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Circular Motion & Highway Curves
Highway Curves: Banked & Unbanked
Case 1 - Unbanked Curve: When a car rounds a curve,
there MUST be a net force toward the circle center (a
Centripetal Force) of which the curve is an arc. If there weren’t
such a force, the car couldn’t follow the curve, but would (by
Newton’s 2nd Law) go in a straight line. On a flat road, this
Centripetal Force is the static friction force.
“Centripetal Force”
No static friction?
 No Centripetal Force
 The Car goes straight!
There is NEVER a
“Centrifugal Force”!!!
=
Centripetal Acceleration: Car on a Curve
• A car rounding a curve travels in an
approximate circle. The radius of this
circle is called the radius of curvature.
• Forces in the y-direction
– Gravity and the normal force
• Forces in the x-direction
– Friction is directed toward the center of the circle
• Since friction is the only force acting in the xdirection, it supplies the centripetal force:
Centripetal Acceleration: Car on a Curve
• A car rounding a curve travels in an
approximate circle. The radius of this
circle is called the radius of curvature.
• Forces in the y-direction
– Gravity and the normal force
• Forces in the x-direction
– Friction is directed toward the center of the circle
• Since friction is the only force acting in the xdirection, it supplies the centripetal force:
• Solving for the maximum velocity at which
the car can safely round the curve gives
Example: Skidding on a Curve
A car, mass m = 1,000 kg car rounds a curve on a Free Body Diagram
flat road of radius r = 50 m at a constant speed v =
14 m/s (50 km/h). Will the car follow the curve, or
will it skid? Assume:
a. Dry pavement with coefficient of static friction μs = 0.6.
b. Icy pavement with μs = 0.25.
Example: Skidding on a Curve
A car, mass m = 1,000 kg car rounds a curve on a Free Body Diagram
flat road of radius r = 50 m at a constant speed v =
14 m/s (50 km/h). Will the car follow the curve, or
will it skid? Assume:
a. Dry pavement with coefficient of static friction μs = 0.6.
b. Icy pavement with μs = 0.25.
Newton’s 2nd Law: ∑F = ma
x: ∑Fx = max  Ffr = maR = m(v2/r)
y: ∑Fy = may = 0  FN - mg = 0, FN = mg
Centripetal Force: m(v2/r) = 3900 N
Example: Skidding on a Curve
A car, mass m = 1,000 kg car rounds a curve on a Free Body Diagram
flat road of radius r = 50 m at a constant speed v =
14 m/s (50 km/h). Will the car follow the curve, or
will it skid? Assume:
a. Dry pavement with coefficient of static friction μs = 0.6.
b. Icy pavement with μs = 0.25.
Newton’s 2nd Law: ∑F = ma
x: ∑Fx = max  Ffr = maR = m(v2/r)
y: ∑Fy = may = 0  FN - mg = 0, FN = mg
Centripetal Force: m(v2/r) = 3900 N
The maximum static friction is Ffr = μsFN
a. μs = 0.6. Maximum Ffr = 5900 N
(It stays on the curve!)
Example: Skidding on a Curve
A car, mass m = 1,000 kg car rounds a curve on a Free Body Diagram
flat road of radius r = 50 m at a constant speed v =
14 m/s (50 km/h). Will the car follow the curve, or
will it skid? Assume:
a. Dry pavement with coefficient of static friction μs = 0.6.
b. Icy pavement with μs = 0.25.
Newton’s 2nd Law: ∑F = ma
x: ∑Fx = max  Ffr = maR = m(v2/r)
y: ∑Fy = may = 0  FN - mg = 0, FN = mg
Centripetal Force: m(v2/r) = 3900 N
The maximum static friction is Ffr = μsFN
a. μs = 0.6. Maximum Ffr = 5900 N
(It stays on the curve!)
b. μs = 0.25. Maximum Ffr = 2500 N
(It skids off the curve!)
If the friction force isn’t sufficient, the car will tend to move more
nearly in a straight line (Newton’s 1st Law) as the skid marks show.
As long as the tires don’t slip, the
friction is static. If the tires start to
slip, the friction is kinetic, which is
bad in 2 ways!!
1. The kinetic friction force is
smaller than the static friction
force.
2. The static friction force
points toward the circle center,
but the kinetic friction force
opposes the direction of motion,
making it difficult to regain
control of the car & continue
around the curve.
Example: Car on Banked Curve
• The maximum speed can be increased by
banking the curve. Assume no friction
between tires & the road. So, the only forces
on the car are gravity & the normal force.
