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Question-22872




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 .
Idisc=MR22M=2IR2=2×0.16(1/25)kg=8kgTheacc.ofropeonleftofdiscis=a=a2+αRαR=a2.(i)T2mg=ma(ii)(T1T2)R=Iα(iii)MgT1T2=Ma2(iv)From(iii):T1=T2+IαRR2=T2+Ia2R2from(iv):MgT2T2Ia2R2=Ma2Mg2mg2maIa2R2=Ma2a=(M2m)g(M2+2m+I2R2)=(82)(10)(4+2+2)m/s2=152m/s230x=152orx=4.
Commented by Physics lover last updated on 24/Oct/17
but the answer given is 4.  ???
buttheanswergivenis4.???
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.
ihavgotit.Anyways,thankyousir.Withouturhintsicudnthavsolvedit.
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:T2=1(a+g)i&8gT1T2=8(a2)T1=7g5aiiRotation:T1RT2R=IαT1R2T2R2=IαR=I(a2)T1T2=2aiiisubstitutingT2fromeqiandT1fromeqiiineqiii7g5aag=2a6g8=a=6(10)8=304x=4

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