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 equation is $x^{2}$ =4ay. The wire frame is fixed and the bead can slide on it without friction. The bead is released from the point $y = 4 a$ on the wire frame from rest. The tangential acceleration of the bead when it reaches the position given by $y = a$ is (Acceleration due to gravity is g)

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Solution

The correct option is C

$x^{2} = 4 a y D i f f e r e n t i a t i n g w . r . t y, w e g e t \frac{d y}{d x} = \frac{x}{2 a} A t (2 a, a), \frac{d y}{d x} = 1 h e n c e q = 45^{0} The component of weight along tangential directions is mgsinq. H e n c e t a n g e n t i a l a c c e l e r a t i o n i s g s i n q = \frac{g}{\sqrt{2}}$

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A piece of wire is bent in the shape of a parabola $y = k x^{2}$ ( $y$ -axis vertical), with a bead of mass $m$ placed on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the positive $x$ -axis with a constant acceleration $a$ . The distance of new equilibrium position of the bead from $y$ -axis, where the bead can stay at rest with respect to the wire will be: