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Question-23748




Question Number 23748 by mrW1 last updated on 05/Nov/17
Commented by mrW1 last updated on 05/Nov/17
The same question as Q23707,  but the pulley C has a mass 1 kg and  radius 5 cm.
$$\mathrm{The}\:\mathrm{same}\:\mathrm{question}\:\mathrm{as}\:\mathrm{Q23707}, \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{pulley}\:\mathrm{C}\:\mathrm{has}\:\mathrm{a}\:\mathrm{mass}\:\mathrm{1}\:\mathrm{kg}\:\mathrm{and} \\ $$$$\mathrm{radius}\:\mathrm{5}\:\mathrm{cm}. \\ $$
Answered by ajfour last updated on 05/Nov/17
Tension on left(A side)=T_1   T_1 −mg=2mA                 ...(i)  F−T_1 −T_2 −mg=mA                   ....(ii)  (T_2 −T_1 )r =(((mr^2 )/2))((A/r))    ...(iii)  ⇒  T_2 =T_1 +((mA)/2)  using in (ii):  F−2T_1 =((3mA)/2)+mg                .....(iv)  Block A is lifted when T_1 =mg  so,     F=3mg       (A =0 till then)  ⇒     20t=30    or   t=(3/2)s  From (iv) and (i):  F−2(mg+2mA)=((3mA)/2)+mg  or   20t−30=((11A)/2)  A=((40)/(11))(t−(3/2))  this is valid till T_2  ≤2mg  Eliminating T_1  from (i) and(iii)  T_1 −mg=2mA ;  T_2 =T_1 +((mA)/2)+mg  T_2 =mg+2mA+((mA)/2)  when block B is lifted,  T_2 =2mg ⇒   ((5mA)/2)=mg  A=((2g)/5)   or   ((40)/(11))(t−(3/2))=((2g)/5)                    t=((11)/(10))+(3/2) =2.6s  s_p  = ∫_(3/2) ^(  2.6) v_p dt  v_p = ∫_(3/2) ^(  t) Adt = ((40)/(11))∫_(3/2) ^(  t) (t−(3/2))dt       =((20)/(11))(t−(3/2))^2         s_p =((20)/(11))∫_(3/2) ^(  2.6) (t−(3/2))^2 dt      =((20)/(11))×(1/3)×(((11)/(10)))^3 =((121)/(150)) m ≈ 0.8m    (s_p  by the time block B is lifted).    still the same answer..!
$${Tension}\:{on}\:{left}\left({A}\:{side}\right)={T}_{\mathrm{1}} \\ $$$${T}_{\mathrm{1}} −{mg}=\mathrm{2}{mA}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left({i}\right) \\ $$$${F}−{T}_{\mathrm{1}} −{T}_{\mathrm{2}} −{mg}={mA}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({ii}\right) \\ $$$$\left({T}_{\mathrm{2}} −{T}_{\mathrm{1}} \right){r}\:=\left(\frac{{mr}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\frac{{A}}{{r}}\right)\:\:\:\:…\left({iii}\right) \\ $$$$\Rightarrow\:\:{T}_{\mathrm{2}} ={T}_{\mathrm{1}} +\frac{{mA}}{\mathrm{2}} \\ $$$${using}\:{in}\:\left({ii}\right): \\ $$$${F}−\mathrm{2}{T}_{\mathrm{1}} =\frac{\mathrm{3}{mA}}{\mathrm{2}}+{mg}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..