Physics”
Example 5-11 (Estimate)
Estimate the force a person must exert on a string attached to a 0.15-kg ball to
make the ball revolve in a horizontal circle of radius 0.6 m. The ball makes 2
rev/s. Ignore the string mass. m = 0.15 kg, r = 0.6 m, f = 2 rev/s  T = 0.5 s
Assumption: Circular path is  in a horizontal plane, so φ  0  cos(φ)  1
Example 5-11 (Estimate)
Estimate the force a person must exert on a string attached to a 0.15-kg ball to
make the ball revolve in a horizontal circle of radius 0.6 m. The ball makes 2
rev/s. Ignore the string mass. m = 0.15 kg, r = 0.6 m, f = 2 rev/s  T = 0.5 s
Assumption: Circular path is  in a horizontal plane, so φ  0  cos(φ)  1
∑F = ma  FTx = max= maR = m(v2/r)
v = (2πr/T) = 7.54 m/s
So, the tension is (approximately)
FTx  14 N
Newton’s 2nd Law:
Example 5-12: Revolving ball (vertical circle)
A ball, mass m = 0.1 kg on the end of a
(massless) cord of length r = 1.1 m cord is
swung in a vertical circle. Calculate:
a. The minimum speed the ball must
have at the top of its arc so that the
ball continues moving in a circle.
b. The tension in the cord at the
bottom of the arc, assuming that
there the ball is moving at twice the
speed found in part a.
Hint: The minimum speed at the top will
happen for the minimum tension FT1
A Similar Problem
r = 0.72 m, v = 4 m/s = constant
m = 0.3 kg, FT1 = ?, FT2 = ?
Newton’s 2nd Law:
∑F = maR
A Similar Problem
r = 0.72 m, v = 4 m/s = constant
m = 0.3 kg, FT1 = ?, FT2 = ?
Newton’s 2nd Law:
∑F = maR
• At the top of the circle, vertical forces:
(down is positive!)
FT1 + mg = m(v2/r)  FT1 = 3.73 N
• At the bottom of the circle, vertical forces:
(up is positive!)
FT2 - mg = m(v2/r)  FT2 = 9.61 N
Example: F, p. 124
A Ferris wheel rider moves in a
vertical circle of radius r at
constant speed v. Is the normal
force FN1 that the seat exerts on
the rider at the top of the wheel
a. less than, b. more than, or
c. equal to the normal force FN2
that the seat exerts on the rider at
the bottom of the wheel?
Use
Newton’s 2nd Law: ∑F = maR
at top & bottom, solve for normal
force & compare.
1
2
Conceptual Example
A tether ball is hit so that it revolves
around a pole in a circle of radius r at
constant speed v. In what direction is the
acceleration & what force causes it?
Newton’s 2nd Law: ∑F = ma
x:
∑Fx = max
 FTx = maR = m(v2/r)
y: ∑Fy = may = 0
 FTy - mg = 0, FTy = mg
Example 5-13: Conical pendulum
A ball of mass m, suspended by
a cord of length ℓ, revolves in a
circle of radius r = ℓsinθ, where
θ is the angle the string makes
with the vertical.
a. In what direction is the ball’s
acceleration & what causes it?
b. Calculate the speed & period
(time for one revolution) of the ball
in terms of ℓ, θ, g, & m.
Sect. 5-4: Highway Curves: Banked & Unbanked
Case 1: Unbanked Curve. When a car goes around a curve, there
must be a net force toward the center of the circle of which the
curve is an arc. For flat road, that force is supplied by static friction.
“Centripetal Force”
No static friction?
 No Centripetal Force?
 The car goes straight!
Never a “Centrifugal Force”!!!
Example 5-14: Skidding on a curve
A car, mass m = 1,000 kg rounds a curve on a flat
road of radius r = 50 m at a constant speed
v =15 m/s (54 km/h). Will the car follow the curve,
or will it skid? Assume:
a. The pavement is dry & the coefficient of static
friction is μs = 0.6.
b. The pavement is icy & μs = 0.25.
Free Body
Diagram
Example 5-14: Skidding on a curve
A car, mass m = 1,000 kg rounds a curve on a flat
road of radius r = 50 m at a constant speed
v =15 m/s (54 km/h). Will the car follow the curve,
or will it skid? Assume:
a. The pavement is dry & the coefficient of static
friction is μs = 0.6.
b. The pavement is icy & 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
Maximum static friction: Ffr = μsFN
Free Body
Diagram
If the friction force is insufficient, the car will tend to move
more nearly in a straight line, as the skid marks show.
As long as the tires don’t slip, the
friction is static. If the tires do
start to slip, the friction is
kinetic, which is bad in 2 ways:
1. The kinetic friction force is
smaller than the static one.
2. The static friction force can
point toward the center of the
circle, but the kinetic friction
force opposes the direction of
motion, making it very difficult
to regain control of the car &
continue around the curve.
Banked Curves
Banking a curve helps keep cars from skidding.
For every banked curve, there is one speed v at
which the entire “centripetal force” is supplied
by the horizontal component of the normal
force FN & no friction is required.
Newton’s 2nd Law
tells us what speed v this is:
x: ∑Fx = max
 FNx = m(v2/r)
or
(solve for v)
Aa
Also, y: ∑Fy = may = 0 
or
FNcosθ = mg
FN cosθ - mg = 0
(solve for FN)
Example 5-15: 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 5-15: 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
x: ∑Fx = max  FNx = m(v2/r)
or FNsinθ = m(v2/r)
y: ∑Fy = may = 0  FN cosθ - mg = 0 or FNcosθ = mg
tanθ = [(v2)/(rg)]
Putting in the given numbers, tanθ = 0.4 or θ = 22º
Dividing (2) by (1) gives:
(1)
(2)