Engineering Mechanics - Kinematics of Particle (KOP) - Discussion

Discussion :: Kinematics of Particle (KOP) - General Questions (Q.No.5)

5. 

A package is dropped from the plane which is flying with a constant horizontal velocity of vA = 150 ft/s at a height h = 1500 ft. Determine the radius of curvature of the path of the package just after it is released from plane at A.

[A]. 2 = 9860 ft
[B]. 2 = 3000 ft
[C]. 2 = 1500 ft
[D]. 2 = 8510 ft

Answer: Option D

Explanation:

No answer description available for this question.

Raplh said: (Oct 21, 2014)  
Can anyone answer this? Please.

M Kumara Swamy said: (Jun 1, 2016)  
To find the radius of curvature first you have to find the equation of path in terms of x and y in the cartesian plane:

Here x = va * t.
y = 0.5 gt^2.
Overall y = 0.5(g/v^2) * x^2.

Now use formula:

Radius of curvature = ([1 + (dy/dx)^2]^1.5)/second order derivative of y.

I got 8300 as an answer because I took g = 10 for my convenience of calculation.

Minam Tungekar said: (Sep 4, 2020)  
According to me, the solution is;

x-direction equation of motion :
s = ut + 1/2 at^2 ( u = 150 ft/s, a in x-direction is 0 as the plane is moving with a constant velocity, s = x)
x = 150t;
t = x/150;

y-direction Equation of motion :
s = ut + 1/2 at^2 ( here "u" in the y-direction is 0 because as soon as the package is released it will have a velocity tangential to the falling projectile curve it will produce, a is only due to gravitational force g = 32.2 slug/ft^2)
y = 1/2 * 32.2 * t^2 ;
y = (16.1/22,500) * x^2;
y' = (16.1/22,500) * 2x;
y'' = (16.1/22,500) * x;

The radius of curvature, R at x=0( just after release from plane at point A) = [ [1 + {(y')^2}]^(3/2) / y''].
R = 698.75 ft.

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