A beadof mass m is located on a parabolic wire with its axis vertical and vertex at the origin as shown in figure and whose equations is x^(2) = 4 ay . The wire frame is fixed and the beadcan slide on it without friction . The bead is released from the point y = 4a on the wire frame from rest . The tangential acceleration of the bead when it reaches the position given by y = a is :

A beadof mass m is located on a parabolic wire with its axis vertical

1.	A beadof mass m is located on a parabolic wire with its axis vertical and vertex at the origin as shown in figure and whose equations is x^(2) = 4 ay . The wire frame is fixed and the beadcan slide on it without friction . The bead is released from the point y = 4a on the wire frame from rest . The tangential acceleration of the bead when it reaches the position given by y = a is :
Answer» `(g)/(2)` `(sqrt3g)(2)` `(g)/(sqrt2)` g Solution :`x^(2) = 4 AY , (dy)/(dx) = (1)/(2a)x` at point (2a , 2a) , `(dy)/(dx) = 1 , theta = 45^(@)` Component of weight ALONG tangetial direction is mg SIN `theta`. Tangential acceleration `(dv)/(dt) = omega (DR)/(dt` `omega = 4 PI , (dr)/(dt) = 0.5` , a = `2 pi` `T - "mg" = "ma" , T = "mg" + "ma"`