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Question Number 22487 by ajfour last updated on 19/Oct/17
Commented by ajfour last updated on 19/Oct/17
Q. 22473 (solution )
$${Q}.\:\mathrm{22473}\:\left({solution}\:\right) \\ $$
Answered by ajfour last updated on 19/Oct/17
∫_0 ^( △t) fdt=m(usin α−vcos α) =0.1(20×(1/2)−10×((√3)/2)) =(1−((√3)/2) )Ns ∫_0 ^( △t) Ndt=m(ucos α+vsin α) =0.1(20×((√3)/2)+10×(1/2)) =(√3)+(1/2) I△ω = −R∫_0 ^( △t) fdt (1/2)(ω−2)=−(1/2)(1−((√3)/2)) ⇒ ω=2−1+((√3)/2) > 0 ⇒ Immediately after collision ring′s angular velocity is in the same direction. For Translatory motion: let V be final translational velocity of ring, then M(V−v_0 )=−∫_0 ^( △t) (fsin α+Ncos α)dt 2(V−1)=−((1/2)−((√3)/4)+(3/2)+((√3)/4)) ⇒ 2V−2=−2 or V=0 Ring stops. So immediately after collision ring rotates about stationary centre of mass in the same sense as before only little slower, friction acts left. Later it should bring ring into pure rolling condition.
$$\int_{\mathrm{0}} ^{\:\:\bigtriangleup{t}} {fdt}={m}\left({u}\mathrm{sin}\:\alpha−{v}\mathrm{cos}\:\alpha\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0}.\mathrm{1}\left(\mathrm{20}×\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{10}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\right){Ns} \\ $$$$\int_{\mathrm{0}} ^{\:\:\bigtriangleup{t}} {Ndt}={m}\left({u}\mathrm{cos}\:\alpha+{v}\mathrm{sin}\:\alpha\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0}.\mathrm{1}\left(\mathrm{20}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\mathrm{10}×\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$${I}\bigtriangleup\omega\:=\:−{R}\int_{\mathrm{0}} ^{\:\:\bigtriangleup{t}} {fdt} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\omega−\mathrm{2}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\: \\ $$$$\Rightarrow\:\omega=\mathrm{2}−\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:>\:\mathrm{0} \\ $$$$\Rightarrow\:{Immediately}\:{after}\:{collision} \\ $$$${ring}'{s}\:{angular}\:{velocity}\:{is}\:{in}\:{the} \\ $$$${same}\:{direction}. \\ $$$${For}\:{Translatory}\:{motion}:\:{let}\:{V}\:{be} \\ $$$${final}\:{translational}\:{velocity}\:{of} \\ $$$${ring},\:{then} \\ $$$${M}\left({V}−{v}_{\mathrm{0}} \right)=−\int_{\mathrm{0}} ^{\:\:\bigtriangleup{t}} \left({f}\mathrm{sin}\:\alpha+{N}\mathrm{cos}\:\alpha\right){dt} \\ $$$$\mathrm{2}\left({V}−\mathrm{1}\right)=−\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right) \\ $$$$\Rightarrow\:\:\mathrm{2}{V}−\mathrm{2}=−\mathrm{2}\:\:\:{or}\:\:\:{V}=\mathrm{0} \\ $$$${Ring}\:{stops}. \\ $$$${So}\:{immediately}\:{after}\:{collision}\:{ring}\:\:{rotates} \\ $$$${about}\:{stationary}\:{centre}\:{of}\:{mass} \\ $$$${in}\:{the}\:{same}\:{sense}\:{as}\:{before}\:{only} \\ $$$${little}\:{slower},\:\boldsymbol{{friction}}\:\boldsymbol{{acts}}\:\boldsymbol{{left}}. \\ $$$${Later}\:{it}\:{should}\:{bring}\:{ring}\:{into} \\ $$$${pure}\:{rolling}\:{condition}. \\ $$$$ \\ $$
Commented by Sahib singh last updated on 19/Oct/17
Thank you very much sir.
$${Thank}\:{you}\:{very}\:{much} \\ $$$${sir}. \\ $$$$ \\ $$

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