Which force provides centripetal acceleration

Reading the above answers and the pertaining comments I have the impression that there is still some confusion regarding the explanation of this problem. Here I also use an inertial frame of reference like the other answers and I use the notation of D. Ennis' answer. In my prior answer below I used a co-rotational frame of reference.

What provides this force? The centripetal force is provided by the vector component of the gravitational force on the car normal to the hill surface which is in the direction of the center of the circular motion. This normal force minus the centrifugal force is equal to the normal force of the hill back on the car.

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A diagram would be nice. Add a comment. Active Oldest Votes. Here is a force diagram a. Improve this answer. Ennis D. Ennis 1, 6 6 silver badges 10 10 bronze badges.

However the answer doesn't make sense as it is. May be you want to think it over. That doesn't mean that I don't make mistakes, but I don't think that I have here. If the force keeping an object in rotation is broken, such as cutting the string that holds a spinning ball, the object will fly off in a straight line, following the tangential direction. This apparent outward force is described by Newton's Laws of Motion. Newton's First Law states that "a body at rest will remain at rest, and a body in motion will remain in motion unless it is acted upon by an external force.

If a massive body is moving through space in a straight line, its inertia will cause it to continue in a straight line unless an outside force causes it to speed up, slow down or change direction. In order for it to follow a circular path without changing speed, a continuous centripetal force must be applied at a right angle to its path.

Newton's Third Law states that "for every action, there is an equal and opposite reaction. When you are in an accelerating car, the seat exerts a forward force on you just as you appear to exert a backward force on the seat.

In the case of a rotating system, the centripetal force pulls the mass inward to follow a curved path, while the mass appears to push outward due to its inertia. In each of these cases, though, there is only one real force being applied, while the other is only an apparent force. There are many applications that exploit centripetal force. One is to simulate the acceleration of a space launch for astronaut training.

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. The car on this banked curve is moving away and turning to the left. Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked.

This banking, with the aid of tire friction and very stable car configurations, allows the curves to be taken at very high speed. To illustrate, calculate the speed at which a m radius curve banked at We first note that all terms in the expression for the ideal angle of a banked curve except for speed are known; thus, we need only rearrange it so that speed appears on the left-hand side and then substitute known quantities. Tire friction enables a vehicle to take the curve at significantly higher speeds.

Ask a friend or relative to swing a golf club or a tennis racquet. Take appropriate measurements to estimate the centripetal acceleration of the end of the club or racquet. You may choose to do this in slow motion. Move the sun, earth, moon and space station to see how it affects their gravitational forces and orbital paths. Visualize the sizes and distances between different heavenly bodies, and turn off gravity to see what would happen without it!

Figure 7. Two paths around a race track curve are shown. Race car drivers will take the inside path called cutting the corner whenever possible because it allows them to take the curve at the highest speed.

Figure 8. Amusement rides with a vertical loop are an example of a form of curved motion.