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Question-53841

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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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