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Question-217252

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Question Number 217252 by mr W last updated on 07/Mar/25
Commented by mr W last updated on 07/Mar/25
As Q217238, but the other end of the string is connected with a block with mass m which can slide on the table frictionlessly. find the accelerations of the objects and the tension in string.
$${As}\:{Q}\mathrm{217238},\:{but}\:{the}\:{other}\:{end}\:{of} \\ $$$${the}\:{string}\:{is}\:{connected}\:{with}\:{a}\:{block} \\ $$$${with}\:{mass}\:{m}\:{which}\:{can}\:{slide}\:{on}\:{the} \\ $$$${table}\:{frictionlessly}. \\ $$$${find}\:{the}\:{accelerations}\:{of}\:{the}\:{objects} \\ $$$${and}\:{the}\:{tension}\:{in}\:{string}. \\ $$
Commented by Tawa11 last updated on 07/Mar/25
a = ((Mg)/(M + m)) T = ((Mmg)/(M + m)) Sir.
$$\mathrm{a}\:=\:\frac{\mathrm{Mg}}{\mathrm{M}\:+\:\mathrm{m}} \\ $$$$\mathrm{T}\:=\:\frac{\mathrm{Mmg}}{\mathrm{M}\:+\:\mathrm{m}} \\ $$$$\mathrm{Sir}. \\ $$
Commented by mr W last updated on 07/Mar/25
how?
$${how}? \\ $$
Commented by Tawa11 last updated on 07/Mar/25
Is it correct sir?
$$\mathrm{Is}\:\mathrm{it}\:\mathrm{correct}\:\mathrm{sir}? \\ $$
Commented by mr W last updated on 07/Mar/25
no!
$${no}! \\ $$
Answered by mahdipoor last updated on 07/Mar/25
T=mA=mx^(..) { ((F=ma^− ⇒ −T+Mg=Ma=My^(..) )),((ΣM_(CM) =I^− α ⇒TR=Iθ^(..) )) :} y−θR=x ⇒ a−θ^(..) R=A ⇒⇒ T=m(a−θ^(..) R)=m((g−(T/M))−(((TR)/((MR^2 )/2)))R) ⇒T=((Mmg)/(M+3m))=cte ⇒A=(T/m) ⇒a=g−(T/M)=g−(m/M)A
$${T}={mA}={m}\overset{..} {{x}} \\ $$$$\begin{cases}{{F}={m}\overset{−} {{a}}\:\Rightarrow\:−{T}+{Mg}={Ma}={M}\overset{..} {{y}}}\\{\Sigma{M}_{{CM}} =\overset{−} {{I}}\alpha\:\Rightarrow{TR}={I}\overset{..} {\theta}}\end{cases} \\ $$$${y}−\theta{R}={x}\:\Rightarrow\:{a}−\overset{..} {\theta}{R}={A}\: \\ $$$$\Rightarrow\Rightarrow \\ $$$${T}={m}\left({a}−\overset{..} {\theta}{R}\right)={m}\left(\left({g}−\frac{{T}}{{M}}\right)−\left(\frac{{TR}}{\frac{{MR}^{\mathrm{2}} }{\mathrm{2}}}\right){R}\right) \\ $$$$\Rightarrow{T}=\frac{{Mmg}}{{M}+\mathrm{3}{m}}={cte} \\ $$$$\Rightarrow{A}=\frac{{T}}{{m}} \\ $$$$\Rightarrow{a}={g}−\frac{{T}}{{M}}={g}−\frac{{m}}{{M}}{A} \\ $$
Commented by mr W last updated on 07/Mar/25
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Commented by Tawa11 last updated on 08/Mar/25
Thanks sir.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$
Answered by aleks041103 last updated on 07/Mar/25
mA=T Ma=Mg−T ((1/2)MR^2 )ε=TR⇒T=(1/2)MεR εR=a−A ⇒T=(1/2)M(a−A) mA=(1/2)M(a−A)⇒Ma=(2m+M)A⇒a=(1+((2m)/M))A Ma+mA=Mg ⇒(3m+M)A=Mg ⇒A = (M/(3m+M)) g and a = ((2m+M)/(3m+M)) g
$${mA}={T} \\ $$$${Ma}={Mg}−{T} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}{MR}^{\mathrm{2}} \right)\varepsilon={TR}\Rightarrow{T}=\frac{\mathrm{1}}{\mathrm{2}}{M}\varepsilon{R} \\ $$$$\varepsilon{R}={a}−{A} \\ $$$$\Rightarrow{T}=\frac{\mathrm{1}}{\mathrm{2}}{M}\left({a}−{A}\right) \\ $$$$ \\ $$$${mA}=\frac{\mathrm{1}}{\mathrm{2}}{M}\left({a}−{A}\right)\Rightarrow{Ma}=\left(\mathrm{2}{m}+{M}\right){A}\Rightarrow{a}=\left(\mathrm{1}+\frac{\mathrm{2}{m}}{{M}}\right){A} \\ $$$${Ma}+{mA}={Mg} \\ $$$$\Rightarrow\left(\mathrm{3}{m}+{M}\right){A}={Mg} \\ $$$$\Rightarrow{A}\:=\:\frac{{M}}{\mathrm{3}{m}+{M}}\:{g}\:\:\:{and}\:\:\:\:{a}\:=\:\frac{\mathrm{2}{m}+{M}}{\mathrm{3}{m}+{M}}\:{g} \\ $$
Commented by mr W last updated on 07/Mar/25
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Commented by Tawa11 last updated on 08/Mar/25
Thanks sir
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 08/Mar/25
α=angular acceleration of cylinder ( ↶) a=acceleration of cylinder (↓) A=acceleration of block on table (←) we have: a=αR+A I=((MR^2 )/2) αI=TR ((αMR^2 )/2)=TR ⇒T=((αMR)/2) Ma=Mg−T M(αR+A)=Mg−T ⇒T=M(g−αR−A) T=mA ((αMR)/2)=M(g−αR−A) ⇒A=g−((3αR)/2) mA=((αMR)/2) m(g−((3αR)/2))=((αMR)/2) ⇒α=((2g)/(R(3+(M/m)))) ⇒A=g−((3R)/2)×((2g)/(R(3+(M/m))))=(g/(1+((3m)/M))) ⇒a=((2gR)/(R(3+(M/m))))+g(1−(3/(3+(M/m))))=(((1+((2m)/M))g)/(1+((3m)/M))) ⇒T=mg(1−(3/(3+(M/m))))=((Mg)/(3+(M/m))) if m→0: (→free fall) α→0 a→g, A→g T→0 if m→∞: (→ case in Q217238) α→((2g)/(3R)) a→((2g)/3), A→0 T→((Mg)/3)
$$\alpha={angular}\:{acceleration}\:{of}\:{cylinder}\:\left(\:\curvearrowleft\right) \\ $$$${a}={acceleration}\:{of}\:{cylinder}\:\:\left(\downarrow\right) \\ $$$${A}={acceleration}\:{of}\:{block}\:{on}\:{table}\:\:\left(\leftarrow\right) \\ $$$${we}\:{have}: \\ $$$${a}=\alpha{R}+{A} \\ $$$${I}=\frac{{MR}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\alpha{I}={TR} \\ $$$$\frac{\alpha{MR}^{\mathrm{2}} }{\mathrm{2}}={TR} \\ $$$$\Rightarrow{T}=\frac{\alpha{MR}}{\mathrm{2}} \\ $$$${Ma}={Mg}−{T} \\ $$$${M}\left(\alpha{R}+{A}\right)={Mg}−{T} \\ $$$$\Rightarrow{T}={M}\left({g}−\alpha{R}−{A}\right) \\ $$$${T}={mA} \\ $$$$\frac{\alpha{MR}}{\mathrm{2}}={M}\left({g}−\alpha{R}−{A}\right) \\ $$$$\Rightarrow{A}={g}−\frac{\mathrm{3}\alpha{R}}{\mathrm{2}} \\ $$$${mA}=\frac{\alpha{MR}}{\mathrm{2}} \\ $$$${m}\left({g}−\frac{\mathrm{3}\alpha{R}}{\mathrm{2}}\right)=\frac{\alpha{MR}}{\mathrm{2}} \\ $$$$\Rightarrow\alpha=\frac{\mathrm{2}{g}}{{R}\left(\mathrm{3}+\frac{{M}}{{m}}\right)} \\ $$$$\Rightarrow{A}={g}−\frac{\mathrm{3}{R}}{\mathrm{2}}×\frac{\mathrm{2}{g}}{{R}\left(\mathrm{3}+\frac{{M}}{{m}}\right)}=\frac{{g}}{\mathrm{1}+\frac{\mathrm{3}{m}}{{M}}} \\ $$$$\Rightarrow{a}=\frac{\mathrm{2}{gR}}{{R}\left(\mathrm{3}+\frac{{M}}{{m}}\right)}+{g}\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{3}+\frac{{M}}{{m}}}\right)=\frac{\left(\mathrm{1}+\frac{\mathrm{2}{m}}{{M}}\right){g}}{\mathrm{1}+\frac{\mathrm{3}{m}}{{M}}} \\ $$$$\Rightarrow{T}={mg}\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{3}+\frac{{M}}{{m}}}\right)=\frac{{Mg}}{\mathrm{3}+\frac{{M}}{{m}}} \\ $$$$ \\ $$$${if}\:{m}\rightarrow\mathrm{0}:\:\:\left(\rightarrow{free}\:{fall}\right) \\ $$$$\alpha\rightarrow\mathrm{0} \\ $$$${a}\rightarrow{g},\:{A}\rightarrow{g} \\ $$$${T}\rightarrow\mathrm{0} \\ $$$$ \\ $$$${if}\:{m}\rightarrow\infty:\:\:\left(\rightarrow\:{case}\:{in}\:{Q}\mathrm{217238}\right) \\ $$$$\alpha\rightarrow\frac{\mathrm{2}{g}}{\mathrm{3}{R}} \\ $$$${a}\rightarrow\frac{\mathrm{2}{g}}{\mathrm{3}},\:{A}\rightarrow\mathrm{0} \\ $$$${T}\rightarrow\frac{{Mg}}{\mathrm{3}} \\ $$
Commented by Tawa11 last updated on 08/Mar/25
Thanks sir.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$

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