This is merely the result of the passenger's inertia - the tendency to resist acceleration. The passenger's lean is not an acceleration in itself but rather the tendency to maintain the state of motion while the car does the acceleration. The tendency of a passenger's body to maintain its state of rest or motion while the surroundings the car accelerate is often misconstrued as an acceleration.
This becomes particularly problematic when we consider the third possible inertia experience of a passenger in a moving automobile - the left hand turn. Suppose that on the next part of your travels the driver of the car makes a sharp turn to the left at constant speed. During the turn, the car travels in a circular-type path. That is, the car sweeps out one-quarter of a circle.
The friction force acting upon the turned wheels of the car causes an unbalanced force upon the car and a subsequent acceleration. The unbalanced force and the acceleration are both directed towards the center of the circle about which the car is turning. Your body however is in motion and tends to stay in motion. It is the inertia of your body - the tendency to resist acceleration - that causes it to continue in its forward motion. While the car is accelerating inward, you continue in a straight line.
If you are sitting on the passenger side of the car, then eventually the outside door of the car will hit you as the car turns inward. This phenomenon might cause you to think that you are being accelerated outwards away from the center of the circle. In reality, you are continuing in your straight-line inertial path tangent to the circle while the car is accelerating out from under you.
The sensation of an outward force and an outward acceleration is a false sensation. There is no physical object capable of pushing you outwards. You are merely experiencing the tendency of your body to continue in its path tangent to the circular path along which the car is turning. You are once more left with the false feeling of being pushed in a direction that is opposite your acceleration.
Any object moving in a circle or along a circular path experiences a centripetal force. That is, there is some physical force pushing or pulling the object towards the center of the circle. This is the centripetal force requirement. The word centripetal is merely an adjective used to describe the direction of the force.
We are not introducing a new type of force but rather describing the direction of the net force acting upon the object that moves in the circle. Whatever the object, if it moves in a circle, there is some force acting upon it to cause it to deviate from its straight-line path, accelerate inwards and move along a circular path.
Three such examples of centripetal force are shown below. As a car makes a turn, the force of friction acting upon the turned wheels of the car provides centripetal force required for circular motion. As a bucket of water is tied to a string and spun in a circle, the tension force acting upon the bucket provides the centripetal force required for circular motion.
As the moon orbits the Earth, the force of gravity acting upon the moon provides the centripetal force required for circular motion. The centripetal force for uniform circular motion alters the direction of the object without altering its speed. The idea that an unbalanced force can change the direction of the velocity vector but not its magnitude may seem a bit strange. How could that be? There are a number of ways to approach this question.
One approach involves to analyze the motion from a work-energy standpoint. Recall from Unit 5 of The Physics Classroom that work is a force acting upon an object to cause a displacement. The amount of work done upon an object is found using the equation. As the centripetal force acts upon an object moving in a circle at constant speed, the force always acts inward as the velocity of the object is directed tangent to the circle. This would mean that the force is always directed perpendicular to the direction that the object is being displaced.
The angle Theta in the above equation is 90 degrees and the cosine of 90 degrees is 0. Thus, the work done by the centripetal force in the case of uniform circular motion is 0 Joules. Recall also from Unit 5 of The Physics Classroom that when no work is done upon an object by external forces, the total mechanical energy potential energy plus kinetic energy of the object remains constant. So if an object is moving in a horizontal circle at constant speed, the centripetal force does not do work and cannot alter the total mechanical energy of the object.
For this reason, the kinetic energy and therefore, the speed of the object will remain constant. The force can indeed accelerate the object - by changing its direction - but it cannot change its speed.
In fact, whenever the unbalanced centripetal force acts perpendicular to the direction of motion, the speed of the object will remain constant. For an unbalanced force to change the speed of the object, there would have to be a component of force in the direction of or the opposite direction of the motion of the object.
A second approach to this question of why the centripetal force causes a direction change but not a speed change involves vector components and Newton's second law.
