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Question-31894

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Question Number 31894 by Tinkutara last updated on 16/Mar/18
Commented by Tinkutara last updated on 16/Mar/18
Can you explain with diagrams please? I can't understand in which direction forces are balanced.
Commented by mrW2 last updated on 16/Mar/18
m(v cos θ)^2 /r=mg ⇒θ=cos^(−1) (√((gr)/v^2 ))
$${m}\left({v}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} /{r}={mg} \\ $$$$\Rightarrow\theta=\mathrm{cos}^{−\mathrm{1}} \sqrt{\frac{{gr}}{{v}^{\mathrm{2}} }} \\ $$
Commented by mrW2 last updated on 16/Mar/18
Commented by mrW2 last updated on 16/Mar/18
since the surface is smooth, the cylinder can only apply a normal force toward the center line of cylinder.
$${since}\:{the}\:{surface}\:{is}\:{smooth},\:{the} \\ $$$${cylinder}\:{can}\:{only}\:{apply}\:{a}\:{normal}\:{force} \\ $$$${toward}\:{the}\:{center}\:{line}\:{of}\:{cylinder}. \\ $$
Commented by Tinkutara last updated on 16/Mar/18
But vcos θ is not along the direction of radius joining the particle then why ((m(vcos θ)^2 )/r)?
$${But}\:{v}\mathrm{cos}\:\theta\:{is}\:{not}\:{along}\:{the}\:{direction} \\ $$$${of}\:{radius}\:{joining}\:{the}\:{particle}\:{then} \\ $$$${why}\:\frac{{m}\left({v}\mathrm{cos}\:\theta\right)^{\mathrm{2}} }{{r}}? \\ $$
Commented by mrW2 last updated on 16/Mar/18
v cos θ is the velocity of particle in horizontal direction. in horizontal direction the particle makes a circular motion with radius r. the corresponding normal force on cylinder wall is ((m(v cos θ)^2 )/r) which equals mg at the moment.
$${v}\:\mathrm{cos}\:\theta\:{is}\:{the}\:{velocity}\:{of}\:{particle}\:{in}\: \\ $$$${horizontal}\:{direction}.\:{in}\:{horizontal} \\ $$$${direction}\:{the}\:{particle}\:{makes}\:{a} \\ $$$${circular}\:{motion}\:{with}\:{radius}\:{r}.\:{the} \\ $$$${corresponding}\:{normal}\:{force}\:{on} \\ $$$${cylinder}\:{wall}\:{is}\:\frac{{m}\left({v}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} }{{r}}\:{which} \\ $$$${equals}\:{mg}\:{at}\:{the}\:{moment}. \\ $$
Commented by Tinkutara last updated on 16/Mar/18
But why the particle makes a circular turn in the direction of horizontal component of velocity?
Commented by mrW2 last updated on 16/Mar/18
it is said that the particle moves along the inner wall of cylinder. this motion has two components: circular motion with velocity v cos θ and motion parallelly to the axis of cylinder with velocity v sin θ
$${it}\:{is}\:{said}\:{that}\:{the}\:{particle}\:{moves}\:{along} \\ $$$${the}\:{inner}\:{wall}\:{of}\:{cylinder}.\:{this}\:{motion} \\ $$$${has}\:{two}\:{components}:\: \\ $$$${circular}\:{motion}\:{with}\:{velocity}\:{v}\:\mathrm{cos}\:\theta\:{and} \\ $$$${motion}\:{parallelly}\:{to}\:{the}\:{axis}\:{of} \\ $$$${cylinder}\:{with}\:{velocity}\:{v}\:\mathrm{sin}\:\theta \\ $$
Commented by Tinkutara last updated on 17/Mar/18
Thank you very much Sir! I got the answer. ��������

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