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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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