Question

A particle is projected with velocity $u$ at an angle $θ$ with the horizontal.
(i) Find the tangential and normal acceleration of the particle at $t = 0$ and at highest point of its trajectory.
(ii) Find the radius of curvature.
(a) at $O$
(b) at highest position $P .$

Accepted Answer

(i) $(a)$ The direction of tangential acceleration is in the line of velocity and the direction of normal acceleration is perpendicular to velocity direction. The tangential and normal directions at $O$ and $P$ are shown in the figure.

At $O :$ (at $P .$ ) tangential acceleration, $a_{t} = - g \sin θ$ normal acceleration, $a_{n} = g \cos θ$

$(b)$ At $P$ (at highest point)
Tangential acceleration, $a_{t} = 0$ Normal acceleration, $a_{n} = g$

(ii) $(a)$ Let radius of curvature at $O$ be $R_{o} .$ The normal acceleration at $O$ be $g \cos θ .$

$a_{n} = \frac{v^{2}}{R} \Rightarrow R_{o} = \frac{v_{o}^{2}}{a_{n}} = \frac{v_{o}^{2}}{g \cos θ}$

$(b)$ The radius of curvature at $P$ (normal acceleration at $P$ is $a_{n} = g$ ) is,

$\Rightarrow R_{P} = \frac{v^{2}}{a_{n}} = \frac{{(v_{o} \cos θ)}^{2}}{g} = \frac{v_{o}^{2} \cos^{2} θ}{g}$

B M Sharma Solutions for Chapter: Circular Motion, Exercise 19: ILLUSTRATION 9.19

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Circular Motion, Exercise 19: ILLUSTRATION 9.19

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