Transcript 7.2 Applications of Circular Motion

7.2 Applications of Circular Motion
1. Newton's Second law - Revisited
2. Amusement Park Physics
The Centripetal Force is
not a new type of force.
It is the NET force
Examples of centripetal force
As a car makes a turn, the
force of friction acting
upon the turned wheels of
the car provides
centripetal force required
for circular motion.
As a bucket of water is tied to
a string and spun in a circle,
the tension force acting upon
the bucket provides the
centripetal force required for
circular motion.
As the moon orbits the
Earth, the force of
gravity acting upon the
moon provides the
centripetal force
required for circular
motion.
Newton's Second Law - Revisited
Where Fnet is the sum (the resultant) of all forces acting on the object.
Newton's second law was used in combination of circular motion equations to analyze a
variety of physical situations.
Note:
centripetal force is the net force!
Steps in solving problems involving forces
1. Drawing Free-Body Diagrams
2. Determining the Net Force from Knowledge
of Individual Force Values
3. Determining Acceleration from Knowledge
of Individual Force Values
Or Determining Individual Force Values from
Knowledge of the Acceleration
Example 1
• A 945-kg car makes a 180-degree turn with a speed of 10.0 m/s. The
radius of the circle through which the car is turning is 25.0 m. Determine
the force of friction and the coefficient of friction acting upon the car.
Given: m = 945 kg; v = 10.0 m/s; R = 25.0 m
Find: Ffrict = ? μ = ?
Ff = Fnet = m*v2/R
Ff = (945 kg)*(10m/s)2/25m
Ff = 3780 N
Ff = μFNorm; FNorm = mg;
Ff = μmg
3780 N = μ(9270 N);
μ = 0.41
FNorm
Ff
Fgrav
Example 2
• The coefficient of friction acting upon a 945-kg car is 0.850. The car is
making a 180-degree turn around a curve with a radius of 35.0 m.
Determine the maximum speed with which the car can make the
turn.
Given: m = 945 kg; μ = 0.85; R = 35.0 m
Find: v = ? (the minimum speed would be the speed achieved with the
given friction coefficient)
FNorm
Ff = Fnet
Ff = μFN = μFN ;
Ff = (0.85)(9270N) = 7880 N
Fnet = m*v2/R
7880 N = (945 kg)(v2) / (35.0 m); v = 17.1 m/s
Ff
Fgrav
Example 3
• A 1.50-kg bucket of water is tied by a rope and whirled in a circle with a
radius of 1.00 m. At the top of the circular loop, the speed of the bucket is
4.00 m/s. Determine the acceleration, the net force and the individual force
values when the bucket is at the top of the circular loop.
Fgrav = mg = 14.7 N
a = v2 / R = 16 m/s2
Fnet = ma = 24 N, down
Fnet = Fgrav + Ftens
24 N = 14.7 N + Ften
Ftens = 24 N - 14.7 N = 9.3 N
m = 1.5 kg
a = ________ m/s/s
Fnet = _________ N
Example 4
• A 1.50-kg bucket of water is tied by a rope and whirled in a circle
with a radius of 1.00 m. At the bottom of the circular loop, the
speed of the bucket is 6.00 m/s. Determine the acceleration, the
net force and the individual force values when the bucket is at the
bottom of the circular loop.
Fgrav = m • g = 14.7 N
a = v2 / R = 36 m/s2
Fnet = ma = 54 N, up
Fnet = Ften - Fgrav
54 N = Ften – 14.7 N
Ften = 54 N +14.7 N = 68.7 N
m = 1.5 kg
a = ________ m/s/s
Fnet = _________ N
Example 5
• Anna Litical is riding on The Demon at Great America. Anna
experiences a downwards acceleration of 15.6 m/s2 at the top of the
loop and an upwards acceleration of 26.3 m/s2 at the bottom of the
loop. Use Newton's second law to determine the normal force acting
upon Anna's 864 kg roller coaster car at the top and the bottom of
the loop.
Given:
m = 864 kg
atop = 15.6 m/s2 , down
abottom = 26.3 m/s2 , up
Fgrav = m•g = 8476 N.
Example 6
• Anna Litical is riding on The American Eagle at Great America. Anna is
moving at 18.9 m/s over the top of a hill which has a radius of
curvature of 12.7 m. Use Newton's second law to determine the
magnitude of the applied force of the track pulling down upon Anna's
621 kg roller coaster car.
Given:
m = 621 kg; v = 18.9 m/s; R=12.7 m
Fgrav = m * g = 6086 N
a = v2 / R = (18.9 m/s)2 / (12.7 m) = 28.1 m/s2
Find:
Fapp at top of hill