Question
Define angular velocity and angular acceleration. The total speed $V_1$ of a projectile at its greatest height is $\sqrt{\frac{6}{7}}$ of its speed $V_2$ when it is at half its greatest height. Show that the angle of projection is $30°$.
✓
Answer
Angular Velocity: Angular velocity of an object in circular motion is defined as the time rate of change of is angular displacement. It is denoted by $\omega$ and is measured in radians per second $(rad.s^{-1})$.$\omega=\frac{\text{angular displacement}}{\text{Time}}=\frac{\theta}{\text{t}}=\frac{\text{d}\theta}{\text{dt}}$
Angular Acceleration: Angular acceleration of an object in circular motion is defined as the time rate of change of its angular velocity. It is denoted by 'a’ and measured in rad $s^{-2}$.
$\alpha=\frac{\text{angular velocity change}}{\text{time taken}}=\frac{\text{d}\omega}{\text{dt}}$
Numerical: Velocity at highest point $=\text{u}\cos\theta=\text{V}_1\ (\text{given})$
$\text{h}_\text{max}=\frac{\text{u}^2\sin^2\theta}{2\text{g}}$
Vertical velocity at $\frac{\text{h}_\text{max}}{2}=\text{V}_{2\text{y}}=\sqrt{\text{u}^2\sin^2\theta-2\text{g}\frac{\text{h}_\text{max}}{2}}$
$\text{V}_{2\text{y}}=\sqrt{\text{u}^2\sin^2\theta\Big(1-\frac{1}{2}\Big)}=\frac{\text{u}\sin\theta}{\sqrt{2}}$
$\text{V}_{2\text{x}}=\text{u}\cos\theta$
$\text{V}_2=\sqrt{\text{V}^2_{2\text{x}}+\text{V}^2_{2\text{y}}}$
$=\sqrt{\text{u}^2\cos^2\theta+\frac{\text{u}^2\sin^2\theta}{2}}$
Given, $\text{V}_1=\sqrt{\frac{6}{7}}\text{V}_2$
$\therefore\ \frac{\text{u}\cos\theta}{\text{u}\sqrt{\cos^2\theta+\frac{\sin^2\theta}{2}}}=\sqrt{\frac{6}{7}}$
Squaring both the sides
$\frac{\cos^2\theta}{\cos^2\theta+\frac{\sin^2\theta}{2}}=\frac{6}{7}$
$\frac{1}{1+\frac{\tan^2\theta}{2}}=\frac{6}{7}$
$1+\frac{\tan^2\theta}{2}=\frac{7}{6}\Rightarrow\ \frac{\tan^2\theta}{2}=\frac{7}{6}-1=\frac{1}{6}$
$\tan\theta=\sqrt{\frac{2}{6}}=\frac{1}{\sqrt{3}}$
$\therefore\ \theta=\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)=30^\circ.$
Angular Acceleration: Angular acceleration of an object in circular motion is defined as the time rate of change of its angular velocity. It is denoted by 'a’ and measured in rad $s^{-2}$.
$\alpha=\frac{\text{angular velocity change}}{\text{time taken}}=\frac{\text{d}\omega}{\text{dt}}$
Numerical: Velocity at highest point $=\text{u}\cos\theta=\text{V}_1\ (\text{given})$
$\text{h}_\text{max}=\frac{\text{u}^2\sin^2\theta}{2\text{g}}$
Vertical velocity at $\frac{\text{h}_\text{max}}{2}=\text{V}_{2\text{y}}=\sqrt{\text{u}^2\sin^2\theta-2\text{g}\frac{\text{h}_\text{max}}{2}}$
$\text{V}_{2\text{y}}=\sqrt{\text{u}^2\sin^2\theta\Big(1-\frac{1}{2}\Big)}=\frac{\text{u}\sin\theta}{\sqrt{2}}$
$\text{V}_{2\text{x}}=\text{u}\cos\theta$
$\text{V}_2=\sqrt{\text{V}^2_{2\text{x}}+\text{V}^2_{2\text{y}}}$
$=\sqrt{\text{u}^2\cos^2\theta+\frac{\text{u}^2\sin^2\theta}{2}}$
Given, $\text{V}_1=\sqrt{\frac{6}{7}}\text{V}_2$
$\therefore\ \frac{\text{u}\cos\theta}{\text{u}\sqrt{\cos^2\theta+\frac{\sin^2\theta}{2}}}=\sqrt{\frac{6}{7}}$
Squaring both the sides
$\frac{\cos^2\theta}{\cos^2\theta+\frac{\sin^2\theta}{2}}=\frac{6}{7}$
$\frac{1}{1+\frac{\tan^2\theta}{2}}=\frac{6}{7}$
$1+\frac{\tan^2\theta}{2}=\frac{7}{6}\Rightarrow\ \frac{\tan^2\theta}{2}=\frac{7}{6}-1=\frac{1}{6}$
$\tan\theta=\sqrt{\frac{2}{6}}=\frac{1}{\sqrt{3}}$
$\therefore\ \theta=\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)=30^\circ.$
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