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A-small-bead-is-slipped-on-a-horizontal-rod-of-length-l-The-rod-starts-moving-with-a-horizontal-acceleration-a-in-a-direction-making-an-angle-with-the-length-of-the-rod-Assuming-that-initially-the

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Question Number 20764 by Tinkutara last updated on 02/Sep/17
A small bead is slipped on a horizontal rod of length l. The rod starts moving with a horizontal acceleration a in a direction making an angle α with the length of the rod. Assuming that initially the bead is in the middle of the rod, find the time elapsed before the bead leaves the rod. Coefficient of friction between the bead and the rod is μ. (Neglect gravity).
$${A}\:{small}\:{bead}\:{is}\:{slipped}\:{on}\:{a}\:{horizontal} \\ $$$${rod}\:{of}\:{length}\:{l}.\:{The}\:{rod}\:{starts}\:{moving} \\ $$$${with}\:{a}\:{horizontal}\:{acceleration}\:{a}\:{in}\:{a} \\ $$$${direction}\:{making}\:{an}\:{angle}\:\alpha\:{with}\:{the} \\ $$$${length}\:{of}\:{the}\:{rod}.\:{Assuming}\:{that} \\ $$$${initially}\:{the}\:{bead}\:{is}\:{in}\:{the}\:{middle}\:{of} \\ $$$${the}\:{rod},\:{find}\:{the}\:{time}\:{elapsed}\:{before} \\ $$$${the}\:{bead}\:{leaves}\:{the}\:{rod}.\:{Coefficient}\:{of} \\ $$$${friction}\:{between}\:{the}\:{bead}\:{and}\:{the}\:{rod} \\ $$$${is}\:\mu.\:\left({Neglect}\:{gravity}\right). \\ $$
Answered by ajfour last updated on 02/Sep/17
Commented by ajfour last updated on 02/Sep/17
N=masin α let a_(r ) be relative acceleration of bead along rod. macos α−μN=ma_r ⇒ ma_r =macos α−μ(masin α) a_r =a(cos α−μsin α) (l/2)=(1/2)a_r t^2 ⇒ t =(√(l/a_r )) =(√(l/(a(cos α−μsin α)))) .
$${N}={ma}\mathrm{sin}\:\alpha \\ $$$${let}\:{a}_{{r}\:} {be}\:{relative}\:{acceleration}\:{of} \\ $$$${bead}\:{along}\:{rod}. \\ $$$$\:\:{ma}\mathrm{cos}\:\alpha−\mu{N}={ma}_{{r}} \\ $$$$\Rightarrow\:\:\:\:{ma}_{{r}} ={ma}\mathrm{cos}\:\alpha−\mu\left({ma}\mathrm{sin}\:\alpha\right) \\ $$$$\:\:\:\:{a}_{{r}} \:={a}\left(\mathrm{cos}\:\alpha−\mu\mathrm{sin}\:\alpha\right) \\ $$$$\:\:\:\frac{{l}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}{a}_{{r}} {t}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:{t}\:=\sqrt{\frac{{l}}{{a}_{{r}} }}\:\:=\sqrt{\frac{{l}}{{a}\left(\mathrm{cos}\:\alpha−\mu\mathrm{sin}\:\alpha\right)}}\:. \\ $$
Commented by Tinkutara last updated on 02/Sep/17
Why the masses of rod and bead same?
$$\mathrm{Why}\:\mathrm{the}\:\mathrm{masses}\:\mathrm{of}\:\mathrm{rod}\:\mathrm{and}\:\mathrm{bead}\:\mathrm{same}? \\ $$
Commented by ajfour last updated on 02/Sep/17
i have not assumed any such thing..
$${i}\:{have}\:{not}\:{assumed}\:{any}\:{such}\:{thing}.. \\ $$
Commented by Tinkutara last updated on 02/Sep/17
In 4^(th) line, you have assumed ma_r and macosα, is m the mass of the bead and rod, both?
$$\mathrm{In}\:\mathrm{4}^{\mathrm{th}} \:\mathrm{line},\:\mathrm{you}\:\mathrm{have}\:\mathrm{assumed}\:{ma}_{{r}} \:\mathrm{and} \\ $$$${ma}\mathrm{cos}\alpha,\:\mathrm{is}\:{m}\:\mathrm{the}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bead}\:\mathrm{and} \\ $$$$\mathrm{rod},\:\mathrm{both}? \\ $$
Commented by ajfour last updated on 02/Sep/17
rod is given to be accelerating, i have marked forces on bead. ma is the pseudo force on bead from frame of rod. m is the mass of bead only.
$${rod}\:{is}\:{given}\:{to}\:{be}\:{accelerating}, \\ $$$${i}\:{have}\:{marked}\:{forces}\:{on}\:{bead}. \\ $$$${ma}\:{is}\:{the}\:{pseudo}\:{force}\:{on}\:{bead} \\ $$$${from}\:{frame}\:{of}\:{rod}.\:{m}\:{is}\:\:{the} \\ $$$${mass}\:{of}\:{bead}\:{only}. \\ $$
Commented by Tinkutara last updated on 02/Sep/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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