a car goes round the corner; (a)which provide the centripetal force ? .(B) is more force or less force needed if the car is 1 . heavier 2. faster 3. going round a tighter curve?. (c)what would happen to the car if there were no centripetal force?
Answers
Answer:
Any force or combination of forces can cause a centripetal or radial acceleration. 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 net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton’s second law of motion, net force is mass times acceleration: net F = ma. For uniform circular motion, the acceleration is the centripetal acceleration—a = ac. Thus, the magnitude of centripetal force Fc is Fc = mac.
By using the expressions for centripetal acceleration ac from
a
c
=
v
2
r
;
a
c
=
r
ω
2
, we get two expressions for the centripetal force Fc in terms of mass, velocity, angular velocity, and radius of curvature:
F
c
=
m
v
2
r
;
F
c
=
m
r
ω
2
.
You may use whichever expression for centripetal force is more convenient. Centripetal force Fc is always perpendicular to the path and pointing to the center of curvature, because ac is perpendicular to the velocity and pointing to the center of curvature.
Note that if you solve the first expression for r, you get
r
=
m
v
2
F
c
.
This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.
The given figure consists of two semicircles, one over the other. The top semicircle is bigger and the one below is smaller. In both the figures, the direction of the path is given along the semicircle in the counter-clockwise direction. A point is shown on the path, where the radius from the circle, r, is shown with an arrow from the center of the circle. At the same point, the centripetal force is shown in the opposite direction to that of radius arrow. The velocity, v, is shown along this point in the left upward direction and is perpendicular to the force. In both the figures, the velocity is same, but the radius is smaller and centripetal force is larger in the lower figure.
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 Fc, the smaller the radius of curvature r and the sharper the curve. The second curve has the same v, but a larger Fc produces a smaller r′.
EXAMPLE 1. WHAT COEFFICIENT OF FRICTION DO CAR TIRES NEED ON A FLAT CURVE?
Calculate the centripetal force exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 m/s.
Assuming an unbanked curve, find the minimum static coefficient of friction, between the tires and the road, static friction being the reason that keeps the car from slipping (see Figure 2).
Strategy and Solution for Part 1
We know that
F
c
=
m
v
2
r
. Thus,
F
c
=
m
v
2
r
=
(
900
kg
)
(
25.0
m/s
)
2
(
500
m
)
=
1125
N
.
Strategy for Part 2
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. We know that the maximum static friction (at which the tires roll but do not slip) is μsN, where μs is the static coefficient of friction and N is the normal force. The normal force equals the car’s weight on level ground, so that N=mg. Thus the centripetal force in this situation is
Fc = f = μsN = μsmg.
Now we have a relationship between centripetal force and the coefficient of friction. Using the first expression for Fc from the equation
{
F
c
=
m
v
2
r
F
c
=
m
r
ω
2
,
m
v
2
r
=
μ
s
m
g
We solve this for μs, noting that mass cancels, and obtain
μ
s
=
v
2
r
g
.
Solution for Part 2
Substituting the knowns,
μ
s
=
(
25.0
m/s
)
2
(
500
m
)
(
9.80
m/s
2
)
=
0.13
.
(Because coefficients of friction are approximate, the answer is given to only two digits.)
Discussion
We could also solve Part 1 using the first expression in
{
F
c
=
m
v
2
r
F
c
=
m
r
ω
2
because m,v, and r are given. 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 0.13, because static friction is a responsive force, being able to assume a value less than but no more than μsN. 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. 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