A uniform rod 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 an angle of 37* with the vertical
using COE
Ui = Mgl(1-0.5xcos60)
Ki=0
Uf = Mgl(1-0.5xcos37)
Kf = 0.5 x ((Ml^2)/3) x w^2
you will get w^2 to be (9xg)/(10xl)
force in the tip of the rod
is dm x l x w^2
ans -> ((9 x (dm) x g)/10)
Please tell me if the answer is right.
Ui = Mgl(1-0.5xcos60)
Ki=0
Uf = Mgl(1-0.5xcos37)
Kf = 0.5 x ((Ml^2)/3) x w^2
you will get w^2 to be (9xg)/(10xl)
force in the tip of the rod
is dm x l x w^2
ans -> ((9 x (dm) x g)/10)
Please tell me if the answer is right.
sorry i forgot the translatory kinetic energy in the equation
Kf will thus have an additional factor of 0.5xMx(lw/2)square
thus w^2 comes out to be 18g/35l
Kf will thus have an additional factor of 0.5xMx(lw/2)square
thus w^2 comes out to be 18g/35l
the actual answer is(0.9 dm g 2^0.5)
Undetermined error: @@U_i = mg(l/2)(1-cos60)@@
Undetermined error: @@K_i=0@@
Undetermined error: @@U_f = mg(l/2)(1-cos37)@@
Undetermined error: @@K_f = (1/2)Iomega^2@@
u get
Undetermined error: @@omega^2 = (0.9g)/(l)@@
now the mass dm at the tip has forces
dm.g and centrifugal force of dm.l.Undetermined error: @@omega^2@@
angle between the forces is 37degrees
on calculating net force you get
1.803dmg
i do not know what is wrong.
Undetermined error: @@K_i=0@@
Undetermined error: @@U_f = mg(l/2)(1-cos37)@@
Undetermined error: @@K_f = (1/2)Iomega^2@@
u get
Undetermined error: @@omega^2 = (0.9g)/(l)@@
now the mass dm at the tip has forces
dm.g and centrifugal force of dm.l.Undetermined error: @@omega^2@@
angle between the forces is 37degrees
on calculating net force you get
1.803dmg
i do not know what is wrong.