Sum

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* = *u*cos*θ*

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.

\[\Rightarrow r = \frac{u^2 \cos^2 \theta}{g}\]

Concept: Circular Motion

Is there an error in this question or solution?

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