Question Number 28339 by Tinkutara last updated on 24/Jan/18
Answered by mrW2 last updated on 24/Jan/18
$${when}\:{the}\:{ring}\:{is}\:{released},\:{it}\:{falls}\:{down} \\ $$$${with}\:{the}\:{max}.\:{acceleration}\:\left({g}\right).\:{due}\:{to} \\ $$$${the}\:{increasing}\:{vertical}\:{component}\:{of} \\ $$$${the}\:{force}\:{in}\:{the}\:{string},\:{the}\:{acceleration} \\ $$$${of}\:{the}\:{ring}\:{decreases}.\: \\ $$$${let}\:\theta={angle}\:{between}\:{string}\:{and}\:{horizont} \\ $$$${let}\:{a}={acceleration}\:{of}\:{ring} \\ $$$${m}_{\mathrm{1}} {g}−{m}_{\mathrm{2}} {g}\:\mathrm{sin}\:\theta={m}_{\mathrm{1}} {a} \\ $$$$\Rightarrow{a}={g}\left(\mathrm{1}−\frac{{m}_{\mathrm{2}} }{{m}_{\mathrm{1}} }\:\mathrm{sin}\:\theta\right) \\ $$$${for}\:{a}=\mathrm{0}, \\ $$$$\mathrm{1}=\frac{{m}_{\mathrm{2}} }{{m}_{\mathrm{1}} }\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{{m}_{\mathrm{1}} }{{m}_{\mathrm{2}} } \\ $$$${that}\:{means}:\:{if}\:{m}_{\mathrm{1}} <{m}_{\mathrm{2}} ,\:{there}\:{is}\:{a} \\ $$$${position}\:\theta_{\mathrm{1}} =\mathrm{sin}^{−\mathrm{1}} \frac{{m}_{\mathrm{1}} }{{m}_{\mathrm{2}} }\:{where}\:{the}\:{ring}'{s} \\ $$$${acceleration}\:{is}\:{zero}\:{and}\:{its}\:{speed} \\ $$$${reaches}\:{the}\:{maximum}.\:{but}\:{if}\:{m}_{\mathrm{1}} \:\geqslant{m}_{\mathrm{2}} , \\ $$$${the}\:{ring}'{s}\:{acceleration}\:{remains} \\ $$$${always}\:>\mathrm{0}\:{and}\:{it}\:{falls}\:{down}\: \\ $$$${continuously}. \\ $$$${Therefore}\:{we}\:{have}\:{following}\:{cases}: \\ $$$${By}\:\left({A}\right)\:{and}\:\left({B}\right),\:{since}\:{m}_{\mathrm{1}} \geqslant{m}_{\mathrm{2}} ,\:{the}\:{ring} \\ $$$${falls}\:{continuously}\:{as}\:{long}\:{as}\:{the} \\ $$$${length}\:{of}\:{string}\:{allows}. \\ $$$${By}\:\left({C}\right)\:{with}\:{m}_{\mathrm{1}} =\frac{{m}_{\mathrm{2}} }{\mathrm{2}},\:{we}\:{get}\:\theta_{\mathrm{1}} =\mathrm{30}°. \\ $$$${The}\:{motion}\:{of}\:{the}\:{ring}\:{is}\:{like}\:{this}: \\ $$$${Released}\:{at}\:\theta=\mathrm{0},\:{the}\:{ring}\:{falls}\:{with} \\ $$$${a}_{{max}} ={g},\:{at}\:\theta=\mathrm{30}°\:{its}\:{a}=\mathrm{0}\:{and}\:{reaches} \\ $$$${v}_{{max}} ,\:{then}\:{it}\:{obtains}\:{a}\:{negative}\: \\ $$$${acceleration}\:{and}\:{its}\:{speed}\:{gets}\:{smaller} \\ $$$${till}\:{it}\:{reaches}\:{the}\:{lowest}\:{point}.\:{then} \\ $$$${the}\:{ring}\:{begins}\:{to}\:{move}\:{upwards}\:{back} \\ $$$${to}\:{the}\:{original}\:{position}\:\theta=\mathrm{0}.\:{this} \\ $$$${cycle}\:{will}\:{repeat}\:{and}\:{repeat}. \\ $$$${it}\:{means}\:{if}\:{m}_{\mathrm{1}} <{m}_{\mathrm{2}} \:{the}\:{ring}\:{will} \\ $$$${make}\:{an}\:{oscillation}\:{motion}. \\ $$
Commented by Tinkutara last updated on 25/Jan/18
Is it 30° or 60° for horizontal?
Commented by ajfour last updated on 24/Jan/18
$${thank}\:{you}\:{Sir}\:. \\ $$
Commented by Tinkutara last updated on 25/Jan/18
Thank you very much Sir! I got the answer.
Commented by mrW2 last updated on 25/Jan/18
$$\mathrm{30}° \\ $$$${the}\:{answer}\:{in}\:{your}\:{book}\:{is}\:\mathrm{60}°,\:{but}\:{that}'{s} \\ $$$${wrong}. \\ $$