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Centripetal Force and Circular Motion: Exit Ramps and Loop-the-Loop, Study Guides, Projects, Research of Physics

The concept of centripetal force, which keeps a body in uniform circular motion. It discusses the magnitude and direction of centripetal force, and the forces acting on a car moving on a curved exit ramp and an aircraft performing a loop-the-loop maneuver. The document also covers the maximum velocity before slippage on an exit ramp and the stall condition for an aircraft.

What you will learn

  • What forces act on a car moving on a curved exit ramp, and how does the frictional force prevent slippage?
  • What is centripetal force and how does it keep a body in circular motion?
  • What forces does an aircraft experience during a loop-the-loop maneuver, and how does the pilot avoid a stall?

Typology: Study Guides, Projects, Research

2021/2022

Uploaded on 09/12/2022

eekanath
eekanath 🇺🇸

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CHAPTER 6 CIRCULAR MOTION
CENTRIPETAL FORCE
The force F necessary to keep a body in uniform circular motion is defined as the
centripetal force. The magnitude of the force is F = m v2/r and it is directed to the center
of rotation. If F were not present object m would move along it’s velocity vector v.
F can be produced by gravitational attraction,
a string, or a roadbed pushing on a tire.
EXIT RAMP (see examples 6-4 and 6-5)
A curved exit ramp is normally inclined to facilitate a higher speed of exit with no
slippage. Consider a car of mass m moving with velocity v along a curved path of radius
R. There is static friction µk between the tire and road. This frictional force f = µk N must
oppose the outward m v2/r force. Then the maximum velocity before slippage is
mv2/r = f = µk N = µk mg
vmax = ( r g µk) 1/2
If the exit ramp is inclined the forces
opposing slippage can be increased.
mv2/r = N sinθ + f cosθ
mg = N cosθ
We can determine the banking angle under extreme conditions f = 0 (ice!!), thus not
relying on frictional forces.
m v2/r = N sinθ
m g = N cosθ
tanθ = v2 /r g
or
vmax = ( r g tan θ ) 1/2
v
F
r
m
r
v
m
ˆF=mv2/r
f
mv2/r
mg
f
N
θ
N cosθ
N sinθ
f cosθ
pf2

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CHAPTER 6 CIRCULAR MOTION

CENTRIPETAL FORCE

The force F necessary to keep a body in uniform circular motion is defined as the centripetal force. The magnitude of the force is F = m v^2 /r and it is directed to the center of rotation. If F were not present object m would move along it’s velocity vector v.

F can be produced by gravitational attraction, a string, or a roadbed pushing on a tire.

EXIT RAMP (see examples 6-4 and 6-5) A curved exit ramp is normally inclined to facilitate a higher speed of exit with no slippage. Consider a car of mass m moving with velocity v along a curved path of radius R. There is static friction μk between the tire and road. This frictional force f = μk N must

oppose the outward m v^2 /r force. Then the maximum velocity before slippage is

mv^2 /r = f = μk N = μk mg

vmax = ( r g μk) 1/

If the exit ramp is inclined the forces opposing slippage can be increased.

mv^2 /r = N sinθ + f cosθ mg = N cosθ

We can determine the banking angle under extreme conditions f = 0 (ice!!), thus not relying on frictional forces.

m v^2 /r = N sinθ m g = N cosθ

tanθ = v^2 /r g or vmax = ( r g tan θ ) 1/

v

F

r

m

r

v

m

f ˆF=mv^2 /r

mv^2 /r

mg

f

N

θ

N cosθ

f cos θ N sinθ

LOOP-THE-LOOP (see example 6-7) Consider a pilot making a vertical loop in an airplane at speed constant speed v. What forces does he aircraft experience? How does he avoid a stall at position A? Let P be the force of lift on the aircraft wing ( the same as the normal force on the pilot in example 6-7). Zero net force insures uniform motion at each location. P changes direction to balance the weight and centripetal force.

A: P + mg = mv^2 /r P = mv^2 /r – mg (lightest) B: P = ( (mv^2 /r )^2 + (mg)^2 )1/ C: P = mv^2 /r + mg P = mv^2 /r + mg (heaviest) D: P = ( (mv^2 /r )^2 + (mg)^2 )1/

A stall condition occurs If P=0 (freefall) at the top (A). Then mv^2 /r =mg

v > (rg)1/2^ or stall

A

B

C

P mg

P

mg

mv^2 /r

mv^2 /r

P

mv^2 /r

mg

P

mg

mv^2 /r

D