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 and angle `theta/2` with the horizontal?
A. `(u^2 cos^2 sec^3 (theta/2))/(g)`
B. `(u^2 cos^2 sec^3 (theta/2))/(2 g)`
C. `(2 u^2 cos^2 theta sec^3 (theta/2))/(g)`
D. `(u^2 cos^2 theta sec^3 (theta/2))/(sqrt(3 g))`

Answer 1

Correct Answer - A
(a) We know that `a_c = (v^2)/(r ) rArr r = (v^2)/(a_c)`, where `r` is known as radius of curvature. At the highest point, `v = u cos theta`,
`a_c = g rArr r = (u^2 cos^2 theta)/(g)`
Similarly, find the velocity and `a_c` at the given point and then find `r. a_c` will be component of `g` perpendicular to velocity at that point (Fig. S5.93).
`v cos(theta//2) = u cos theta`
`rArr v = u cos theta sec (theta//2)`
`a_c = g cos (theta//2)`
Now find `r = (v^2)/(a_c)`
`h = H//2, v cos phi = u cos theta` ...(i)
`v^2 sin^2 phi = u^2 sin^2 theta - 2g (H)/(2)`...(ii)
Squaring Eq. (i) and adding in Eq. (ii), we get
`v^2 = u^2 - gH` where `H = (u^2 sin^ theta)/(2 g)`
`a_c = g cos phi = (g u cos theta)/(v)`
Now `r = (v^2)/(a_c)`.
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