Question

A particle is fired with initial speed 'u=40 m/s' at an angle of 53° with the horizontal, then find out the radius of curvature of the particle at the instant the particles velocity becomes perpendicular to the initial velocity.

Answer 1

The radius of curvature of the particle is r = 112.5 m/s

Explanation:

Initial velocity "u" = 40 m/s

Angle "θ" = 53°

μx = μcosθ ( Horizontal component)

μy = μsinθ  ( Vertical component)

Velocity becomes equal to initial velocity.

θ = 90° - θ   ( Downward direction with horizontal)

μcos(90 - θ) = μcosθ

μsinθ = μcosθ

μ = μcosθ / sinθ

v = μcot

Centripetal acceleration.

a = v^2 / r

where gsinθ = a

r = v^2 /a

r = u^2 cot^2 θ

r = 1600 x (0.75)^2 / 9.8 x (0.7986)

r = 1600 x 0.5625 / 9.8 x 0.7986

r = 112.5 m/s

Hence the radius of curvature of the particle is r = 112.5 m/s

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