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Kinematics of a Particle in a Plane (using intrinsic coordinates

Example: Sally is walking left to right with a constant speed, v_o, in the x-direction along the

curve shown below. The "hill" she walks up is given by y(x) = 8a³ / ( 4a² + x² ) . Find Sally's acceleration when she reaches the top of the hill. i.e. at (0, 2a).

Strategy: Apply the definitions for velocity and acceleration using intrinsic coordinates.

v(t) = v e_t and a(t) = ( dv/dt) e_t + ( v²/ρ) e_n Note: At (0, 2a) e_t = i and e_n = ˗ j

v(top) = v i and a(top) = ( dv/dt) i ˗ ( v_o²/a) j = 0 i ˗ ( v_o²/a) j (result)

Check using rectangular coordinates: v(t) = dx/dt i + dy/dt j Now dx/dt = v_o and d²x/dt² = 0

dy/dt = (dy/dx) (dx/dt) and (d²y/dt²) = (d²y/dx²)(dx/dt)(dx/dt) + (dy/dx) (d²x/dt²)

(d²y/dt²) = (d²y/dx²)(dx/dt)² + (dy/dx) (d²x/dt²) = (d²y/dx²)(dx/dt)²

Now y(x) = 8a³ / ( 4a² + x² ) So dy/dx = [ ˗ 8a³ ( 4a² + x² ) ^˗2 ] 2x = ˗ 16 a³ x ( 4a² + x² ) ^˗2

d²y/dx² = [ ˗ 16a³ ( 4a² + x² ) ^˗2 ] + 32 a³ x ( 4a² + x² ) ^˗3 (2x)

d²y/dx² = [ ˗ 16a³ ( 4a² + x² ) ^˗2 ] + 64 a³ x² ( 4a² + x² ) ^˗3

d²y/dt² = { [ ˗ 16a³ ( 4a² + x² ) ^˗2 ] + 64 a³ x² ( 4a² + x² ) ^˗3 } v_o²

At the top of the hill x = 0 d²y/dt² = ˗ 16a³ ( 4a² ) ^˗2 v_o² = ˗ v_o² / a

and a(top) = d²x/dt² i + d²y/dt² j = ˗ v_o² / a j (same result)

Click here for discussion of center of curvature and the "osculating" circle.

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