Lifting Force

Now consider Figure 07.05.02 which illustrates the effects of imposing higher rotational speeds on a mass. The radius of 24 cm (R24) and of 16 cm (R16) are now each propelled at the higher rate of 6 m/s (V6). The inward "centripetal" acceleration is correspondingly greater and is given by the equation A = 62 / 0.24 which works out at about 150 m/s2 (A150) and about 225 m/s2 (A225) respectively.

In both of these cases, the centrifugal force is substantially greater than the gravitational force (shown as the short green near-vertical vector marked as G9.8) and so the resulting net forces (shown in blue in the diagram) are much closer to the horizontal than before. These masses will therefore rotate at a constant height when moving along the inner face of a cone which has much steeper walls (shown in grey).

The lowest diagram of Figure 07.05.02 shows the situation where these forces press against a less steeply sloping wall (shown in grey). The wall resists this pressure by pressing back at right angles to its surface (dark green vectors). Consequently, the remainder of the nearly horizontal centrifugal force produces an upward component (H20 and H30, shown in red), parallel to the sloping face of the wall. Depending on the speed of the mass and the angle of inclination of the wall, this upward force causes an acceleration of the mass, upwards along the wall. In these examples, that acceleration is about 20 to 30 m/s . In our example of coffee being stirred in a cup, the faster the stirring and the more angled the sides of the cup, the larger the amount of coffee which spills over the lip of the cup. Notice that part of this centrifugal force becomes a component which acts in a direction opposite to gravity. In our example, the 6 m/s (six revolutions per second or 360 rpm), produces a lifting-force which is much greater than the force of gravity.

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