Question

Using the conditions of the foregoing problem, draw the approximate time dependence of moduli of the normal $w_{n}$ and tangent $w_{τ}$ acceleration vectors, as well as of the projection of the total acceleration vector $w_{v}$ on the velocity vector direction.

Foregoing problem:- A body is thrown from the surface of the Earth at an angle

α

to the horizontal with the initial velocity

v_{0}

. Assuming the air drag to be negligible, find,

(a)

the time of motion,

(b)

the maximum height of ascent and the horizontal range; at what value of the angle

α

they will be equal to each other,

(c)

the equation of trajectory

y (x)

, where

y

and

x

are displacements of the body along the vertical and the horizontal respectively,

(d)

the curvature radii of trajectory at its initial point and at its peak.

Accepted Answer

A projectile has a constant acceleration $\vec{w} = \vec{g}$ .

The velocity of the particle after time $t$ is given by,

$α$

$\Rightarrow \vec{v} = (v_{0} \cos α \hat{i} + v_{0} \sin α \hat{j}) - g t \hat{j}$

$\Rightarrow \vec{v} = v_{0} \cos α \hat{i} + (v_{0} \sin α - g t) \hat{j} . . . (1)$

Hence, $x$ -component of velocity is,

$v_{x} = v_{0} \cos α . . . (2)$

And, $v_{0}$ -component of velocity is,

$v_{y} = v_{0} \sin α - g t . . . (3)$

The speed $v$ of the projectile is the magnitude of velocity. Hence, the speed is given by,

$∴ v = |\vec{v}| \Rightarrow v = \sqrt{{(v_{0} \cos α)}^{2} + {(v_{0} \sin α - g t)}^{2}} \Rightarrow v^{2} = {v_{0}}^{2} - 2 g t v_{0} \sin α + g^{2} t^{2} . . . (4)$

Now, differentiating both sides of the equation $(4)$ with respect to time $t$ , we get,

$2 v \frac{d v}{d t} = 0 - 2 g v_{0} \sin α + 2 g^{2} t \Rightarrow \frac{d v}{d t} = \frac{- g}{v} (v_{0} \sin α - g t) \Rightarrow \frac{d v}{d t} = \frac{- g}{v} v_{y} . . . (5)$

[using the equation $(3)$ ]

As the tangential acceleration $w_{t}$ is the time rate of change of speed, so we have,

$w_{t} = \frac{d v}{d t} ∴ |w_{t}| = \frac{g}{v} |v_{y}| . . . (6) \Rightarrow |w_{t}| = \frac{g (v_{0} \sin α - g t)}{\sqrt{{v_{0}}^{2} - 2 g t v_{0} \sin α + g^{2} t^{2}}} . . . (7)$

[using $(3)$ and $(4)]$

Hence, $|w_{t}|$ vs $t$ plot is as shown in the figure.

Let $w_{n}$ be the normal acceleration.

As the net acceleration is also given by $w = \sqrt{w_{t}^{2} + w_{n}^{2}}$ , so the normal acceleration is given by,

$w_{n} = \sqrt{w^{2} - w_{t}^{2}} \Rightarrow w_{n} = \sqrt{g^{2} - g^{2} \frac{v_{y}^{2}}{v^{2}}} [∵ w = g & |w_{t}| = \frac{g}{v} |v_{y}|]$

$w_{n} = \frac{g}{v} (\sqrt{v^{2} - v_{y}^{2}}) \Rightarrow w_{n} = \frac{g v_{x}}{v} [∵ v = \sqrt{v_{x}^{2} + v_{y}^{2}}] \Rightarrow w_{n} = \frac{g v_{0} \cos α}{\sqrt{{v_{0}}^{2} - 2 g t v_{0} \sin α + g^{2} t^{2}}} . . . (8)$

[using $(2)$ and $(4)]$

Hence, $|w_{n}|$ vs $t$ plot is as shown in the figure.

The projection of the total acceleration on the velocity vector direction is the same as the tangential acceleration. Hence, we have,

$w_{v} = w_{t} \Rightarrow w_{v} = - g \frac{v_{y}}{v} \Rightarrow w_{v} = - \frac{g (v_{0} \sin α - g t)}{\sqrt{{v_{0}}^{2} - 2 g t v_{0} \sin α + g^{2} t^{2}}} . . . (9)$

[using $(7)]$
Hence, $w_{v}$ vs $t$ plot is as shown in the figure.

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