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Accelerating or Braking on Turns / 27 Accelerating or· Braking on Turns The physics of driving a car around a corner has been discussed from one point of view in the section on "Going Around Corners and Friction", but some Qther interesting physics are associated with the forces exerted on cars on turns. A car must have a centripetal force exerted on it by friction to make it go around the corner. If the car accelerates or decelerates on a tum, a force parallel to the direction of travel must also be exerted. This force must be added to the centripetal force to find the total force that friction must exert to prevent a skid. The direction of centripetal force is towards the center of a circle, or at right angles to the direction of trauel. The forces needed to brake or to accelerate the car are directed parallel to the direction of travel. Because centripetal force and the force of acceleration are at right angles to each other, the total force is a vector sum. Direction is important when adding vectors, but the addition of these two vectors is easy because the vectors are directed at right angles to each other. The centripetal force is given by the equation: f= m(v2)/r, wherefis the force, m is the mass, vis velocity and ris the radius:. The centripetal force is supplied by friction; the frictional force has a maximum value given by: f = Ilmg, where f is the frictional force, mg is the gravitational force directed perpendicular to the .,. ground, and Il is the friction coefficient. The acceleration force from either braking or putting a foot on the accelerator is given by: f~ma, fc fe Ag. 6.1 Addition of linear and centripetal forces ,. where/is the force, m is the mass, and a is the accelera tion caused either by put~ing the foot on the brake or on the accelerator. When these forces are added together, their sum is lar-ger than either of the "component" forces. (See Figure 6.1.) If this sum is larger than the maximum frictional force that can be exerted, then the car will skid on the turn and may run off the road. p. is the coefficient of friction, and is relative to the road surface and the condition of the tires. Because the tires and the road surface can't be changed, if the car goes around a corner too fast, or brakes or accelerates too much, the friction will not be great enough to exert the entire force. The car will skid! Therefore, it is dangerous to brake on a turn if you are traveling fast to start with. You should brake before entering a turn or the total force needed to keep the car on the turn may be more than friction can provide, and you will be in trouble! @ 1985 J. Weston Walch, Publisher , Going Around Corners and Friction Everyone has experienced centripetal force in one way or another. Centripetal force allows you, or forces you, in certain circumstances, to go around corners. Without centripetal force, or a force towards the center of a circle, you would instead travel in a straight line. The centripetal force for a car going around a corner is provided by the frictional force. Friction is also needed in a "rotor" machine to prevent you from slipping out, while centripetal force is needed to make you go in a circle. (A rotor is a ride found in many amusement parks. A large cylinder turns on a vertical axis. You get into it and stand next to the wall. The rotor starts turning and when it is spinning fast enough, the floor drops and you are, "stuck" to the wall.) Some force is needed to make any object move in a circle, because velocity changes with direction. Velocity has both direction and magnitude. As a car goes around a corner, because its direction changes, the velocity must change. A change in velocity requires acceleration and it is centripetal force which causes this "centripetal" acceleration. Centripetal force is needed to make anything go in a circle, but it does not exist by itself; something must cause it. When a car goes around a corner, friction causes the centripetal force. Without friction (on an icy road, for example), a car will not go around a turn. The equation for centripetal force is given as: f =mv2jr, where f is the force, m is the mass, u is the velocity and r is the radius of the circle. Increasing the velocity or decreasing the radius of the turn makes it harder to go around a corner without skidding-your experience should tell you this. The frictional force, which provides the sentripetal forc£, is directed parallel to the sur face of the road. Friction is caused by a force perpendicular to the surfaces, and is given by the equation: where the frictional force, f, equals the coeffi cient of friction (Jl) times the perpendicular force. The coefficient of friction is a constant which is related to the surfaces which are in con tact with one another. (See Figure 5. l.)