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Question-28113




Question Number 28113 by ajfour last updated on 20/Jan/18
Commented by ajfour last updated on 20/Jan/18
If system of disc, spring, and  block of mass m, be released from  rest with spring in natural length,  (friction being sufficent to  prevent slipping) find maximum  deformation in spring.  Also find the minimum coefficient  of friction required to prevent  slipping of disc on the fixed  platform.
$${If}\:{system}\:{of}\:{disc},\:{spring},\:{and} \\ $$$${block}\:{of}\:{mass}\:\boldsymbol{{m}},\:{be}\:{released}\:{from} \\ $$$${rest}\:{with}\:{spring}\:{in}\:{natural}\:{length}, \\ $$$$\left({friction}\:{being}\:{sufficent}\:{to}\right. \\ $$$$\left.{prevent}\:{slipping}\right)\:{find}\:{maximum} \\ $$$${deformation}\:{in}\:{spring}. \\ $$$${Also}\:{find}\:{the}\:{minimum}\:{coefficient} \\ $$$${of}\:{friction}\:{required}\:{to}\:{prevent} \\ $$$${slipping}\:{of}\:{disc}\:{on}\:{the}\:{fixed} \\ $$$${platform}. \\ $$
Answered by mrW2 last updated on 21/Jan/18
x_1 =displacement of block  a_1 =acceleration of block=(d^2 x_1 /dt^2 )  x_2 =displacement of disc  a_2 =acceleration of disc=(d^2 x_2 /dt^2 )  T=force in spring  e=deformation in spring=x_1 −x_2     Spring:  T=ke=k(x_1 −x_2 )  ⇒(d^2 T/dt^2 )=k(a_1 −a_2 )    Block:  mg−T=ma_1   ⇒a_1 =g−(T/m)    Disc:  TR=I_2 α_2 =(((mR^2 )/2)+mR^2 )×(a_2 /R)=((3mR)/2)×a_2   ⇒a_2 =((2T)/(3m))    ⇒(d^2 T/dt^2 )=k(a_1 −a_2 )=k(g−(T/m)−((2T)/(3m)))=(k/m)(mg−((5T)/3))  ⇒(d^2 T/dt^2 )=(k/m)(mg−((5T)/3))  let S=mg−((5T)/3)  (d^2 S/dt^2 )=−(5/3)×(d^2 T/dt^2 )  ⇒−(3/5)(d^2 S/dt^2 )=(k/m)S  ⇒(d^2 S/dt^2 )+((5k)/(3m))S=0  ⇒S=c_1  sin ωt+c_2  cos ωt  with ω=(√((5k)/(3m)))  ⇒T=(3/5)(mg−S)=(3/5)(mg−c_1  sin ωt−c_2  cos ωt)  ⇒(dT/dt)=(3/5)(−c_1  cos ωt+c_2  sin ωt)ω  at t=0: T=0 and (dT/dt)=k(v_1 −v_2 )=0  mg−c_2 =0⇒c_2 =mg  ⇒c_1 =0  ⇒T=((3mg)/5)(1− cos ωt)  ⇒T_(max) =((6mg)/5) at ωt=(2n−1)π or t=(2n−1)π(√((3m)/(5k)))  ⇒e_(max) =((6mg)/(5k))    T−f=ma_2   f=T−ma_2 =T−((2T)/3)=(T/3)≤μmg  ⇒μ≥(T_(max) /(3mg))=((6mg)/(3×5mg))=(2/5)=0.4    a_1 =g−((3g)/5)(1−cos ωt)=(g/5)(2+3cos ωt)  v_1 =(g/5)(2t+(3/ω)sin ωt)  ⇒x_1 =(g/5)[t^2 +(3/ω^2 )(1−cos ωt)]    a_2 =((2g)/5)(1−cos ωt)  ⇒v_2 =((2g)/5)(t−(1/ω)sin ωt)  ⇒x_2 =(g/5)[t^2 −(2/ω^2 )(1−cos ωt)]    we see the system is in vibration. the  force in spring varies between 0 and  ((6mg)/5), about the mean value ((3mg)/5).  after the vibration is vanished, the  force in spring remains constant,  the spring acts as normal string,  block and disc have the same  acceleration:  a_1 =a_2   g−(T/m)=((2T)/(3m))  ⇒T=((3mg)/5)=mean value of vibration  ⇒e=((3mg)/5)
