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