Question Number 20916 by Tinkutara last updated on 07/Sep/17 $$\mathrm{A}\:\mathrm{body}\:\mathrm{starts}\:\mathrm{rotating}\:\mathrm{about}\:\mathrm{a} \\ $$$$\mathrm{stationary}\:\mathrm{axis}\:\mathrm{with}\:\mathrm{an}\:\mathrm{angular} \\ $$$$\mathrm{acceleration}\:{b}\:=\:\mathrm{2}{t}\:\mathrm{rad}/\mathrm{s}^{\mathrm{2}} .\:\mathrm{How}\:\mathrm{soon} \\ $$$$\mathrm{after}\:\mathrm{the}\:\mathrm{beginning}\:\mathrm{of}\:\mathrm{rotation}\:\mathrm{will}\:\mathrm{the} \\ $$$$\mathrm{total}\:\mathrm{acceleration}\:\mathrm{vector}\:\mathrm{of}\:\mathrm{an}\:\mathrm{arbitrary} \\ $$$$\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{body}\:\mathrm{forms}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{60}° \\ $$$$\mathrm{with}\:\mathrm{its}\:\mathrm{velocity}\:\mathrm{vector}? \\ $$$$\left(\mathrm{1}\right)\:\left(\mathrm{2}\sqrt{\mathrm{3}}\right)^{\mathrm{1}/\mathrm{3}}…
Question Number 20891 by Tinkutara last updated on 06/Sep/17 $$\mathrm{The}\:\mathrm{Figure}\:\mathrm{shows}\:\mathrm{a}\:\mathrm{system}\:\mathrm{consisting} \\ $$$$\mathrm{of}\:\left({i}\right)\:\mathrm{a}\:\mathrm{ring}\:\mathrm{of}\:\mathrm{outer}\:\mathrm{radius}\:\mathrm{3}{R}\:\mathrm{rolling} \\ $$$$\mathrm{clockwise}\:\mathrm{without}\:\mathrm{slipping}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{horizontal}\:\mathrm{surface}\:\mathrm{with}\:\mathrm{angular}\:\mathrm{speed} \\ $$$$\omega\:\mathrm{and}\:\left({ii}\right)\:\mathrm{an}\:\mathrm{inner}\:\mathrm{disc}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{2}{R} \\ $$$$\mathrm{rotating}\:\mathrm{anti}-\mathrm{clockwise}\:\mathrm{with}\:\mathrm{angular} \\ $$$$\mathrm{speed}\:\omega/\mathrm{2}.\:\mathrm{The}\:\mathrm{ring}\:\mathrm{and}\:\mathrm{disc}\:\mathrm{are} \\ $$$$\mathrm{separated}\:\mathrm{by}\:\mathrm{frictionless}\:\mathrm{ball}\:\mathrm{bearing}. \\…
Question Number 20842 by Tinkutara last updated on 04/Sep/17 $$\mathrm{Acceleration}\:\mathrm{of}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{which}\:\mathrm{is}\:\mathrm{at} \\ $$$$\mathrm{rest}\:\mathrm{at}\:{x}\:=\:\mathrm{0}\:\mathrm{is}\:\overset{\rightarrow} {{a}}\:=\:\left(\mathrm{4}\:−\:\mathrm{2}{x}\right)\:\overset{\wedge} {{i}}.\:\mathrm{Select} \\ $$$$\mathrm{the}\:\mathrm{correct}\:\mathrm{alternative}\left(\mathrm{s}\right). \\ $$$$\left({a}\right)\:\mathrm{Maximum}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{is} \\ $$$$\mathrm{4}\:\mathrm{units} \\ $$$$\left({b}\right)\:\mathrm{Particle}\:\mathrm{further}\:\mathrm{comes}\:\mathrm{to}\:\mathrm{rest}\:\mathrm{at} \\ $$$${x}\:=\:\mathrm{4} \\…
I-think-it-will-be-0-pi-4-dx-1-tanx-0-pi-4-dx-1-x-4-1-2-1-2-4-2-1-3-2-4-1-3-4-3-1-3-5-2-4-6-1-4-4-4-
Question Number 86365 by Prithwish Sen 1 last updated on 28/Mar/20 $$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{will}\:\mathrm{be} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}+\mathrm{tanx}}}\:\approx\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}+\mathrm{x}}}\: \\ $$$$=\frac{\boldsymbol{\pi}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}}.\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}.\mathrm{4}}.\frac{\mathrm{1}}{\mathrm{3}}.\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)^{\mathrm{3}} −\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}.\mathrm{4}.\mathrm{6}}.\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)^{\mathrm{4}} +…. \\ $$…
