**Determine the radius of curvature - motion of a particle:**

The motion of a particle in XOY plane is defined by the equation

r (t ) = 3t i^ + (4t - 3t ^{2} ) j^

The distances are in metres. Determine its radius of curvature and its acceleration while it crosses the x axis again.

**Solution**

We have x = 3t, y = (4t - 3t ^{2} )

∴ t = x/3 , ∴ y = 4 x/3 - x ^{2}/3

∴ The equation is a second degree curve and if we equate it to zero, we shall get two values of x.

The path crosses the x axis at x = 0, y = 0 and t = 0 second, x = 4, y = 0, t = 4 /3 second as shown in Figure.

The radius of curvature is attained as below.

1/ ρ = ± (d ^{2} y/ d x ^{2}) / [1+ (dy/dx)^{2}] ^{(3/2)}

y = 4 /3 x - x ^{2} /3

dy / dx = 4/3 - (2/3) x

and

d^{2 } y /dx^{2}= - 2/3

∴ 1/ ρ = ± (2/3) / [ 1+ ((4/3)-(2/3))^{2}]^{(3/2)}

at x = 0 or at x = 4 m

∴ 1/ ρ = ± (2/3) / [ 1+ ((4/3)^{ 2}]^{(3/2) }∴ 1/ ρ = ± (2/3) / [ 1+ ((-4/3)^{ 2}]^{(3/2)}

±( 2 /3) /(25/9)^{(3/2)}; ±( 2 /3) /(25/9)^{(3/2)}

= 18/125 ; = 18/ 125

ρ = 6.94 m ρ = 6.94 m

We have, x = 3t y = 4t - 3t^{ 2}

∴ v_{x} = 3 m / sec ∴ v _{y } = 4 - 6t m/ sec.

∴ for t = 0, v_{x} = 3 m/sec., v_{y} = 4 m/sec.

Differentiating further, we obtain

d ^{2} x/dt^{2} = a_{x} = 0, d ^{2} y /dt ^{2 } = a_{ y} = - 6

The total acceleration is constant and equal in magnitude to 6 m/sec^{2}.

At both of instants t = 0 and t = 4/3 seconds. The normal acceleration may be found as

a _{n} = v^{2} / ρ = 25 /6.94 = 3.6 m / sec^{2}

and tangential acceleration