IIT JEE , AIEEE Forum - Physics , Mechanics

nishi.tiwari

Q ;A uniform rod is pivoted at its upper end hangs vertically .it is displaced through an angle of 60 and then released.Find the magnitude of the force acting on a particle of mass (dm) at the tip of the rod when the rod makes the angle of 37 with the vertical?????

nishi.tiwari

plssssssss koi to solve karooo

armaan

Do it using energy consevation .!

nishantsingh89

L be length of rod
two force acting on it ((dm)g downwards

nd (dm)v^2/r , radially outwards,

by energy conservation

P.E of dm at 60 degree angle = KE at 37 degree + PE at 37 degree

dmgL(1-cos60) = dm v^2/2 + dmgL(1-cos37)

dmv^2/2=dmgL(cos37-cos60)

dmV^2/L = 2dmg( 4/5 - 1/2)
dmV^2/L = 3dmg/5 this is radially outward

now angle between dmg and dmV^2/L is 37 degree

F = sqrt( ((dmg)^2 + (3dmg/5)^2 + 2*(3dmg/5)^dmg*cos37 ))

solve it to get answer
you shud get 1/5* dmg*sqrt(58)

check for calculation mistakes i might have committed a few, :),

akshay.khare91

look the force acting on particle is F= dmv^2 / r ----1

now for finding velocity use energy conservation ..

potential energy at 60 degress = P.E. at 37 + Rotational energy at 37

dmgL - dmglcos60 = dmgl - dmglcos37 + 1/2 I

^2

dmgl(1-cos60) = dmgl(1-cos37) + 1/2 I

^2

now write

= v/r

and solve for v

and put value in 1 to get ans...

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