Question Number 86356 by MASANJAJ last updated on 28/Mar/20 Commented by MASANJAJ last updated on 28/Mar/20 $${please}\:{help}\:{me}\:{to}\:{find}\:{inqualities}\:{of}\: \\ $$$${this}\:{question} \\ $$ Answered by MJS last…
Question Number 20786 by Tinkutara last updated on 02/Sep/17 $$\mathrm{Two}\:\mathrm{particles}\:{A}\:\mathrm{and}\:{B}\:\mathrm{start}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{position}\:\mathrm{along}\:\mathrm{the}\:\mathrm{circular}\:\mathrm{path}\:\mathrm{of} \\ $$$$\mathrm{radius}\:\mathrm{0}.\mathrm{5}\:\mathrm{m}\:\mathrm{with}\:\mathrm{a}\:\mathrm{speed}\:{v}_{{A}} \:=\:\mathrm{1}\:\mathrm{ms}^{−\mathrm{1}} \\ $$$$\mathrm{and}\:{v}_{{B}} \:=\:\mathrm{1}.\mathrm{2}\:\mathrm{ms}^{−\mathrm{1}} \:\mathrm{in}\:\mathrm{opposite}\:\mathrm{direction}. \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{time}\:\mathrm{before}\:\mathrm{they}\:\mathrm{collide}. \\ $$ Answered by…
Question Number 20777 by Tinkutara last updated on 02/Sep/17 $${In}\:{the}\:{figure}\:{shown},\:{mass}\:'{m}'\:{is} \\ $$$${placed}\:{on}\:{the}\:{inclined}\:{surface}\:{of}\:{a} \\ $$$${wedge}\:{of}\:{mass}\:{M}.\:{All}\:{the}\:{surfaces} \\ $$$${are}\:{smooth}.\:{Find}\:{the}\:{acceleration}\:{of} \\ $$$${the}\:{wedge}. \\ $$ Commented by Tinkutara last updated…
Question Number 20764 by Tinkutara last updated on 02/Sep/17 $${A}\:{small}\:{bead}\:{is}\:{slipped}\:{on}\:{a}\:{horizontal} \\ $$$${rod}\:{of}\:{length}\:{l}.\:{The}\:{rod}\:{starts}\:{moving} \\ $$$${with}\:{a}\:{horizontal}\:{acceleration}\:{a}\:{in}\:{a} \\ $$$${direction}\:{making}\:{an}\:{angle}\:\alpha\:{with}\:{the} \\ $$$${length}\:{of}\:{the}\:{rod}.\:{Assuming}\:{that} \\ $$$${initially}\:{the}\:{bead}\:{is}\:{in}\:{the}\:{middle}\:{of} \\ $$$${the}\:{rod},\:{find}\:{the}\:{time}\:{elapsed}\:{before} \\ $$$${the}\:{bead}\:{leaves}\:{the}\:{rod}.\:{Coefficient}\:{of} \\…
Question Number 20761 by Tinkutara last updated on 02/Sep/17 $${A}\:\mathrm{5}\:{kg}\:{block}\:{B}\:{is}\:{suspended}\:{from}\:{a} \\ $$$${cord}\:{attached}\:{to}\:{a}\:\mathrm{40}\:{kg}\:{cart}\:{A}.\:{Find} \\ $$$${the}\:{accelerations}\:{of}\:{both}\:{the}\:{block}\:{and} \\ $$$${cart}.\:\left({All}\:{surfaces}\:{are}\:{frictionless}\right) \\ $$$$\left({g}\:=\:\mathrm{10}\:{m}/{s}^{\mathrm{2}} \right) \\ $$ Commented by Tinkutara last…
Question Number 20745 by Tinkutara last updated on 02/Sep/17 $$\mathrm{Consider}\:\mathrm{a}\:\mathrm{disc}\:\mathrm{rotating}\:\mathrm{in}\:\mathrm{the}\:\mathrm{horizontal} \\ $$$$\mathrm{plane}\:\mathrm{with}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{angular}\:\mathrm{speed}\:\omega \\ $$$$\mathrm{about}\:\mathrm{its}\:\mathrm{centre}\:{O}.\:\mathrm{The}\:\mathrm{disc}\:\mathrm{has}\:\mathrm{a} \\ $$$$\mathrm{shaded}\:\mathrm{region}\:\mathrm{on}\:\mathrm{one}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{diameter} \\ $$$$\mathrm{and}\:\mathrm{an}\:\mathrm{unshaded}\:\mathrm{region}\:\mathrm{on}\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{side}\:\mathrm{as}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\:\mathrm{Figure}.\:\mathrm{When}\:\mathrm{the} \\ $$$$\mathrm{disc}\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:\mathrm{orientation}\:\mathrm{as}\:\mathrm{shown},\:\mathrm{two} \\ $$$$\mathrm{pebbles}\:{P}\:\mathrm{and}\:{Q}\:\mathrm{are}\:\mathrm{simultaneously} \\…