Question

A man M slides down a curved frictionless track, starting from rest. The curve obeys the equation $y=\frac{x^2}{2}$. The tangential acceleration of man is:

• A. g
• B. $\frac{gx}{\sqrt{x^2+4}}$
• C. $\frac{g}{2}$
• D. $\frac{gx}{\sqrt{x^2+1}}$

Answer 1

The correct option is B $\frac{gx}{\sqrt{x^2+4}}$

$a_{tan}=g\sin \alpha \tan \alpha =\frac{dy}{dx}=\frac{x}{2}\frac{\sin \alpha }{\cos \alpha }=\frac{\sin \alpha }{\sqrt{1-{sin}^2\alpha }}=\frac{x}{2}\Rightarrow \frac{{sin}^2\alpha }{1-{sin}^2\alpha }=\frac{x^2}{4}\Rightarrow 4{sin}^2\alpha =x^2-x^2{sin}^2\alpha \Rightarrow {sin}^2\alpha =\frac{x^2}{4+x^2}\Rightarrow sin\alpha =\frac{x}{\sqrt{4+x^2}}\Rightarrow a_{tan}=\frac{gx}{\sqrt{4+x^2}}$