Menu Close

Question-22487




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.
$${Thank}\:{you}\:{very}\:{much} \\ $$$${sir}. \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *