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 equation is x² = 4ay. The wire frame is fixed and the bead can 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. $\frac{\text{g}}{2}$
• B. $\frac{\sqrt{3}\text{g}}{2}$
• C. $\frac{\text{g}}{\sqrt{2}}$
• D. $\frac{\text{g}}{\sqrt{5}}$

Answer 1

Answer: (C)

x² = 4ay
Differentiating w.r.t. y, we get
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{2\text{a}}$

$\therefore \text{At}\left(2\text{ a, a}\right)\text{,}\frac{\text{dy}}{\text{dx}}=1\Rightarrow$ hence $\theta =45^{{}^{\circ }}$
the component of weight along tangential direction is mg sin $\theta$ .
hence tangential acceleration is $\text{g sin }\theta =\frac{\text{g}}{\sqrt{2}}$