

Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
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.
Typology: Study Guides, Projects, Research
1 / 2
This page cannot be seen from the preview
Don't miss anything!


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
r
m
r
v
m
mv^2 /r
mg
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
P mg
mg
mv^2 /r
mv^2 /r
mv^2 /r
mg
mg
mv^2 /r