Question
What is the radius of curvature of the parabola traced out by the projectile in the previous problem at a point where the particle velocity makes an angle $\frac{\theta}{2}$ with the horizontal?
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Answer
Let ‘u’ the velocity at the pt where it makes an angle $\frac{\theta}{2}$ with horizontal. The horizontal component remains unchanged So, $\text{v}\cos\frac{\theta}{2}=\omega\cos\theta$$\Rightarrow\text{v}=\frac{\text{u}\cos\theta}{\cos\Big(\frac{\theta}{2}\Big)}\ \dots(1)$
From figure
$\text{mg}\cos\Big(\frac{\theta}{2}\Big)=\frac{\text{mv}^2}{\text{r}}$
$\Rightarrow\text{r}=\frac{\text{v}^2}{\text{g}\cos\Big(\frac{\theta}{2}\Big)}$
putting the value of ‘v’ from equn(1)$\text{r}=\frac{\text{u}^2\cos^2\theta}{\text{g}\cos^3\Big(\frac{\theta}{2}\Big)}$
From figure
$\text{mg}\cos\Big(\frac{\theta}{2}\Big)=\frac{\text{mv}^2}{\text{r}}$
$\Rightarrow\text{r}=\frac{\text{v}^2}{\text{g}\cos\Big(\frac{\theta}{2}\Big)}$
putting the value of ‘v’ from equn(1)$\text{r}=\frac{\text{u}^2\cos^2\theta}{\text{g}\cos^3\Big(\frac{\theta}{2}\Big)}$
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