Question

A particle is projected horizontally from the top of a tower with a velocity. If ${\upsilon }_0$ be its velocity at any instant, then the radius of curvature of the path of the particle at that instant is directly proportional to

• A. ${\text{v}}^2$
• B. $\frac{1}{\text{v}}$
• C. $\text{v}$
• D. ${\text{v}}^3$

Answer 1

The correct option is D ${\text{v}}^3$

Draw a diagram of given problem.


Its components will be described with the help of

Calculate centripetal acceleration.
Formula used: ${\text{a}}_c=\frac{{\upsilon }^2}{\text{R}}$
As centripetal acceleration is given by
${\text{a}}_c=\frac{{\upsilon }^2}{\text{R}}$
With the help of diagram,
$\text{g}\text{sin}\theta =\frac{{\upsilon }^2}{\text{R}}$
$\therefore \sin \theta =\frac{{\upsilon }_0}{\upsilon }$
Hence,
$R=\frac{{\upsilon }^2}{g\sin \theta }$
$R=\frac{{\upsilon }^2}{g\frac{{\upsilon }_0}{\upsilon }}$
$R\propto {\upsilon }^3$
Final answer: (a)