The following imaginary scenario will be used to help illustrate the point. Suppose at the local ice factory, a block of ice slides out of the freezer and a mechanical arm exerts a force to accelerate it across the icy, friction free surface. Last week, the mechanical arm was malfunctioning and exerting pushes in a randomly directed fashion. The various direction of forces applied to the moving block of ice are shown below.
For each case, observe the force in comparison to the direction of motion of the ice block and predict whether the force will speed up, slow down or not affect the speed of the block. Use vector components to make your predictions. Then check your answers by clicking on the button.
There is an unbalanced force in the same direction as the block's motion. Centripetal force F c is always perpendicular to the path and pointing to the center of curvature, because a c is perpendicular to the velocity and pointing to the center of curvature.
This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve. Figure 1. The frictional force supplies the centripetal force and is numerically equal to it. Centripetal force is perpendicular to velocity and causes uniform circular motion.
The larger the F c , the smaller the radius of curvature r and the sharper the curve. Figure 2 shows the forces acting on the car on an unbanked level ground curve. Friction is to the left, keeping the car from slipping, and because it is the only horizontal force acting on the car, the friction is the centripetal force in this case.
Thus the centripetal force in this situation is. Now we have a relationship between centripetal force and the coefficient of friction. Using the first expression for F c from the equation. The coefficient of friction found in Part 2 is much smaller than is typically found between tires and roads. The car will still negotiate the curve if the coefficient is greater than 0. Note that mass cancels, implying that in this example, it does not matter how heavily loaded the car is to negotiate the turn.
Mass cancels because friction is assumed proportional to the normal force, which in turn is proportional to mass. If the surface of the road were banked, the normal force would be less as will be discussed below.
Figure 2. This car on level ground is moving away and turning to the left. The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway.
Let us now consider banked curves , where the slope of the road helps you negotiate the curve. See Figure 3. Race tracks for bikes as well as cars, for example, often have steeply banked curves. For ideal banking , the net external force equals the horizontal centripetal force in the absence of friction.
The components of the normal force N in the horizontal and vertical directions must equal the centripetal force and the weight of the car, respectively. In cases in which forces are not parallel, it is most convenient to consider components along perpendicular axes—in this case, the vertical and horizontal directions. Figure 3 shows a free body diagram for a car on a frictionless banked curve. The only two external forces acting on the car are its weight w and the normal force of the road N.
A frictionless surface can only exert a force perpendicular to the surface—that is, a normal force. Because this is the crucial force and it is horizontal, we use a coordinate system with vertical and horizontal axes. Only the normal force has a horizontal component, and so this must equal the centripetal force—that is,. Because the car does not leave the surface of the road, the net vertical force must be zero, meaning that the vertical components of the two external forces must be equal in magnitude and opposite in direction.
That is, roads must be steeply banked for high speeds and sharp curves. Friction helps, because it allows you to take the curve at greater or lower speed than if the curve is frictionless. Figure 3. In both of these examples, if you take away the centripetal force break the string or drive on ice , the object would move in a straight line.
Most textbooks use the centripetal force term and use it correctly. However, it really isn't needed. They may also instead refer to centripetal acceleration as the acceleration of an object moving in a circle. Most textbooks write it like this:. This form does not include the vector nature of the acceleration. The direction of this acceleration should always be pointing towards the center of the circle. This term is much less common in introductory physics texts.
It is not a real force according to my definition of real above. It does, however, have a useful role in some physics. Let me start with an example. Suppose a boy or girl, it really could be a girl too was sitting on a stationary merry-go-round with a ball. He then rolls the ball across to a girl on the other side. No problem, right? Speaking of merry-go-round, they just don't put these things in parks anymore do they?
Anyway, here is a diagram. In this case, with the merry-go-round not rotating, the ball goes straight across to the girl. No problem. Now suppose the merry-go-round is spinning. If the boy tries to roll the ball straight across to the girl, the ball will indeed still go in a straight line assume small frictional forces as seen from someone not on the merry-go-round.
The ball will not make it to the girl because she will rotate out of the way.
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