$${x}_{\mathrm{1}} ={displacement}\:{of}\:{block} \\ $$$${a}_{\mathrm{1}} ={acceleration}\:{of}\:{block}=\frac{{d}^{\mathrm{2}} {x}_{\mathrm{1}} }{{dt}^{\mathrm{2}} } \\ $$$${x}_{\mathrm{2}} ={displacement}\:{of}\:{disc} \\ $$$${a}_{\mathrm{2}} ={acceleration}\:{of}\:{disc}=\frac{{d}^{\mathrm{2}} {x}_{\mathrm{2}} }{{dt}^{\mathrm{2}} } \\ $$$${T}={force}\:{in}\:{spring} \\ $$$${e}={deformation}\:{in}\:{spring}={x}_{\mathrm{1}} −{x}_{\mathrm{2}} \\ $$$$ \\ $$$${Spring}: \\ $$$${T}={ke}={k}\left({x}_{\mathrm{1}} −{x}_{\mathrm{2}} \right) \\ $$$$\Rightarrow\frac{{d}^{\mathrm{2}} {T}}{{dt}^{\mathrm{2}} }={k}\left({a}_{\mathrm{1}} −{a}_{\mathrm{2}} \right) \\ $$$$ \\ $$$${Block}: \\ $$$${mg}−{T}={ma}_{\mathrm{1}} \\ $$$$\Rightarrow{a}_{\mathrm{1}} ={g}−\frac{{T}}{{m}} \\ $$$$ \\ $$$${Disc}: \\ $$$${TR}={I}_{\mathrm{2}} \alpha_{\mathrm{2}} =\left(\frac{{mR}^{\mathrm{2}} }{\mathrm{2}}+{mR}^{\mathrm{2}} \right)×\frac{{a}_{\mathrm{2}} }{{R}}=\frac{\mathrm{3}{mR}}{\mathrm{2}}×{a}_{\mathrm{2}} \\ $$$$\Rightarrow{a}_{\mathrm{2}} =\frac{\mathrm{2}{T}}{\mathrm{3}{m}} \\ $$$$ \\ $$$$\Rightarrow\frac{{d}^{\mathrm{2}} {T}}{{dt}^{\mathrm{2}} }={k}\left({a}_{\mathrm{1}} −{a}_{\mathrm{2}} \right)={k}\left({g}−\frac{{T}}{{m}}−\frac{\mathrm{2}{T}}{\mathrm{3}{m}}\right)=\frac{{k}}{{m}}\left({mg}−\frac{\mathrm{5}{T}}{\mathrm{3}}\right) \\ $$$$\Rightarrow\frac{{d}^{\mathrm{2}} {T}}{{dt}^{\mathrm{2}} }=\frac{{k}}{{m}}\left({mg}−\frac{\mathrm{5}{T}}{\mathrm{3}}\right) \\ $$$${let}\:{S}={mg}−\frac{\mathrm{5}{T}}{\mathrm{3}} \\ $$$$\frac{{d}^{\mathrm{2}} {S}}{{dt}^{\mathrm{2}} }=−\frac{\mathrm{5}}{\mathrm{3}}×\frac{{d}^{\mathrm{2}} {T}}{{dt}^{\mathrm{2}} } \\ $$$$\Rightarrow−\frac{\mathrm{3}}{\mathrm{5}}\frac{{d}^{\mathrm{2}} {S}}{{dt}^{\mathrm{2}} }=\frac{{k}}{{m}}{S} \\ $$$$\Rightarrow\frac{{d}^{\mathrm{2}} {S}}{{dt}^{\mathrm{2}} }+\frac{\mathrm{5}{k}}{\mathrm{3}{m}}{S}=\mathrm{0} \\ $$$$\Rightarrow{S}={c}_{\mathrm{1}} \:\mathrm{sin}\:\omega{t}+{c}_{\mathrm{2}} \:\mathrm{cos}\:\omega{t} \\ $$$${with}\:\omega=\sqrt{\frac{\mathrm{5}{k}}{\mathrm{3}{m}}} \\ $$$$\Rightarrow{T}=\frac{\mathrm{3}}{\mathrm{5}}\left({mg}−{S}\right)=\frac{\mathrm{3}}{\mathrm{5}}\left({mg}−{c}_{\mathrm{1}} \:\mathrm{sin}\:\omega{t}−{c}_{\mathrm{2}} \:\mathrm{cos}\:\omega{t}\right) \\ $$$$\Rightarrow\frac{{dT}}{{dt}}=\frac{\mathrm{3}}{\mathrm{5}}\left(−{c}_{\mathrm{1}} \:\mathrm{cos}\:\omega{t}+{c}_{\mathrm{2}} \:\mathrm{sin}\:\omega{t}\right)\omega \\ $$$${at}\:{t}=\mathrm{0}:\:{T}=\mathrm{0}\:{and}\:\frac{{dT}}{{dt}}={k}\left({v}_{\mathrm{1}} −{v}_{\mathrm{2}} \right)=\mathrm{0} \\ $$$${mg}−{c}_{\mathrm{2}} =\mathrm{0}\Rightarrow{c}_{\mathrm{2}} ={mg} \\ $$$$\Rightarrow{c}_{\mathrm{1}} =\mathrm{0} \\ $$$$\Rightarrow{T}=\frac{\mathrm{3}{mg}}{\mathrm{5}}\left(\mathrm{1}−\:\mathrm{cos}\:\omega{t}\right) \\ $$$$\Rightarrow{T}_{{max}} =\frac{\mathrm{6}{mg}}{\mathrm{5}}\:{at}\:\omega{t}=\left(\mathrm{2}{n}−\mathrm{1}\right)\pi\:{or}\:{t}=\left(\mathrm{2}{n}−\mathrm{1}\right)\pi\sqrt{\frac{\mathrm{3}{m}}{\mathrm{5}{k}}} \\ $$$$\Rightarrow{e}_{{max}} =\frac{\mathrm{6}{mg}}{\mathrm{5}{k}} \\ $$$$ \\ $$$${T}−{f}={ma}_{\mathrm{2}} \\ $$$${f}={T}−{ma}_{\mathrm{2}} ={T}−\frac{\mathrm{2}{T}}{\mathrm{3}}=\frac{{T}}{\mathrm{3}}\leqslant\mu{mg} \\ $$$$\Rightarrow\mu\geqslant\frac{{T}_{{max}} }{\mathrm{3}{mg}}=\frac{\mathrm{6}{mg}}{\mathrm{3}×\mathrm{5}{mg}}=\frac{\mathrm{2}}{\mathrm{5}}=\mathrm{0}.\mathrm{4} \\ $$$$ \\ $$$${a}_{\mathrm{1}} ={g}−\frac{\mathrm{3}{g}}{\mathrm{5}}\left(\mathrm{1}−\mathrm{cos}\:\omega{t}\right)=\frac{{g}}{\mathrm{5}}\left(\mathrm{2}+\mathrm{3cos}\:\omega{t}\right) \\ $$$${v}_{\mathrm{1}} =\frac{{g}}{\mathrm{5}}\left(\mathrm{2}{t}+\frac{\mathrm{3}}{\omega}\mathrm{sin}\:\omega{t}\right) \\ $$$$\Rightarrow{x}_{\mathrm{1}} =\frac{{g}}{\mathrm{5}}\left[{t}^{\mathrm{2}} +\frac{\mathrm{3}}{\omega^{\mathrm{2}} }\left(\mathrm{1}−\mathrm{cos}\:\omega{t}\right)\right] \\ $$$$ \\ $$$${a}_{\mathrm{2}} =\frac{\mathrm{2}{g}}{\mathrm{5}}\left(\mathrm{1}−\mathrm{cos}\:\omega{t}\right) \\ $$$$\Rightarrow{v}_{\mathrm{2}} =\frac{\mathrm{2}{g}}{\mathrm{5}}\left({t}−\frac{\mathrm{1}}{\omega}\mathrm{sin}\:\omega{t}\right) \\ $$$$\Rightarrow{x}_{\mathrm{2}} =\frac{{g}}{\mathrm{5}}\left[{t}^{\mathrm{2}} −\frac{\mathrm{2}}{\omega^{\mathrm{2}} }\left(\mathrm{1}−\mathrm{cos}\:\omega{t}\right)\right] \\ $$$$ \\ $$$${we}\:{see}\:{the}\:{system}\:{is}\:{in}\:{vibration}.\:{the} \\ $$$${force}\:{in}\:{spring}\:{varies}\:{between}\:\mathrm{0}\:{and} \\ $$$$\frac{\mathrm{6}{mg}}{\mathrm{5}},\:{about}\:{the}\:{mean}\:{value}\:\frac{\mathrm{3}{mg}}{\mathrm{5}}. \\ $$$${after}\:{the}\:{vibration}\:{is}\:{vanished},\:{the} \\ $$$${force}\:{in}\:{spring}\:{remains}\:{constant}, \\ $$$${the}\:{spring}\:{acts}\:{as}\:{normal}\:{string}, \\ $$$${block}\:{and}\:{disc}\:{have}\:{the}\:{same} \\ $$$${acceleration}: \\ $$$${a}_{\mathrm{1}} ={a}_{\mathrm{2}} \\ $$$${g}−\frac{{T}}{{m}}=\frac{\mathrm{2}{T}}{\mathrm{3}{m}} \\ $$$$\Rightarrow{T}=\frac{\mathrm{3}{mg}}{\mathrm{5}}={mean}\:{value}\:{of}\:{vibration} \\ $$$$\Rightarrow{e}=\frac{\mathrm{3}{mg}}{\mathrm{5}} \\ $$
Commented by ajfour last updated on 21/Jan/18
Thank you sir.  ENTIRELY  Beautiful Sir !
$${Thank}\:{you}\:{sir}. \\ $$$$\mathcal{ENTIRELY}\:\:\mathcal{B}{eautiful}\:\mathcal{S}{ir}\:! \\ $$

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