Question

A particle is projected at angle $\theta$ with horizontal with velocity $v_0$ at $t=0$. Find
a. tangential and normal acceleration of the particle at $t=0$ and at highest point of its trajectory.
b. the radius of curvature at $t=0$ and highest point.

Answer 1

The direction of tangential acceleration is in the line of velocity and the direction perpendicular to velocity direction. The tangential and normal directions at $O$ and $P$ are shown in Figs. $161$ and $162$, respectively.

The net acceleration of the particle during motion is acceleration due to gravity. i.e., $g$ is acting vertically downward.

$AtO(att=0)$:

Tangential acceleration.

$a_t=-gsin\theta$

Normal acceleration.

$a_n=-gcos\theta$

$AtP\left(athighestpoint\right):$

Tangential acceleration.

$a_t=0$ and $a_n=g$

b. Let the radius of curvature at $O$ be $R_0$. The normal acceleration at $O$ be $gcos\theta$.

$a_n=\frac{v^2}{R}\Rightarrow R_0=\frac{v_0^2}{a_n}=\frac{v_0^2}{gcos\theta }$

Radius of curvature at $P$:

Normal acceleration at $P$ is $g$. Hence.

$R_p=\frac{v^2}{a_n}=\frac{(v_{0}cos\theta {)}^2}{g}=\frac{v_0^{2}co{s}^2\theta }{g}$