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P 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.

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प्रश्न

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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उत्तर

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.

\[\Rightarrow r = \frac{u^2 \cos^2 \theta}{g}\]
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पाठ 7: Circular Motion - Exercise [पृष्ठ ११५]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 7 Circular Motion
Exercise | Q 25 | पृष्ठ ११५

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