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A Particle Moves in X-y Plane with Acceleration Components Ax = -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 - Engineering Mechanics

Answer in Brief

A particle moves in x-y plane with acceleration components ax = -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.

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Solution

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

Given, ax= -3 m/s2 and ay = -16t m/s2
Integrating, Vx = -3t + c1 and Vy = -8t2 + c2
At t=0
c1 = 40.96 and c2 = 28.68

Now,
Vx = -3t + 40.96 and Vy = -8t2 + 28.68
At t=2sec
Vx = -3(2) + 40.96 and Vy = -8(22) + 28.68
Vx = 34.96 m/s and Vy = -3.32 m/s
ax = -3 m/s2 and ay = -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`

Concept: Tangentialand Normal Component of Acceleration
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