Rolling

Example: A hula hoop (shown below) of radius, R, rolls to the right with a constant velocity, v_C i ft/sec. At the instant shown point B on the hoop is to the right of the center, C, of the hula hoop.

Point A is the point of contact of the hula hoop with the horizontal surface. Find | dv_B/dt | , the

radius of curvature of point B at the instant shown, and the location of the center of curvature.

Strategy: Apply the relative velocity and relative acceleration equations between points C and A

and between points C and B. Find the unit tangential vector at B from the velocity of point B. Use

it to find the tangential component of acceleration.

Solution: v_C = v_A + v_C/A = 0 + ˗ ω k x R j = R ω i So v_C = R ω , ω = v_C / R

Since v_C = constant, a_C = dv_C/dt = 0 and also α = dω/dt = 0

v_B = v_C + v_B/C = v_C i + ˗ ω k x R i = v_C i + ˗ R ω j = v_C i + ˗ v_C j

So v_B = v_C ( i ˗ j ) Then e_t = v_B / | v_B | = ( i ˗ j ) / √ 2 (unit tangential vector at B)

a_B = a_C + a_B/C |_n + a_B/C |_t = 0 ˗ ω² R i + α k x R i = ˗ ω² R i

Now a_B|_t = | dv_B/dt | = a_B • e_t = ˗ ω² R i • ( i ˗ j ) / √ 2 = ˗ R ω² / √ 2 (result)

By vector addition, a_B = a_{B n} + a_{B t}

a_B² = a_Bn² + a_Bt² = (˗ R ω² )² = ( v_B² / ρ )² + (˗ R ω² / √ 2 )²

( 2v_C ² )²/ ρ² = R² ω⁴ / 2 but v_C = R ω So 4 (Rω)⁴ / ρ² = R² ω⁴ / 2

4 ( R / ρ )² = 1/2 Therefore ρ = √ 8 R = ( 2√ 2) R (result)

Note: The center of curvature for point B, at this instant, is a distance of √ 8 R from B along the

line from B to and beyond A.

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