Question

A bead of mass m is located on a parabolic wire with its axis vertical and vertex at the origin as shown in figure and whose equastion is `x^(2) =4ay`. The wire frame is fixed and 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. `(g)/(2)`
B. `(sqrt(3)g)/(2)`
C. `(g)/(sqrt(2))`
D. `(g)/(sqrt(5))`

Answer 1

Correct Answer - C
`x^(2)=4ay` Differenctiating `w.r.t. y,` we get
`(dy)/(dx)=(x)/(2a)`
`:. ` At `(2aa,a),(dy)/(dx)=1`
`rArr `hence `theta = 45^(@) `

the component of weight along tangential direction is `mg sin theta`
hence tangential acceleration is `g sin theta =(g)/( sqrt(2))`