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

d²  y /dx²= - 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/dt²  = a_x  = 0,                                             d ² y /dt ²  = a _y  = - 6

The total acceleration is constant and equal in magnitude to 6 m/sec².

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

a _n   = v² / ρ =   25 /6.94  = 3.6 m / sec²

and tangential acceleration