You feel as if you are thrown (that is, forced) toward the left relative to the car. An even more common experience occurs when you make a tight curve in your car-say, to the right ( Figure 6.24). Yet a physicist would say that you tend to remain stationary while the seat pushes forward on you. When taking off in a jet, most people would agree it feels as if you are being pushed back into the seat as the airplane accelerates down the runway. What do taking off in a jet airplane, turning a corner in a car, riding a merry-go-round, and the circular motion of a tropical cyclone have in common? Each exhibits inertial forces-fictitious forces that merely seem to arise from motion, because the observer’s frame of reference is accelerating or rotating. Inertial Forces and Noninertial (Accelerated) Frames: The Coriolis Force This simplified model of a carousel demonstrates this force. Only the normal force has a horizontal component, so this must equal the centripetal force, that is,Ī circular motion requires a force, the so-called centripetal force, which is directed to the axis of rotation. Because this is the crucial force and it is horizontal, we use a coordinate system with vertical and horizontal axes. (A frictionless surface can only exert a force perpendicular to the surface-that is, a normal force.) These two forces must add to give a net external force that is horizontal toward the center of curvature and has magnitude m v 2 / r. The only two external forces acting on the car are its weight w → w → and the normal force of the road N →. If the angle θ θ is ideal for the speed and radius, then the net external force equals the necessary centripetal force. 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.įigure 6.22 shows a free-body diagram for a car on a frictionless banked curve. 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. If the surface of the road were banked, the normal force would be greater, as discussed next.įigure 6.22 The car on this banked curve is moving away and turning to the left.įor ideal banking, the net external force equals the horizontal centripetal force in the absence of friction. Mass cancels because friction is assumed proportional to the normal force, which in turn is proportional to mass. Note that mass cancels, implying that, in this example, it does not matter how heavily loaded the car is to negotiate the turn. A higher coefficient would also allow the car to negotiate the curve at a higher speed, but if the coefficient of friction is less, the safe speed would be less than 25 m/s. The car still negotiates the curve if the coefficient is greater than 0.13, because static friction is a responsive force, able to assume a value less than but no more than μ s N. SignificanceThe coefficient of friction found in Figure 6.21(b) is much smaller than is typically found between tires and roads. (Because coefficients of friction are approximate, the answer is given to only two digits.) Thus, the magnitude of centripetal force F c F c is For uniform circular motion, the acceleration is the centripetal acceleration. According to Newton’s second law of motion, net force is mass times acceleration: F net = m a. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. Any net force causing uniform circular motion is called a centripetal force. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, friction between roller skates and a rink floor, a banked roadway’s force on a car, and forces on the tube of a spinning centrifuge. Any force or combination of forces can cause a centripetal or radial acceleration. This acceleration acts along the radius of the curved path and is thus also referred to as a radial acceleration.Īn acceleration must be produced by a force. Angular velocity gives the rate at which the object is turning through the curve, in units of rad/s.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |