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Car in a Curve
Introduction
Lauren feels like she is being pushed to the right when she rounds a curve
to the left at a high speed. Instead of experiencing some mysterious centrifugal
force pulling on her, she is trying to move in a straight path as the car turns.
To make her go in a curved path, the seat (friction) or the door (direct
contact) exerts a force on her.
The car itself must have an inward force exerted on it to move in a curve.
On a flat roads this force is supplied by the sum of friction forces acting on
each tire. This is static friction as long a tires are not slipping. If the
tires are slipping, it is kinetic friction. On a banked curve, the force is sum
of friction and force parallel due to gravity.
If the driver locks the brakes, the kinetic instead of static coefficient
must be used. The kinetic coefficient of friction is lower than the static
coefficient of friction.
Table of Contents
Problems
| 1. |
A 2500. kg car rounds a flat curve of radius
75.0 m at a speed of 80.0 km/h. Will the car make the turn, or will it
skid if (a) the pavement is dry and the coefficient of static friction is
0.60; (b) the pavement is icy and
Coefficient of static friction is 0. 25?
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| 2. |
The same car rounds a banked curve at an
angle of 30.0o. What speed can the curve be driven without any
friction? This speed is called the "design speed." Radius of the
curve is 58.0 m.
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| 3. |
What is the coefficient of friction
necessary if the car in the previous problem is traveling 30.0 km/h
greater than the design speed?
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| 4. |
A 2000. kg car rounds a banked curve with a
radius of 150. m at 120. km/h. How much must the curve be banked to
keep the car from skidding without any friction? |
Table of Contents
Answers
| 1. |
A 2500. kg car rounds a flat curve of radius
75.0 m at a speed of 80.0 km/h. Will the car make the turn, or will it
skid if (a) the pavement is dry and the coefficient of static friction is
0.60; (b) the pavement is icy and
Coefficient of static friction is 0. 25?
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80.0 km
1 h 1000 m
----------- x ----------- x ------------- = 22.2 m s -1
h
3600 s 1 km
mv2
1500. kg x (22.2 m s -1)2
Fc = -------- = ---------------------------------- = 9860
N
r
75.0 m
(a) Ffr =  mg
= 0.60 x 2500. kg x 9.80 m s -2 = 14,700
N Since Ffr >
Fc the car will make the turn.
(b) Ffr = mg
= 0.25 x 2500. kg x 9.80 m s -2 = 6130
N
Since Ffr < Fc the car will skid.
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| 2. |
The same car rounds a banked curve at an
angle of 30.0o. What speed can the curve be driven without any
friction? This speed is called the "design speed." Radius of the
curve is 58.0 m.
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FP = mg x Sin 30.0o
Since there is no friction FP
= FC
m v2
FC = ---------- = mg x Sin 30.0o
m cancels
r
v = ( rg Sin 30.0o ) 1/2
= ( 58.0 m x 9.80 m s -2 x Sin 30.0o ) 1/2
v = 16.9 m/s = 60.8 km/hr
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| 3. |
What is the coefficient of friction
necessary if the car in the previous problem is traveling 30.0 km/h
greater than the design speed?
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FP = 2500. kg x 9.80 m s -2 x Sin 30.0o =
21200 N
FN = 2500. kg x 9.80 m s -2
x Sin 30.0o = 12300 N
60.8 km/h + 30.0 km/h = 90.8 km/h = 22.4 m/s
m v2 2500. kg x
(25.2 m s -1)2
FC = ---------- = ---------------------------------- = 27400 N
r
58.0 m
FC = FP + Ffr
=> Ffr = FC
- FP = 27400 N - 21200 N =
6200 N
Ffr 6200 N
Ffr = FN
=> =
----- = -------------- = 0.50
FN 12300 N
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| 4. |
A 2000. kg car rounds a banked curve with a
radius of 150. m at 120. km/h. How much must the curve be banked to
keep the car from skidding without any friction?
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FP = mg x Sin X
120. km 1
h 1000 m
------------ x ----------- x ------------ = 33.3 m/s
1
h 3600
s 1 km
Since there is no friction FP
= FC
m v2
FC = ---------- = mg x Sin
X m
cancels
r
v2
(33.3 m s -1)2
X = Sin -1 ( ------) = Sin -1 (
--------------------------- ) = 49.0o
rg
150. m x 9.80 m s -2
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Table of Contents
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