Question

A mass M slides down a curved frictionless track in vertical plane, starting from rest The curve obeys the equation y = x^2/2. The tangential acceleration of the mass is

Answer 1

Answer:

The tangential acceleration is $\frac{gx}{\sqrt{x^2+1}}$

Explanation:

If at any point on the curve $y=\frac{x^2}{2}$ the angle made by the slope with the horizontal is $\theta$ then

The slope $=\frac{dy}{dx}=\tan\theta$

$\frac{dy}{dx}=\frac{2x}{2}=x$

Thus, $\tan\theta=x$

Therefore, $\sin\theta=\frac{x}{\sqrt{x^2+1}}$

Since the acceleration $g$ acts vertically downwards, the component of $g$ in the direction of the tangent will be $g\sin\theta$

The tangential acceleration is given by

$a_t=g\sin\theta$

$\implies a_t=g\times\frac{x}{\sqrt{x^2+1}}$

$\implies a_t=\frac{gx}{\sqrt{x^2+1}}$

Hope this helps.