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Question Number 22872 by Physics lover last updated on 23/Oct/17

Commented by ajfour last updated on 23/Oct/17

Commented by ajfour last updated on 24/Oct/17

I_(disc) =((MR^2 )/2) ⇒ M=((2I)/R^2 ) =((2×0.16)/((1/25))) kg = 8kg The acc. of rope on left of disc is = a =(a/2)+αR ⇒ 𝛂R=(a/2) ....(i) T_2 −mg=ma ...(ii) (T_1 −T_2 )R=I𝛂 ...(iii) Mg−T_1 −T_2 =((Ma)/2) ...(iv) From (iii): ⇒ T_1 =T_2 +((IαR)/R^2 ) =T_2 +((Ia)/(2R^2 )) from (iv): Mg−T_2 −T_2 −((Ia)/(2R^2 )) =((Ma)/2) ⇒ Mg−2mg−2ma−((Ia)/(2R^2 ))=((Ma)/2) ⇒ a=(((M−2m)g)/(((M/2)+2m+(I/(2R^2 ))))) =(((8−2)(10))/((4+2+2))) m/s^2 =((15)/2)m/s^2 ⇒ ((30)/x) = ((15)/2) or x=4 .

$${I}_{{disc}} =\frac{{MR}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\:{M}=\frac{\mathrm{2}{I}}{{R}^{\mathrm{2}} }\:=\frac{\mathrm{2}×\mathrm{0}.\mathrm{16}}{\left(\mathrm{1}/\mathrm{25}\right)}\:{kg}\:=\:\mathrm{8}{kg} \\ $$$${The}\:{acc}.\:{of}\:{rope}\:{on}\:{left}\:{of}\:{disc} \\ $$$${is}\:=\:{a}\:=\frac{{a}}{\mathrm{2}}+\alpha{R} \\ $$$$\Rightarrow\:\:\:\:\:\:\boldsymbol{\alpha{R}}=\frac{\boldsymbol{{a}}}{\mathrm{2}}\:....\left({i}\right) \\ $$$$\:\:\:\:\boldsymbol{{T}}_{\mathrm{2}} −\boldsymbol{{mg}}=\boldsymbol{{ma}}\:\:\:...\left({ii}\right) \\ $$$$\:\:\:\:\left(\boldsymbol{{T}}_{\mathrm{1}} −\boldsymbol{{T}}_{\mathrm{2}} \right)\boldsymbol{{R}}=\boldsymbol{{I}\alpha}\:\:\:\:...\left({iii}\right) \\ $$$$\:\:\:\:\:\boldsymbol{{Mg}}−\boldsymbol{{T}}_{\mathrm{1}} −\boldsymbol{{T}}_{\mathrm{2}} =\frac{\boldsymbol{{Ma}}}{\mathrm{2}}\:\:...\left({iv}\right) \\ $$$${From}\:\left({iii}\right): \\ $$$$\Rightarrow\:\:{T}_{\mathrm{1}} ={T}_{\mathrm{2}} +\frac{{I}\alpha{R}}{{R}^{\mathrm{2}} }\:={T}_{\mathrm{2}} +\frac{{Ia}}{\mathrm{2}{R}^{\mathrm{2}} } \\ $$$${from}\:\left({iv}\right): \\ $$$${Mg}−{T}_{\mathrm{2}} −{T}_{\mathrm{2}} −\frac{{Ia}}{\mathrm{2}{R}^{\mathrm{2}} }\:=\frac{{Ma}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:{Mg}−\mathrm{2}{mg}−\mathrm{2}{ma}−\frac{{Ia}}{\mathrm{2}{R}^{\mathrm{2}} }=\frac{{Ma}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\boldsymbol{{a}}=\frac{\left(\boldsymbol{{M}}−\mathrm{2}\boldsymbol{{m}}\right)\boldsymbol{{g}}}{\left(\frac{\boldsymbol{{M}}}{\mathrm{2}}+\mathrm{2}\boldsymbol{{m}}+\frac{{I}}{\mathrm{2}\boldsymbol{{R}}^{\mathrm{2}} }\right)} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{8}−\mathrm{2}\right)\left(\mathrm{10}\right)}{\left(\mathrm{4}+\mathrm{2}+\mathrm{2}\right)}\:{m}/{s}^{\mathrm{2}} \:=\frac{\mathrm{15}}{\mathrm{2}}\boldsymbol{{m}}/\boldsymbol{{s}}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\frac{\mathrm{30}}{{x}}\:=\:\frac{\mathrm{15}}{\mathrm{2}}\:\:{or}\:\:\:\boldsymbol{{x}}=\mathrm{4}\:. \\ $$$$ \\ $$

Commented by Physics lover last updated on 24/Oct/17

but the answer given is 4. ???

$${but}\:{the}\:{answer}\:{given}\:{is}\:\mathrm{4}. \\ $$$$??? \\ $$

Commented by Physics lover last updated on 24/Oct/17

i hav got it.Anyways,thank you sir. Without ur hints i cudnt hav solved it.

$${i}\:{hav}\:{got}\:{it}.{Anyways},{thank}\:{you}\:{sir}. \\ $$$${Without}\:{ur}\:{hints}\:{i}\:{cudnt}\:{hav}\:{solved} \\ $$$${it}. \\ $$

Answered by Physics lover last updated on 24/Oct/17

Translation: T_(2 ) = 1(a+g) ...i & 8g −T_1 −T_2 =8 ((a/2)) ⇒T_(1 ) = 7g − 5a ...ii Rotation: T_1 R−T_2 R = I α ⇒T_1 R^2 −T_2 R^2 =IαR=I((a/2)) ⇒T_(1 ) − T_(2 ) = 2a ...iii substituting T_(2 ) from eq ...i and T_1 from eq ...ii in eq ...iii ⇒7g−5a−a−g=2a ⇒((6g)/(8 )) = a = ((6(10))/8) = ((30)/4) ⇒ x = 4

$${Translation}: \\ $$$${T}_{\mathrm{2}\:} =\:\mathrm{1}\left({a}+{g}\right)\:\:...{i} \\ $$$$\& \\ $$$$\mathrm{8}{g}\:−{T}_{\mathrm{1}} −{T}_{\mathrm{2}} \:=\mathrm{8}\:\left(\frac{{a}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{T}_{\mathrm{1}\:} =\:\mathrm{7}{g}\:−\:\mathrm{5}{a}\:...{ii} \\ $$$${Rotation}: \\ $$$${T}_{\mathrm{1}} {R}−{T}_{\mathrm{2}} {R}\:=\:{I}\:\alpha \\ $$$$\Rightarrow{T}_{\mathrm{1}} {R}^{\mathrm{2}} −{T}_{\mathrm{2}} {R}^{\mathrm{2}} \:={I}\alpha{R}={I}\left(\frac{{a}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{T}_{\mathrm{1}\:} −\:{T}_{\mathrm{2}\:} =\:\mathrm{2}{a}\:...{iii} \\ $$$${substituting}\:{T}_{\mathrm{2}\:} {from}\:{eq}\:\:\:\:...{i} \\ $$$${and}\:{T}_{\mathrm{1}} \:{from}\:{eq}\:...{ii}\:\:{in}\:{eq}\:...{iii} \\ $$$$ \\ $$$$\Rightarrow\mathrm{7}{g}−\mathrm{5}{a}−{a}−{g}=\mathrm{2}{a} \\ $$$$\Rightarrow\frac{\mathrm{6}{g}}{\mathrm{8}\:}\:=\:{a}\:=\:\frac{\mathrm{6}\left(\mathrm{10}\right)}{\mathrm{8}}\:=\:\frac{\mathrm{30}}{\mathrm{4}} \\ $$$$\Rightarrow\:{x}\:=\:\mathrm{4} \\ $$$$ \\ $$

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