Question

A particle moves in x-y plane with acceleration components a_x = -3m/s² and ay = -16t m/s². If its initial velocity is V₀ = 50m/s directed at 35⁰ to the x–axis, compute the radius of curvature of the path at t = 2 sec.

Answer 1

At t=0
V₀ = 50 m/s at 35⁰ to the x-axis
V_x = 50cos(35) = 40.96 m/s
V_y = 50sin(35) = 28.68 m/s

Given, a_x= -3 m/s² and ay = -16t m/s²
Integrating, V_x = -3t + c1 and V_y = -8t² + c₂
At t=0
c₁ = 40.96 and c₂ = 28.68

Now,
V_x = -3t + 40.96 and V_y = -8t² + 28.68
At t=2sec
V_x = -3(2) + 40.96 and Vy = -8(2²) + 28.68
V_x = 34.96 m/s and V_y = -3.32 m/s
a_x = -3 m/s₂ and a_y = -32 m/s²

`V= sqrt(Vx^2 +Vy^2 )= sqrt(34.96^2+(-3.32)^2` = 35.12m/s

Radius of curvature at t = 2sec,

`R= V^3/|Vxay - Vyax|  = (35.12^3)/|(34.96 X-32)-(-3.32 X-3) |  = 38.38m`