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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.
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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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