Ch. 6: Circular Motion & Other
Applications of Newton’s Laws
Recall From Ch. 4:
Acceleration of Mass Moving in Circle (Const. Speed)
Particle moving in a circle, radius r,
speed v (= constant). The velocity is
tangent to the circle. The centripetal
acceleration, a = ac is radially inward.
 ac  v always
ac = (v2/r)
Newton’s Laws + Circular Motion
ac = (v2/r)  v
Newton’s 1st Law:
There must be a force acting.
Newton’s 2nd Law:
∑F = ma = mac
= m(v2/r) (magnitude)
Direction: The total force must
be radially inward.
• A particle moving in uniform circular motion, radius
r (speed v = constant). The acceleration: ac = (v2/r), ac 
v always!! ac is radially inward always!
• Newton’s 1st Law: There must be a force acting!
• Newton’s 2nd Law:
∑F = ma  Fr = mac= m(v2/r)
The total force must be radially inward always!
 The Force entering 2nd Law  Centripetal Force Fr
(Center directed force)
• NOT a new kind of force. Could be string tension,
gravity, etc. The right side of ∑F = ma, not the left
side! (The form of ma, above, for circular motion)
Example: A ball twirled on a string in a
circle at constant speed. The centripetal
force Fr is the tension in the string.
MISCONCEPTION!!
The force on the ball is NEVER outward
(“centrifugal force”). The force on the ball
is ALWAYS inward (centripetal force).
An outward force (“centrifugal”) is NOT
a valid concept! The force ON THE
BALL is inward (centripetal).
What happens when the ball is
released? (Fr = 0). Newton’s 1st
 Law says it should move off in
a straight line at constant v.
Example 6.1: Conical Pendulum
A ball, mass m, is suspended
from a string of length L. It
revolves with constant speed v
in a horizontal circle of radius r.
The angle L makes with the
horizontal is θ. Find an expression
for v.
T ≡ tension in the string. Fig. (b) shows horizontal & vertical
components of T:
Tx = Tsinθ, Ty = Tcosθ.
Newton’s 2nd Law: ∑Fx = Tsinθ = mac= m(v2/r)
(1)
∑Fy = Tcosθ – mg = 0; Tcosθ = mg
(2)
Dividing (1) by (2) gives: tanθ = [v2/(rg)] , or v = (rg tanθ)½
From trig, r = L sinθ so, v = (Lg sinθ tanθ)½
(Reminder: ½ power means the square root)
Example 6.2: Car Around a Curve
Curve radius: r = 35 m. Static friction coefficient
between tires & road: μs = 0.523. The centripetal
force that keeps the car on the road is the static
friction force fs between the tires & the road.
Calculate the maximum speed vmax for the car to
stay on the curve. Free body diagram is (b).
Newton’s 2nd Law (let + x be to left) is:
∑Fx = fs = mac = m(v2/r) (1)
∑Fy = 0 = n – mg; n = mg (2)
The maximum static friction force is (using (2)) :
fs(max) = μsn = μsmg
(3)
 If
m(v2/r) > fs(max), so vmax is the
solution to μsmg = m[(vmax)2/r]
Or,
vmax = (μsgr)½
Putting in numbers gives: vmax = 13.4 m/s
Example 6.4: Banked Curves
Engineers design curves which are banked (tilted
towards the inside of the curve) to keep cars on the
road. If r = 35 m & we need v = 13.4 m/s,
calculate the angle θ of banking needed (without
friction). From free body diagram, the horizontal
(radial) & vertical components of the force n
normal to the surface are:
nx = n sinθ, ny = n cosθ,
Newton’s 2nd Law
∑Fx = n sinθ = m(v2/r)
(1)
∑Fy = 0 = n cosθ – mg; n cosθ = mg (2)
Dividing (1) by (2) gives: tanθ = [(v2)/(gr)]
Putting in numbers gives: tanθ = 0.523 or
θ = 27.6°
Example 6.5: “Loop-the-Loop”!
A pilot, mass m, in a jet does a “loop-theloop. The plane, Fig. (a), moves in a
vertical circle, radius r = 2.7 km = 2,700 m
at a constant speed v = 225 m/s.
a) Calculate the force, nbot (normal force),
exerted by the seat on the pilot at the
bottom of the circle, Fig. (b).
b) Calculate this force, ntop, at the top of the
circle, Fig. (c).
TOP: Fig. (b). Newton’s 2nd Law in the radial (y) direction (up is “+”).
∑Fy = nbot – mg = m(v2/r) so nbot = m(v2/r) + mg or
nbot = mg[1 + (v2/rg)] = 2.91 mg (putting in numbers) he feels “heavier”.
BOTTOM: Fig. (c). Newton’s 2nd Law in the radial (y) direction (down is “+”).
∑Fy = ntop + mg = m(v2/r) so ntop = m(v2/r) - mg or
ntop = mg[(v2/rg) - 1] = 0.913 mg (putting in numbers) he feels “lighter”.
Example (Estimate)
m = 0.15 kg, r = 0.6 m, f = 2 rev/s  T = 0.5 s
Assumption: Circular path is  in horizontal
plane, so θ  0  cos(θ)  1
∑F = ma 
FTx = max= mac = m(v2/r)
v =(2πr/T) = 7.54 m/s
FTx = 14 N (tension)
Example
Problem
r = 0.72 m, v = 4 m/s
m = 0.3 kg
• Use: ∑F = mac
• Top of circle:
Vertical forces:
(down is positive!)
FT1 + mg = m(v2/r)
FT1 = 3.73 N
• Bottom of circle:
Vertical forces:
(up is positive)
FT2 - mg = m(v2/r)
• FT2 = 9.61 N
Example
n
n