Question Number 19223 by Tinkutara last updated on 07/Aug/17
$$\mathrm{A}\:\mathrm{particle}\:{P}\:\mathrm{is}\:\mathrm{sliding}\:\mathrm{down}\:\mathrm{a}\:\mathrm{frictionless} \\ $$$$\mathrm{hemispherical}\:\mathrm{bowl}.\:\mathrm{It}\:\mathrm{passes}\:\mathrm{the}\:\mathrm{point} \\ $$$${A}\:\mathrm{at}\:{t}\:=\:\mathrm{0}.\:\mathrm{At}\:\mathrm{this}\:\mathrm{instant}\:\mathrm{of}\:\mathrm{time},\:\mathrm{the} \\ $$$$\mathrm{horizontal}\:\mathrm{component}\:\mathrm{of}\:\mathrm{its}\:\mathrm{velocity}\:\mathrm{is} \\ $$$${v}.\:\mathrm{A}\:\mathrm{bead}\:{Q}\:\mathrm{of}\:\mathrm{the}\:\mathrm{same}\:\mathrm{mass}\:\mathrm{as}\:{P}\:\mathrm{is} \\ $$$$\mathrm{ejected}\:\mathrm{from}\:{A}\:\mathrm{at}\:{t}\:=\:\mathrm{0}\:\mathrm{along}\:\mathrm{the} \\ $$$$\mathrm{horizontal}\:\mathrm{direction},\:\mathrm{with}\:\mathrm{the}\:\mathrm{speed}\:{v}. \\ $$$$\mathrm{Friction}\:\mathrm{between}\:\mathrm{the}\:\mathrm{bead}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{string}\:\mathrm{may}\:\mathrm{be}\:\mathrm{neglected}.\:\mathrm{Let}\:{t}_{{P}} \:\mathrm{and}\:{t}_{{Q}} \\ $$$$\mathrm{be}\:\mathrm{the}\:\mathrm{respective}\:\mathrm{times}\:\mathrm{taken}\:\mathrm{by}\:{P}\:\mathrm{and} \\ $$$${Q}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{the}\:\mathrm{point}\:{B}.\:\mathrm{Then} \\ $$$$\left({a}\right)\:{t}_{{P}} \:<\:{t}_{{Q}} \\ $$$$\left({b}\right)\:{t}_{{P}} \:=\:{t}_{{Q}} \\ $$$$\left({c}\right)\:{t}_{{P}} \:>\:{t}_{{Q}} \\ $$$$\left({d}\right)\:\frac{{t}_{{P}} }{{t}_{{Q}} }\:=\:\frac{\mathrm{length}\:\mathrm{of}\:\mathrm{at}\:\mathrm{arc}\:{ACB}}{\mathrm{length}\:\mathrm{of}\:\mathrm{chord}\:{AB}} \\ $$
Commented by Tinkutara last updated on 07/Aug/17
Commented by Tinkutara last updated on 08/Aug/17
$$\mathrm{Help}. \\ $$
Commented by ajfour last updated on 08/Aug/17
$$\int\frac{\mathrm{dx}}{\:\sqrt{\mathrm{a}+\mathrm{cos}\:\mathrm{x}}}\:=\:? \\ $$$$\int_{\mathrm{0}} ^{\:\:\mathrm{t}_{\mathrm{P}} } \mathrm{dt}_{\mathrm{P}} =\int_{−\alpha} ^{\:\:\alpha} \frac{\mathrm{Rd}\theta}{\:\sqrt{\mathrm{v}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{2gR}\left(\mathrm{cos}\:\theta−\mathrm{cos}\:\alpha\right)}} \\ $$
Commented by Tinkutara last updated on 08/Aug/17
$$\mathrm{This}\:\mathrm{integral}\:\mathrm{is}\:\mathrm{undefined}\:\mathrm{except}\:\mathrm{for}\:\mathrm{a} \\ $$$$=\:\mathrm{1}\:\mathrm{as}\:\mathrm{I}\:\mathrm{put}\:\mathrm{this}\:\mathrm{expression}\:\mathrm{in}\:\mathrm{Geogebra} \\ $$$$\mathrm{by}\:\mathrm{taking}\:\mathrm{many}\:\mathrm{values}. \\ $$
Commented by Tinkutara last updated on 08/Aug/17
$$\mathrm{So}\:{t}_{{P}} \:\mathrm{can}'\mathrm{t}\:\mathrm{be}\:\mathrm{evaluated}? \\ $$
Commented by ajfour last updated on 08/Aug/17
$$\mathrm{I}\:\mathrm{cannot}. \\ $$
Commented by ajfour last updated on 08/Aug/17
$$\mathrm{i}\:\mathrm{came}\:\mathrm{across}\:\mathrm{it}\:\mathrm{in}\:\mathrm{trying}\:\mathrm{to}\:\: \\ $$$$\mathrm{evaluate}\:\mathrm{t}_{\mathrm{P}} \:. \\ $$
Answered by ajfour last updated on 08/Aug/17
$$\left(\mathrm{a}\right)\:\mathrm{t}_{\mathrm{P}} <\:\mathrm{t}_{\mathrm{Q}} \\ $$
Commented by Tinkutara last updated on 08/Aug/17
$$\mathrm{Explain}. \\ $$
Commented by ajfour last updated on 08/Aug/17
Commented by ajfour last updated on 08/Aug/17
$$\mathrm{v}>\mathrm{v}_{\mathrm{0}} \:\:\mathrm{and}\:\:\mathrm{cos}\:\theta\:>\:\mathrm{cos}\:\alpha \\ $$$$\mathrm{v}_{\mathrm{x}} =\mathrm{vcos}\:\theta\:\:\:\:\mathrm{u}=\mathrm{v}_{\mathrm{0}} \mathrm{cos}\:\alpha \\ $$$$\Rightarrow\:\:\mathrm{v}_{\mathrm{x}} >\mathrm{v}_{\mathrm{0}} \mathrm{cos}\:\alpha\:\left(=\mathrm{u}\right)\:. \\ $$
Commented by Tinkutara last updated on 08/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$