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Question-53841




Question Number 53841 by rahul 19 last updated on 26/Jan/19
Commented by rahul 19 last updated on 27/Jan/19
Mrw sir, 1)when do we apply formula  K.E= (1/2)Iω^2  ?     2)) what is the diff.  between circular & rotational motion   for a point  object having mass m ??
$$\left.{Mrw}\:{sir},\:\mathrm{1}\right){when}\:{do}\:{we}\:{apply}\:{formula} \\ $$$$\left.{K}\left..{E}=\:\frac{\mathrm{1}}{\mathrm{2}}{I}\omega^{\mathrm{2}} \:?\:\:\:\:\:\mathrm{2}\right)\right)\:{what}\:{is}\:{the}\:{diff}. \\ $$$${between}\:{circular}\:\&\:{rotational}\:{motion}\: \\ $$$${for}\:{a}\:{point}\:\:{object}\:{having}\:{mass}\:{m}\:?? \\ $$
Commented by rahul 19 last updated on 26/Jan/19
ΔU + ΔK=0   Now ΔU=−mg(2R)   and ΔK= (1/2) Iω^2 = (1/2)(mR^2 +((mR^2 )/2))((v/R))^2   ⇒ΔK= (3/4)mv^2 .  ⇒(3/4) mv^2 = 2mgR  ⇒ v= (√((8/3)gR)).
$$\Delta{U}\:+\:\Delta{K}=\mathrm{0} \\ $$$$\:{Now}\:\Delta{U}=−{mg}\left(\mathrm{2}{R}\right)\: \\ $$$${and}\:\Delta{K}=\:\frac{\mathrm{1}}{\mathrm{2}}\:{I}\omega^{\mathrm{2}} =\:\frac{\mathrm{1}}{\mathrm{2}}\left({mR}^{\mathrm{2}} +\frac{{mR}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\frac{{v}}{{R}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\Delta{K}=\:\frac{\mathrm{3}}{\mathrm{4}}{mv}^{\mathrm{2}} . \\ $$$$\Rightarrow\frac{\mathrm{3}}{\mathrm{4}}\:{mv}^{\mathrm{2}} =\:\mathrm{2}{mgR} \\ $$$$\Rightarrow\:{v}=\:\sqrt{\frac{\mathrm{8}}{\mathrm{3}}{gR}}. \\ $$
Commented by rahul 19 last updated on 26/Jan/19
My doubt: I ′ve studied that we can  apply the formula : (1/2)Iω^2  only when  the system performs pure rotation.  But here bead is performing circular  motion (only disc is performing  pure rotation), then why the above  method is giving correct answer ?
$${My}\:{doubt}:\:{I}\:'{ve}\:{studied}\:{that}\:{we}\:{can} \\ $$$${apply}\:{the}\:{formula}\::\:\frac{\mathrm{1}}{\mathrm{2}}{I}\omega^{\mathrm{2}} \:{only}\:{when} \\ $$$${the}\:{system}\:{performs}\:{pure}\:{rotation}. \\ $$$${But}\:{here}\:{bead}\:{is}\:{performing}\:{circular} \\ $$$${motion}\:\left({only}\:{disc}\:{is}\:{performing}\right. \\ $$$$\left.{pure}\:{rotation}\right),\:{then}\:{why}\:{the}\:{above} \\ $$$${method}\:{is}\:{giving}\:{correct}\:{answer}\:? \\ $$
Commented by mr W last updated on 26/Jan/19
each single point on the disc performs  circular motion. rotation of an object=  circular motion of all its points about  the same axis!  bead is here a normal point on the disc  like every other point on it, they all  make circular motion about the same  point. the difference is that the disc  with the bead has an other I than the  disc without bead. but you have taken  this into consideration, therefore  you got the right result.
$${each}\:{single}\:{point}\:{on}\:{the}\:{disc}\:{performs} \\ $$$${circular}\:{motion}.\:{rotation}\:{of}\:{an}\:{object}= \\ $$$${circular}\:{motion}\:{of}\:{all}\:{its}\:{points}\:{about} \\ $$$${the}\:{same}\:{axis}! \\ $$$${bead}\:{is}\:{here}\:{a}\:{normal}\:{point}\:{on}\:{the}\:{disc} \\ $$$${like}\:{every}\:{other}\:{point}\:{on}\:{it},\:{they}\:{all} \\ $$$${make}\:{circular}\:{motion}\:{about}\:{the}\:{same} \\ $$$${point}.\:{the}\:{difference}\:{is}\:{that}\:{the}\:{disc} \\ $$$${with}\:{the}\:{bead}\:{has}\:{an}\:{other}\:{I}\:{than}\:{the} \\ $$$${disc}\:{without}\:{bead}.\:{but}\:{you}\:{have}\:{taken} \\ $$$${this}\:{into}\:{consideration},\:{therefore} \\ $$$${you}\:{got}\:{the}\:{right}\:{result}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 26/Jan/19
moment of inertia=M.I of disc+M.I of welded bead    =((mR^2 )/2)+mR^2
$${moment}\:{of}\:{inertia}={M}.{I}\:{of}\:{disc}+{M}.{I}\:{of}\:{welded}\:{bead} \\ $$$$\:\:=\frac{{mR}^{\mathrm{2}} }{\mathrm{2}}+{mR}^{\mathrm{2}} \\ $$$$ \\ $$
Commented by mr W last updated on 26/Jan/19
rotation of a body =Σ circular motion of  all its masses  K.E.=∫(1/2)dmv^2 =(1/2)∫r^2 ω^2 dm=(1/2)(∫r^2 dm)ω^2   =(1/2)Iω^2
$${rotation}\:{of}\:{a}\:{body}\:=\Sigma\:{circular}\:{motion}\:{of} \\ $$$${all}\:{its}\:{masses} \\ $$$${K}.{E}.=\int\frac{\mathrm{1}}{\mathrm{2}}{dmv}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\int{r}^{\mathrm{2}} \omega^{\mathrm{2}} {dm}=\frac{\mathrm{1}}{\mathrm{2}}\left(\int{r}^{\mathrm{2}} {dm}\right)\omega^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{I}\omega^{\mathrm{2}} \\ $$
Commented by rahul 19 last updated on 27/Jan/19
Thank you Sir!  Q.2 ??
$${Thank}\:{you}\:{Sir}! \\ $$$${Q}.\mathrm{2}\:?? \\ $$
Commented by mr W last updated on 27/Jan/19
again: circular motion=rotation  a point mass is also an object.  KE=(1/2)mv^2 =(1/2)mr^2 ω^2 =(1/2)Iω^2   I=mr^2 =IOM of the mass  the physics for circular motion and  rotation is the same. this is my  understanding.
$${again}:\:{circular}\:{motion}={rotation} \\ $$$${a}\:{point}\:{mass}\:{is}\:{also}\:{an}\:{object}. \\ $$$${KE}=\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{mr}^{\mathrm{2}} \omega^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{I}\omega^{\mathrm{2}} \\ $$$${I}={mr}^{\mathrm{2}} ={IOM}\:{of}\:{the}\:{mass} \\ $$$${the}\:{physics}\:{for}\:{circular}\:{motion}\:{and} \\ $$$${rotation}\:{is}\:{the}\:{same}.\:{this}\:{is}\:{my} \\ $$$${understanding}. \\ $$

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