\:\:{L}} \:\frac{{dx}}{\:\sqrt{{x}}}\:=\:\sqrt{\frac{\mathrm{4}{g}\mathrm{sin}\:\theta}{\mathrm{3}}}\int_{\mathrm{0}} ^{\:\:{t}} {dt} \\ $$$$\mathrm{2}\sqrt{{L}}\:=\:\left(\sqrt{\frac{\mathrm{4}{g}\mathrm{sin}\:\theta}{\mathrm{3}}}\right)\:{t} \\ $$$${or}\:\:\:\:\:\:\:{t}=\sqrt{\frac{\mathrm{3}{L}}{{g}\mathrm{sin}\:\theta}}\:\:. \\ $$
Commented by rahul 19 last updated on 08/Mar/18
sir, why I_(cm) =(3/2)mr^2 ? assuming it to be disc we get, ((mr^2 )/4)+mr^2 = (5/4)mr^2 . but u have taken as ((mr^2 )/2)+mr^2 ?? ( i am taking torque about the surface ).
$${sir},\:{why}\:{I}_{{cm}} =\frac{\mathrm{3}}{\mathrm{2}}{mr}^{\mathrm{2}} ? \\ $$$${assuming}\:{it}\:{to}\:{be}\:{disc}\:{we}\:{get}, \\ $$$$\frac{{mr}^{\mathrm{2}} }{\mathrm{4}}+{mr}^{\mathrm{2}} =\:\frac{\mathrm{5}}{\mathrm{4}}{mr}^{\mathrm{2}} .\:{but}\:{u}\:{have}\:{taken} \\ $$$${as}\:\frac{{mr}^{\mathrm{2}} }{\mathrm{2}}+{mr}^{\mathrm{2}} ??\:\left(\:{i}\:{am}\:{taking}\:{torque}\right. \\ $$$$\left.{about}\:{the}\:{surface}\:\right). \\ $$
Commented by mrW2 last updated on 08/Mar/18
for solid disc ((mr^2 )/2)+mr^2 is correct sir.
$${for}\:{solid}\:{disc}\:\frac{{mr}^{\mathrm{2}} }{\mathrm{2}}+{mr}^{\mathrm{2}} \:{is}\:{correct}\:{sir}. \\ $$
Commented by rahul 19 last updated on 08/Mar/18
where am i wrong ? can′t i take torque along the surface .( slant surface of wedge ) ?
$${where}\:\:{am}\:{i}\:\:{wrong}\:?\:{can}'{t}\:{i}\:{take}\: \\ $$$${torque}\:{along}\:{the}\:{surface}\:.\left(\:{slant}\:\right. \\ $$$$\left.{surface}\:{of}\:{wedge}\:\right)\:? \\ $$
Commented by mrW2 last updated on 08/Mar/18
I_0 =((mr^2 )/2) for solid disc.
$${I}_{\mathrm{0}} =\frac{{mr}^{\mathrm{2}} }{\mathrm{2}}\:{for}\:{solid}\:{disc}. \\ $$
Commented by rahul 19 last updated on 08/Mar/18
yes sir, it is when axis is perpendicular to disc but i am taking along surface.
$${yes}\:{sir},\:{it}\:{is}\:{when}\:{axis}\:{is}\:{perpendicular} \\ $$$${to}\:{disc}\:{but}\:{i}\:{am}\:{taking}\:{along}\:{surface}. \\ $$

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