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Accelerating or Braking on Turns /
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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!
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1985 J. Weston Walch, Publisher
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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 /
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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
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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
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1985 J. Weston Walch. Publisher
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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
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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?
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