the velocity vector, be it a change in speed or a change in direction, requires a force to create the change.
Consider a ball rolling along a straight line. One could constantly tap the ball with a stick, forcing it to move in a circular path. The tapping force, always pointing to the center, is changing the ball’s velocity vector’s direction. The ball is moving at a constant speed but changing direction; the ball is said to be accelerating toward the center of the circle.
Consider a 1-lb (0.45-kg) block rotating on the end of a 4-foot (1.2-m) string in a horizontal plane at a constant speed of 20 ft/sec (6 m/s). The direction of the velocity vector, always perpendicular to the string, is constantly changing and creating acceleration toward the center of rotation (Figure 7.1).
Acceleration for circular motion equals the velocity squared divided by the radius of the circle, or 100 ft/sec2 (30.5 m/s2). Per Newton’s Law, the string must exert a force on the block equal to m × a, or a force of 3.1 lbs (13.8 N) toward the center of rotation. (Recall that to correctly calculate f m × a the weight must be converted into a mass by dividing by the acceleration of gravity—32.2 ft/sec2 [9.8 m/s2].) The force the string exerts on the block is called the centripetal, or center-seeking, force. The block exerts an inertial load on the string, keeping it tight. The so-called centrifugal force is not a force; it’s the block’s inertial resistance to the centripetal acceleration. The 1-lb (0.45-kg) block resists the acceleration imposed by the string just as the person in the elevator
resists the upward acceleration. The term centrifugal force is incorrect. We will use the term centrifugal inertial loading. But, of course, the so-called centrifugal force feels like a force when holding on to the string attached to the rotating block.
A locomotive moving around a curve is similar to a rotating block on the end of a string. Both experience acceleration toward the center of rotation. The inertial loading keeps the string tight and creates a lateral force on the locomotive at the wheels. Lateral forces between the wheels and the rail must react against the centrifugal inertial loading to keep the train on the tracks.
If the centrifugal inertial loading is excessive, the locomotive begins to tip. The ﬂange of the wheel catches on the rail and the locomotive starts to rotate, as shown in Figure 7.2. In fact, that’s why the ﬂanges are on the inside of the wheels. If the ﬂanges were on the outside, the slightest bit of wheel lift would slide the locomotive off the tracks.
In the 1947 Pennsylvania Railroad overturning accident, the locomotive weighed 320,000 lbs (145,150 kg). The centrifugal inertial loading of the locomotive moving at 88 ft/sec (60 mph [97km/h]) on a curve with a radius of 675 feet (206 m) is:
Horrible number photo
The centrifugal inertial loading is trying to tip the locomotive clock-wise about the pivot point (the bottom of the right wheel). This rotation is resisted by the weight of the locomotive (also acting through its center of gravity), which tries to rotate the locomotive counterclockwise.
The locomotive’s weight and inertial load both exert a torque. A torque is a twisting force applied to the end of a lever arm that tries to tighten a nut. A 10-lb (44.5-N) force on the end of a 9-inch (23-cm)-long wrench exerts a torque of 10 × 9 = 90 inch lbs of torque (10 Nm).
The Pennsylvania locomotive had a center of gravity 80 inches (2 m) above the rail. The centrifugal inertial loading tries to rotate the locomotive with a clockwise torque equal to 114,000 lbs × 80 inches—more than 9 million inch lbs of torque (6.3 × 106 Nm).