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Question Number 23346 by ajfour last updated on 29/Oct/17

Commented by Physics lover last updated on 29/Oct/17

Nice trick sir.

$${Nice}\:{trick}\:{sir}. \\ $$

Commented by ajfour last updated on 29/Oct/17

The centre of mass velocity of the conveyor belt vehicle of mass M rolling along a level road is v_0 .Find its kinetic energy.

$${The}\:{centre}\:{of}\:{mass}\:{velocity}\:{of} \\ $$$${the}\:{conveyor}\:{belt}\:\:{vehicle}\:{of}\: \\ $$$${mass}\:{M}\:{rolling}\:{along}\:{a}\:{level}\: \\ $$$${road}\:{is}\:\boldsymbol{{v}}_{\mathrm{0}} .\boldsymbol{{F}}{ind}\:{its}\:{kinetic}\:{energy}. \\ $$

Commented by Physics lover last updated on 29/Oct/17

Commented by Physics lover last updated on 29/Oct/17

Sorry,Mr Ajfour , I flagged this question by mistake. Did not intend to do so at all.

$${Sorry},{Mr}\:{Ajfour}\:,\:{I}\:{flagged}\:{this}\: \\ $$$${question}\:{by}\:{mistake}. \\ $$$${Did}\:{not}\:\:{intend}\:{to}\:{do}\:{so}\:{at}\:{all}. \\ $$

Commented by mrW1 last updated on 29/Oct/17

M=M_R +2M_L M_R =mass of circular parts(=DA^(⌢) +BC^(⌢) ) M_L =mass of parallel part (=AB=DC) KE_R =(1/2)M_R v_0 ^2 +(1/2)Iω^2 =(1/2)M_R v_0 ^2 +(1/2)M_R R^2 ((v_0 /R))^2 =M_R v_0 ^2 KE_L =(1/2)M_L (2v_0 )^2 +(1/2)M_L 0^2 =2M_L v_0 ^2 total KE=KE_R +KE_L =M_R v_0 ^2 +2M_L v_0 ^2 =(M_R +2M_L )v_0 ^2 =Mv_0 ^2

$$\mathrm{M}=\mathrm{M}_{\mathrm{R}} +\mathrm{2M}_{\mathrm{L}} \\ $$$$\mathrm{M}_{\mathrm{R}} =\mathrm{mass}\:\mathrm{of}\:\mathrm{circular}\:\mathrm{parts}\left(=\overset{\frown} {\mathrm{DA}}+\overset{\frown} {\mathrm{BC}}\right) \\ $$$$\mathrm{M}_{\mathrm{L}} =\mathrm{mass}\:\mathrm{of}\:\mathrm{parallel}\:\mathrm{part}\:\left(=\mathrm{AB}=\mathrm{DC}\right) \\ $$$$ \\ $$$$\mathrm{KE}_{\mathrm{R}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{M}_{\mathrm{R}} \mathrm{v}_{\mathrm{0}} ^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{I}\omega^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{M}_{\mathrm{R}} \mathrm{v}_{\mathrm{0}} ^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{M}_{\mathrm{R}} \mathrm{R}^{\mathrm{2}} \left(\frac{\mathrm{v}_{\mathrm{0}} }{\mathrm{R}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{M}_{\mathrm{R}} \mathrm{v}_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{KE}_{\mathrm{L}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{M}_{\mathrm{L}} \left(\mathrm{2v}_{\mathrm{0}} \right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{M}_{\mathrm{L}} \mathrm{0}^{\mathrm{2}} \\ $$$$=\mathrm{2M}_{\mathrm{L}} \mathrm{v}_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{total}\:\mathrm{KE}=\mathrm{KE}_{\mathrm{R}} +\mathrm{KE}_{\mathrm{L}} \\ $$$$=\mathrm{M}_{\mathrm{R}} \mathrm{v}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{2M}_{\mathrm{L}} \mathrm{v}_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$=\left(\mathrm{M}_{\mathrm{R}} +\mathrm{2M}_{\mathrm{L}} \right)\mathrm{v}_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$=\mathrm{Mv}_{\mathrm{0}} ^{\mathrm{2}} \\ $$

Commented by ajfour last updated on 29/Oct/17

thanks again, Sir.

$${thanks}\:{again},\:{Sir}. \\ $$

Commented by ajfour last updated on 29/Oct/17

If we assume small radius, or height, then K=(1/2)((M/2))(2v_0 )^2 +0+0+0 K =Mv_0 ^2 .

$${If}\:{we}\:{assume}\:{small}\:{radius},\:{or} \\ $$$${height},\:{then} \\ $$$${K}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{M}}{\mathrm{2}}\right)\left(\mathrm{2}{v}_{\mathrm{0}} \right)^{\mathrm{2}} +\mathrm{0}+\mathrm{0}+\mathrm{0} \\ $$$$\:\:\:\boldsymbol{{K}}\:=\boldsymbol{{Mv}}_{\mathrm{0}} ^{\mathrm{2}} \:. \\ $$

