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Question
A particle is projected with a speed u at an angle θ with the horizontal. Consider a small part of its path near the highest position and take it approximately to be a circular arc. What is the radius of this circular circle? This radius is called the radius of curvature of the curve at the point.
Solution
At the highest point, the vertical component of velocity is zero.
So, at the highest point, we have:
velocity = v = ucosθ
Centripetal force on the particle = \[\frac{m v^2}{r}\]
\[\Rightarrow \frac{m v^2}{r} = \frac{m u^2 \cos^2 \theta}{r}\]
At the highest point, we have :
\[mg = \frac{m v^2}{r}\]
Here, r is the radius of curvature of the curve at the point.
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