\left({iv}\right) \\ $$$${Block}\:{A}\:{is}\:{lifted}\:{when}\:{T}_{\mathrm{1}} ={mg} \\ $$$${so},\:\:\:\:\:{F}=\mathrm{3}{mg}\:\:\:\:\:\:\:\left({A}\:=\mathrm{0}\:{till}\:{then}\right) \\ $$$$\Rightarrow\:\:\:\:\:\mathrm{20}{t}=\mathrm{30}\:\:\:\:{or}\:\:\:\boldsymbol{{t}}=\frac{\mathrm{3}}{\mathrm{2}}{s} \\ $$$${From}\:\left({iv}\right)\:{and}\:\left({i}\right): \\ $$$${F}−\mathrm{2}\left({mg}+\mathrm{2}{mA}\right)=\frac{\mathrm{3}{mA}}{\mathrm{2}}+{mg} \\ $$$${or}\:\:\:\mathrm{20}{t}−\mathrm{30}=\frac{\mathrm{11}{A}}{\mathrm{2}} \\ $$$${A}=\frac{\mathrm{40}}{\mathrm{11}}\left({t}−\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$${this}\:{is}\:{valid}\:{till}\:{T}_{\mathrm{2}} \:\leqslant\mathrm{2}{mg} \\ $$$${Eliminating}\:{T}_{\mathrm{1}} \:{from}\:\left({i}\right)\:{and}\left({iii}\right) \\ $$$${T}_{\mathrm{1}} −{mg}=\mathrm{2}{mA}\:;\:\:{T}_{\mathrm{2}} ={T}_{\mathrm{1}} +\frac{{mA}}{\mathrm{2}}+{mg} \\ $$$${T}_{\mathrm{2}} ={mg}+\mathrm{2}{mA}+\frac{{mA}}{\mathrm{2}} \\ $$$${when}\:{block}\:{B}\:{is}\:{lifted}, \\ $$$${T}_{\mathrm{2}} =\mathrm{2}{mg}\:\Rightarrow\:\:\:\frac{\mathrm{5}{mA}}{\mathrm{2}}={mg} \\ $$$${A}=\frac{\mathrm{2}{g}}{\mathrm{5}}\:\:\:{or}\:\:\:\frac{\mathrm{40}}{\mathrm{11}}\left({t}−\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\mathrm{2}{g}}{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{t}=\frac{\mathrm{11}}{\mathrm{10}}+\frac{\mathrm{3}}{\mathrm{2}}\:=\mathrm{2}.\mathrm{6}{s} \\ $$$${s}_{{p}} \:=\:\int_{\mathrm{3}/\mathrm{2}} ^{\:\:\mathrm{2}.\mathrm{6}} {v}_{{p}} {dt} \\ $$$${v}_{{p}} =\:\int_{\mathrm{3}/\mathrm{2}} ^{\:\:{t}} {Adt}\:=\:\frac{\mathrm{40}}{\mathrm{11}}\int_{\mathrm{3}/\mathrm{2}} ^{\:\:{t}} \left({t}−\frac{\mathrm{3}}{\mathrm{2}}\right){dt} \\ $$$$\:\:\:\:\:=\frac{\mathrm{20}}{\mathrm{11}}\left({t}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{s}_{{p}} =\frac{\mathrm{20}}{\mathrm{11}}\int_{\mathrm{3}/\mathrm{2}} ^{\:\:\mathrm{2}.\mathrm{6}} \left({t}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} {dt} \\ $$$$\:\:\:\:=\frac{\mathrm{20}}{\mathrm{11}}×\frac{\mathrm{1}}{\mathrm{3}}×\left(\frac{\mathrm{11}}{\mathrm{10}}\right)^{\mathrm{3}} =\frac{\mathrm{121}}{\mathrm{150}}\:{m}\:\approx\:\mathrm{0}.\mathrm{8}{m} \\ $$$$\:\:\left({s}_{{p}} \:{by}\:{the}\:{time}\:{block}\:{B}\:{is}\:{lifted}\right). \\ $$$$\:\:{still}\:{the}\:{same}\:{answer}..! \\ $$
Commented by mrW1 last updated on 05/Nov/17
I checked again. The answer ((121)/(150)) is  correct.  Can you also find the displacement of  pulley at t=5 sec?
$${I}\:{checked}\:{again}.\:{The}\:{answer}\:\frac{\mathrm{121}}{\mathrm{150}}\:{is} \\ $$$${correct}. \\ $$$${Can}\:{you}\:{also}\:{find}\:{the}\:{displacement}\:{of} \\ $$$${pulley}\:{at}\:{t}=\mathrm{5}\:{sec}? \\ $$

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