Auniform rod of length l rests on a frictionless horizontal surface. the rod pivots about a fixed frictionless axis at one end. the rod is initially at rest. a bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. the mass of the bullet is one-fourth the mass of the rod. what is thefinal angular speed of the rod?

a) 0.315 (V/L)

Explanation:

From Conservation of angular momentum, we know that

L1 = L2 ,

Therefore MV L/2 = ( Irod + Ib) x W

M/4 x V x L/2 = (M (L/2)^2 + 1/3xMxL^2) x W

M/8 X VL = (ML^2/16 + ML^2 /3 )

After elimination we have,

V/8 = 19/48 x L x W

W = 48/8 x V/19L = 6/19 x V/L

Therefore W = (0.136)X V/L

the answer would be a

the higher the current the higher the mag field around it the higher the influence on the compass