• The centripetal force is the horizontal
component of the normal force. For every
banked curve, there is one speed v at which the
Centripetal Force is the horizontal component
of the normal force N, so that no friction is
required!! Let the horizontal be the x-direction.
Example: Car on Banked Curve
• The maximum speed can be increased by
banking the curve. Assume no friction
between tires & the road. So, the only forces
on the car are gravity & the normal force.
• The centripetal force is the horizontal
component of the normal force. For every
banked curve, there is one speed v at which the
Centripetal Force is the horizontal component
of the normal force N, so that no friction is
required!! Let the horizontal be the x-direction.
Newton’s 2nd Law
x-direction: ∑Fx = maC  Nx = Nsin θ = m(v2/r) (1)
y-direction: ∑Fy = may = 0  Ny - mg = 0, but Ny = Ncosθ
So,  Ncosθ = mg, which gives N = (mg/cosθ)
(2)
Example: Car on Banked Curve
• The maximum speed can be increased by
banking the curve. Assume no friction
between tires & the road. So, the only forces
on the car are gravity & the normal force.
• The centripetal force is the horizontal
component of the normal force. For every
banked curve, there is one speed v at which the
Centripetal Force is the horizontal component
of the normal force N, so that no friction is
required!! Let the horizontal be the x-direction.
Newton’s 2nd Law
x-direction: ∑Fx = maC  Nx = Nsin θ = m(v2/r) (1)
y-direction: ∑Fy = may = 0  Ny - mg = 0, but Ny = Ncosθ
So,  Ncosθ = mg, which gives N = (mg/cosθ)
(2)
• Put (2) into (1):  g(sinθ/cosθ) = (v2/r) or tanθ = (v2/gr)
So,
Example: Banking angle
a. For a car traveling with speed v around a
curve of radius r, find a formula for the
angle θ at which a road should be banked
so that no friction is required.
b. Calculate this angle for an expressway
off-ramp curve of radius r = 50 m at a
design speed of v = 14 m/s (50 km/h).
Example: Banking angle
a. For a car traveling with speed v around a
curve of radius r, find a formula for the
angle θ at which a road should be banked
so that no friction is required.
b. Calculate this angle for an expressway
off-ramp curve of radius r = 50 m at a
design speed of v = 14 m/s (50 km/h).
Newton’s 2nd Law
∑Fx = max  FNx = m(v2/r)
or FNsinθ = m(v2/r)
(1)
y: ∑Fy = may = 0  FNcosθ - mg = 0
or FNcosθ = mg
(2)
x:
Example: Banking angle
a. For a car traveling with speed v around a
curve of radius r, find a formula for the
angle θ at which a road should be banked
so that no friction is required.
b. Calculate this angle for an expressway
off-ramp curve of radius r = 50 m at a
design speed of v = 14 m/s (50 km/h).
Newton’s 2nd Law
∑Fx = max  FNx = m(v2/r)
or FNsinθ = m(v2/r)
(1)
y: ∑Fy = may = 0  FNcosθ - mg = 0
or FNcosθ = mg
(2)
x:
Dividing (2) by (1) gives: tanθ = [(v2)/(rg)]
Putting in the given numbers gives
tanθ = 0.4 or θ = 22º
Circular Motion Example: Roller Coaster
• The roller coaster’s path is
nearly circular at the minimum
or maximum points on the track.
• When at the top, there is a
maximum speed at which the
coaster will not leave the top of
the track.
Circular Motion Example: Roller Coaster
• The roller coaster’s path is
nearly circular at the minimum
or maximum points on the track.
• When at the top, there is a
maximum speed at which the
coaster will not leave the top of
the track.
Newton’s 2nd Law
• Let down be positive:
∑Fy = maC
mg – N = (mv2/r) (1)
• v is maximized when N = 0.
Solve (1) for v when N = 0.
Circular Motion Example: Roller Coaster
• The roller coaster’s path is
nearly circular at the minimum
or maximum points on the track.
• When at the top, there is a
maximum speed at which the
coaster will not leave the top of
the track.
Newton’s 2nd Law
• Let down be positive:
∑Fy = maC
mg – N = (mv2/r) (1)
• v is maximized when N = 0.
Solve (1) for v when N = 0.
This gives:
If v is greater than this, N
would have to be negative.
This is impossible, so the
coaster would leave the
track.
Circular Motion Example: Artificial Gravity
• Circular motion can be used to
create “artificial gravity”.
• The normal force acting on the
passengers due to the floor
would be
• If N = mg it would feel like the
passengers are experiencing
normal Earth gravity.