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Question-28339




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 (g). due to  the increasing vertical component of  the force in the string, the acceleration  of the ring decreases.   let θ=angle between string and horizont  let a=acceleration of ring  m_1 g−m_2 g sin θ=m_1 a  ⇒a=g(1−(m_2 /m_1 ) sin θ)  for a=0,  1=(m_2 /m_1 ) sin θ  ⇒sin θ=(m_1 /m_2 )  that means: if m_1 <m_2 , there is a  position θ_1 =sin^(−1) (m_1 /m_2 ) where the ring′s  acceleration is zero and its speed  reaches the maximum. but if m_1  ≥m_2 ,  the ring′s acceleration remains  always >0 and it falls down   continuously.  Therefore we have following cases:  By (A) and (B), since m_1 ≥m_2 , the ring  falls continuously as long as the  length of string allows.  By (C) with m_1 =(m_2 /2), we get θ_1 =30°.  The motion of the ring is like this:  Released at θ=0, the ring falls with  a_(max) =g, at θ=30° its a=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 θ=0. this  cycle will repeat and repeat.  it means if m_1 <m_2  the ring will  make an oscillation motion.
$${when}\:{the}\:{ring}\:{is}\:{released},\:{it}\:{falls}\:{down} \\ $$$${with}\:{the}\:{max}.\:{acceleration}\:\left({g}\right).\:{due}\:{to} \\ $$$${the}\:{increasing}\:{vertical}\:{component}\:{of} \\ $$$${the}\:{force}\:{in}\:{the}\:{string},\:{the}\:{acceleration} \\ $$$${of}\:{the}\:{ring}\:{decreases}.\: \\ $$$${let}\:\theta={angle}\:{between}\:{string}\:{and}\:{horizont} \\ $$$${let}\:{a}={acceleration}\:{of}\:{ring} \\ $$$${m}_{\mathrm{1}} {g}−{m}_{\mathrm{2}} {g}\:\mathrm{sin}\:\theta={m}_{\mathrm{1}} {a} \\ $$$$\Rightarrow{a}={g}\left(\mathrm{1}−\frac{{m}_{\mathrm{2}} }{{m}_{\mathrm{1}} }\:\mathrm{sin}\:\theta\right) \\ $$$${for}\:{a}=\mathrm{0}, \\ $$$$\mathrm{1}=\frac{{m}_{\mathrm{2}} }{{m}_{\mathrm{1}} }\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{{m}_{\mathrm{1}} }{{m}_{\mathrm{2}} } \\ $$$${that}\:{means}:\:{if}\:{m}_{\mathrm{1}} <{m}_{\mathrm{2}} ,\:{there}\:{is}\:{a} \\ $$$${position}\:\theta_{\mathrm{1}} =\mathrm{sin}^{−\mathrm{1}} \frac{{m}_{\mathrm{1}} }{{m}_{\mathrm{2}} }\:{where}\:{the}\:{ring}'{s} \\ $$$${acceleration}\:{is}\:{zero}\:{and}\:{its}\:{speed} \\ $$$${reaches}\:{the}\:{maximum}.\:{but}\:{if}\:{m}_{\mathrm{1}} \:\geqslant{m}_{\mathrm{2}} , \\ $$$${the}\:{ring}'{s}\:{acceleration}\:{remains} \\ $$$${always}\:>\mathrm{0}\:{and}\:{it}\:{falls}\:{down}\: \\ $$$${continuously}. \\ $$$${Therefore}\:{we}\:{have}\:{following}\:{cases}: \\ $$$${By}\:\left({A}\right)\:{and}\:\left({B}\right),\:{since}\:{m}_{\mathrm{1}} \geqslant{m}_{\mathrm{2}} ,\:{the}\:{ring} \\ $$$${falls}\:{continuously}\:{as}\:{long}\:{as}\:{the} \\ $$$${length}\:{of}\:{string}\:{allows}. \\ $$$${By}\:\left({C}\right)\:{with}\:{m}_{\mathrm{1}} =\frac{{m}_{\mathrm{2}} }{\mathrm{2}},\:{we}\:{get}\:\theta_{\mathrm{1}} =\mathrm{30}°. \\ $$$${The}\:{motion}\:{of}\:{the}\:{ring}\:{is}\:{like}\:{this}: \\ $$$${Released}\:{at}\:\theta=\mathrm{0},\:{the}\:{ring}\:{falls}\:{with} \\ $$$${a}_{{max}} ={g},\:{at}\:\theta=\mathrm{30}°\:{its}\:{a}=\mathrm{0}\:{and}\:{reaches} \\ $$$${v}_{{max}} ,\:{then}\:{it}\:{obtains}\:{a}\:{negative}\: \\ $$$${acceleration}\:{and}\:{its}\:{speed}\:{gets}\:{smaller} \\ $$$${till}\:{it}\:{reaches}\:{the}\:{lowest}\:{point}.\:{then} \\ $$$${the}\:{ring}\:{begins}\:{to}\:{move}\:{upwards}\:{back} \\ $$$${to}\:{the}\:{original}\:{position}\:\theta=\mathrm{0}.\:{this} \\ $$$${cycle}\:{will}\:{repeat}\:{and}\:{repeat}. \\ $$$${it}\:{means}\:{if}\:{m}_{\mathrm{1}} <{m}_{\mathrm{2}} \:{the}\:{ring}\:{will} \\ $$$${make}\:{an}\:{oscillation}\:{motion}. \\ $$
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 .
$${thank}\:{you}\:{Sir}\:. \\ $$
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
30°  the answer in your book is 60°, but that′s  wrong.
$$\mathrm{30}° \\ $$$${the}\:{answer}\:{in}\:{your}\:{book}\:{is}\:\mathrm{60}°,\:{but}\:{that}'{s} \\ $$$${wrong}. \\ $$

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