(Generally, ff = force of friction fg = force of gravity fe. = centripetal force Fig. 5.1 Car going around a circle © 1985 J. Weston Walch. Publisher Going Around Corners and Friction / 23 the rougher the surface, the higher the friction coefficient. For example, a dry road has a higher coefficient of friction than an icy road.) Because friction must supply the centripetal force for a car rounding a corner, the faster the car goes, the greater the frictional force must be. If more force is needed than friction can supply, the car will go in a circle of larger radius than the driver may wish; Le., it will skid. Because the frictional force must equal the centripetal force, we may write the equation: fc = ff, but fr = Ilfp, so: ~ fc = Ilfp, but fp = fg = mg, so: mv2 = Ilmg; rearranging this, we get: r Il =v2/gr - This is the minimum value of the friction coefficient because there is only a certain maximum force which friction can provide. If more force is needed than friction can provide, the car will skid. In a rotor machine, the rider moves in a circle without slipping down the wall. Friction opposes the grav ity which would otherwise pull the rider down the wall after the floor drops out. Therefore, the force of friction must be equal to the force of gravity, i.e., /g = It- Friction is caused by a force perpendicular to the surfaces. The centripetal force exerted by the wall of the machine on the person's back does just this. (See Figure 5.2.) If the wall of the machine were to break (which is doubtful), the person would continue in a straight line, and go through the wall! Now, the centripetal force can be set equal to the perpendicular force. By substituting into the above equation: f1 = force of friction +p = fw =force of the wall fg =force of gravity Fig. 5.2 Person in a rotor @ 1985 J. Weston Walch. Publisher 24 / Going Around Corners and Friction Dividing by mass, m, the coefficient of friction is: J.I. = gr/v 2 This, too, is a minimum value for the coefficient of friction. Without it, the frictional force will not completely combat gravity and the rider will slip down ... ! In both cases-a car going around a corner and a person in a rotor-think about what would happen if friction were not present. Less friction than a certain value will allow the car or the person to slip. This is why the coefficient of friction must be a minimum value. © 1985 J. Weston Walch. Publisher '- NAME____________________________ DATF~ ________________ Going Around Corners and Friction Questions and Problems 1. If the radius of a turn is 50 meters and a car with a mass of 1,000 kg is going at 20 m/sec (about 40 mph), what is the centripetal force e~erted on the car? 2. What is the minimum coefficient of friction required to enable the car in problem 1. to go around the corner? 3. Given the same radius and speed as in problem 1, what would the minimum coefficient of friction'be for a 2,000 kg truck? 4. Does the amount of mass matter when you are looking for the minimum coefficient of friction needed to go around a corner? (You might try vehicles of other masses if you are in doubt.) 5. How does the minimum coefficient of friction change when the radius of the turn is made larger? (You might try changing the radius given for problem 2 to 100 lneters.) 6. What forces must be equal to enable a rider to stay inside the rotor after the floor has dropped out? 7. If a rotor which has a radius of 5 meters is carrying a person with a 60 kg mass , (about 130 pounds) and is traveling at 5 m/sec, what is the minimum coefficient of friction needed to keep the person up on the wall? 8. What effect does your mass have on your ability to stay in the rotor? NAME __________________________ DATE_______________ Accelerating or Btaking on Turns Questions and Problems 1. A 1,000 kg car is going around a comer with a velocity of 20 m/sec (about 40 mph). What is the centripetal force on the car if the radius of the tum is 100 meters? 2. Rnd the minimum coefficient of friction for the car in problem 1. 3. . What force would the brakes have to exert in order to slow the car in problem 1 in order to decelerate ~t at 4 m/sec 2? '-."--:'-:-.'-. 4. What is the vector sum of the forces that the friction between the tires and the road would have to provide to keep the car of problem 1 going around the comer while . braking at a rate of 4 m/sec 2? ,. S. What would the minimum coefficient of friction have to be to provide the total force required to keep the car on the tum if the gravitational force is given by:! = mg, and the frictional force is given by:! = +mg? -'\ i' 6. What would the minimum coefficient of friction be if only the centripetal force for the car in problem 1 were needed? ' 7. What coefficient of friction is needed to allow a 1,000 kg car to go around a 25-rneter-radius tum at 10 m/sec while accelerating at a rate of 3 m/sec2? . 8. A car like the one in problem 1 is on the moon, where the acceleration due to gravity is about 1.6 m/sec 2 • If it goes around a lOO-meter-radius tum at 20 m/sec, what coefficient of friction would be needed to keep the car on the tum? 9. Compare the results of problem 1 and problem 8. Do these results agree with what you would expect? Why or why not? ~.. I