Question Number 28171 by ajfour last updated on 21/Jan/18
Commented by ajfour last updated on 21/Jan/18
$${Q}.\mathrm{28113}\:\:{another}\:{solution}: \\ $$
Answered by ajfour last updated on 21/Jan/18
![TR=(((3MR^2 )/2))((a_2 /R)) ⇒ (T/a_2 )=((3M)/2) (mass inertia of disc) a_(cm) of ((3M)/2) and M =((Mg)/((5M/2)))=((2g)/5) force on M from centre of mass frame =Mg−((2Mg)/5)=((3Mg)/5) spring constant of section of spring attached to M is =k(((5M/2)/(3M/2))) =((5k)/3) under max. stretching of this section of spring : (1/2)(((5k)/3))e_2 ^2 =(((3Mg)/5))e_2 ⇒ e_2 =((18Mg)/(25k)) max stretching in full spring =(5/3)×((18Mg)/(25k)) =((6Mg)/(5k )) . for minimum friction coefficient between disc and platform so that disc doesnot slip, T_(max) −μ(Mg)=M[((T_(max) R)/((((3MR^2 )/2))))]R with T_(max) =((6Mg)/5) μMg= ((6Mg)/5)−M((2/(3M))×((6Mg)/5)) ⇒ μ=(6/5)−(4/5) =(2/5) .](https://www.tinkutara.com/question/Q28173.png)
$${TR}=\left(\frac{\mathrm{3}{MR}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\frac{{a}_{\mathrm{2}} }{{R}}\right) \\ $$$$\Rightarrow\:\frac{{T}}{{a}_{\mathrm{2}} }=\frac{\mathrm{3}{M}}{\mathrm{2}}\:\:\:\left({mass}\:{inertia}\:{of}\:{disc}\right) \\ $$$${a}_{{cm}} \:{of}\:\:\:\:\frac{\mathrm{3}{M}}{\mathrm{2}}\:{and}\:{M}\:=\frac{{Mg}}{\left(\mathrm{5}{M}/\mathrm{2}\right)}=\frac{\mathrm{2}{g}}{\mathrm{5}} \\ $$$${force}\:{on}\:{M}\:{from}\:{centre}\:{of}\:{mass} \\ $$$${frame}\:={Mg}−\frac{\mathrm{2}{Mg}}{\mathrm{5}}=\frac{\mathrm{3}{Mg}}{\mathrm{5}} \\ $$$${spring}\:{constant}\:{of}\:{section}\:{of} \\ $$$${spring}\:{attached}\:{to}\:{M}\:{is}\: \\ $$$$\:\:\:\:\:\:={k}\left(\frac{\mathrm{5}{M}/\mathrm{2}}{\mathrm{3}{M}/\mathrm{2}}\right)\:=\frac{\mathrm{5}{k}}{\mathrm{3}} \\ $$$${under}\:{max}.\:{stretching}\:{of}\:{this}\: \\ $$$${section}\:{of}\:{spring}\:: \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{5}{k}}{\mathrm{3}}\right){e}_{\mathrm{2}} ^{\mathrm{2}} =\left(\frac{\mathrm{3}{Mg}}{\mathrm{5}}\right){e}_{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\:{e}_{\mathrm{2}} =\frac{\mathrm{18}{Mg}}{\mathrm{25}{k}} \\ $$$${max}\:{stretching}\:{in}\:{full}\:{spring} \\ $$$$\:\:\:\:\:=\frac{\mathrm{5}}{\mathrm{3}}×\frac{\mathrm{18}{Mg}}{\mathrm{25}{k}}\:=\frac{\mathrm{6}{Mg}}{\mathrm{5}{k}\:}\:. \\ $$$${for}\:{minimum}\:{friction}\:{coefficient} \\ $$$${between}\:{disc}\:{and}\:{platform}\:{so}\:{that} \\ $$$${disc}\:{doesnot}\:{slip}, \\ $$$$\:\:\:\:\:\:\:{T}_{{max}} −\mu\left({Mg}\right)={M}\left[\frac{{T}_{{max}} {R}}{\left(\frac{\mathrm{3}{MR}^{\mathrm{2}} }{\mathrm{2}}\right)}\right]{R} \\ $$$${with}\:\:{T}_{{max}} =\frac{\mathrm{6}{Mg}}{\mathrm{5}} \\ $$$$\:\mu{Mg}=\:\frac{\mathrm{6}{Mg}}{\mathrm{5}}−{M}\left(\frac{\mathrm{2}}{\mathrm{3}{M}}×\frac{\mathrm{6}{Mg}}{\mathrm{5}}\right) \\ $$$$\Rightarrow\:\:\mu=\frac{\mathrm{6}}{\mathrm{5}}−\frac{\mathrm{4}}{\mathrm{5}}\:=\frac{\mathrm{2}}{\mathrm{5}}\:. \\ $$
Commented by ajfour last updated on 21/Jan/18
Thank you Sir.
Commented by mrW2 last updated on 21/Jan/18
$${very}\:{tricky}! \\ $$