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Q. A parabolic bowl with its bottom at origin has the shape $y=\frac{x^{2}}{20}$ . Here $x$ and $y$ are in metres. The maximum height at which a small mass $m$ can be placed on the bowl without slipping (coefficient of static friction is $0.5$ ) is

Question

NTA AbhyasNTA Abhyas 2020Laws of Motion

Solution:

Slope, $\frac{d y}{d x}=\frac{x}{10}$
Solution
or $tan \theta =\frac{x}{10}$ ... (i)
Equilibrium of mass in horizontal direction gives the equation,
$\mu Ncos \theta =Nsin ⁡ \theta $
or $tan \theta =\mu =\frac{1}{2}$ ... (ii)
From Equations. (i) and (ii)
$\frac{x}{10}=\frac{1}{2}$
or $x=5m$
$\therefore $ $y=\frac{x^{2}}{20}=\frac{25}{20}=1.25 \, m$