Answered by Physics lover last updated on 29/Oct/17

the particles on AB ard purely translating with velocity 2v_0 And the semicurcular rings AD and BC are pure rolling with velocity v_(0 ) and ω = (v_0 /R) let λ be linear mass density ⇒ λ = (M/(2l+2πR)) Tot K.E. = K.E._(rotational ) + K.E._(translational) ={ (1/(2 ))(2πRλ)R^2 ((v_0 ^2 /R^(2 ) )) } +{ (1/2)(λl)(2v_0 )^2 + (1/2)(2πRλ)v_0 ^2 } ={ πRλv_0 ^(2 ) } +{ 2lλv_0 ^( 2) + πRλv_(0 ) ^( 2) } = 2πRv_0 ^(2 ) +2lλv_0 ^2 = 2λv_0 ^2 (πR + l ) = 2((M/(2l + 2πR)))v_0 ^(2 ) (πR + l ) K.E. = Mv_0 ^2

$${the}\:{particles}\:{on}\:{AB}\:{ard}\:{purely}\: \\ $$$${translating}\:{with}\:{velocity}\:\mathrm{2}{v}_{\mathrm{0}} \\ $$$${And}\:{the}\:{semicurcular}\:{rings}\:{AD} \\ $$$${and}\:{BC}\:{are}\:{pure}\:{rolling}\:{with}\:{velocity} \\ $$$${v}_{\mathrm{0}\:} {and}\:\omega\:=\:\frac{{v}_{\mathrm{0}} }{{R}} \\ $$$${let}\:\lambda\:{be}\:{linear}\:{mass}\:{density} \\ $$$$\Rightarrow\:\lambda\:=\:\frac{{M}}{\mathrm{2}{l}+\mathrm{2}\pi{R}} \\ $$$${Tot}\:{K}.{E}.\:=\:{K}.{E}._{{rotational}\:} +\:{K}.{E}._{{translational}} \\ $$$$\:\:\:=\left\{\:\frac{\mathrm{1}}{\mathrm{2}\:}\left(\mathrm{2}\pi{R}\lambda\right){R}^{\mathrm{2}} \left(\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{{R}^{\mathrm{2}\:} }\right)\:\right\}\:+\left\{\:\frac{\mathrm{1}}{\mathrm{2}}\left(\lambda{l}\right)\left(\mathrm{2}{v}_{\mathrm{0}} \right)^{\mathrm{2}} +\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\pi{R}\lambda\right){v}_{\mathrm{0}} ^{\mathrm{2}} \:\right\} \\ $$$$\:\: \\ $$$$\:=\left\{\:\pi{R}\lambda{v}_{\mathrm{0}} ^{\mathrm{2}\:} \right\}\:+\left\{\:\mathrm{2}{l}\lambda{v}_{\mathrm{0}} ^{\:\mathrm{2}} \:+\:\pi{R}\lambda{v}_{\mathrm{0}\:} ^{\:\mathrm{2}} \right\} \\ $$$$\:=\:\mathrm{2}\pi{Rv}_{\mathrm{0}} ^{\mathrm{2}\:} +\mathrm{2}{l}\lambda{v}_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$=\:\mathrm{2}\lambda{v}_{\mathrm{0}} ^{\mathrm{2}} \:\left(\pi{R}\:+\:{l}\:\right) \\ $$$$=\:\mathrm{2}\left(\frac{{M}}{\mathrm{2}{l}\:+\:\mathrm{2}\pi{R}}\right){v}_{\mathrm{0}} ^{\mathrm{2}\:} \left(\pi{R}\:+\:{l}\:\right) \\ $$$${K}.{E}.\:\:=\:\:{Mv}_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by ajfour last updated on 29/Oct/17

Ans. is K.E. = Mv_0 ^2 .

$${Ans}.\:{is}\:{K}.{E}.\:=\:{Mv}_{\mathrm{0}} ^{\mathrm{2}} \:. \\ $$

Commented by ajfour last updated on 29/Oct/17

Thank you Physics lover.

$${Thank}\:{you}\:{Physics}\:{lover}. \\ $$

Commented by Physics lover last updated on 29/Oct/17

You are welcome sir. But i guess i did it in the most lengthy way.lol

$${You}\:{are}\:{welcome}\:{sir}.\:{But}\:{i}\:{guess} \\ $$$${i}\:{did}\:{it}\:{in}\:{the}\:{most}\:{lengthy}\:{way}.{lol} \\ $$

Commented by ajfour last updated on 29/Oct/17

Your proof is quite general one (answer to a subjective question).

$${Your}\:{proof}\:{is}\:{quite}\:{general}\:{one} \\ $$$$\left({answer}\:{to}\:{a}\:{subjective}\:{question}\right). \\ $$

Commented by Physics lover last updated on 29/Oct/17

Oh thank you sir. :)

$$\left.{Oh}\:{thank}\:{you}\:{sir}.\:\::\right) \\ $$

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