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Section 6.2: Newton’s Laws &
Non-Uniform Circular Motion
Tangential & Radial Acceleration
(Recall from Ch. 4)
• Consider an object moving in a curved path. If
it’s speed |v| is changing, There is an
Acceleration in the Direction of Motion
≡ Tangential Acceleration
at ≡ |dv/dt|
But, There
still is always a radial (centripetal)
acceleration perpendicular to the direction of motion.
≡ Radial (Centripetal) Acceleration
ar = |ac| ≡ (v2/r)
Total Acceleration
• In general, there are two
 vector components of
the acceleration:
Tangential: at = |dv/dt|
Radial: ar = (v2/r)
• The Total Acceleration is the vector sum: a = ar+ at
• So, in general, there also always two  vector components of
the net force. That is, ∑F = ma has 2 vector components.
Tangential: ∑Ft = mat
Radial: ∑Fr = mar = m(v2/r)
The Total Force is the vector sum:
∑F = ∑Ft + ∑Fr = m(ar+ at)
Example 6.6
A sphere, mass m, is attached to a cord, length R.
It is twirled in a vertical circle about a fixed point O.
Find a general expression for the tension T at any
instant when the sphere’s speed is v & the angle the
cord makes with the vertical is θ. Forces acting are
gravity, Fg = mg, & cord tension, T.
Solution
1. Resolve Fg into components tangent to the path, Fgt = mg sinθ,
& perpendicular to path (radial), Fgr = mg cosθ.
2. Apply Newton’s 2nd Law separately in the two directions:
∑Ft = mg sinθ = mat, so at, = mg sinθ
∑Fr = T - mg cosθ = mac = m(v2/R) so
T = mg [v2/(Rg) + cosθ]
 The general expression needed!
3. Evaluate T at top (θ = 180°, cosθ = -1) & at bottom: (θ = 0°, cosθ = 1)
Ttop = mg [(vtop)2/(Rg) - 1]
Tbot = mg [(vbot)2/(Rg) + 1]
Sect. 6.3: Motion in Accelerated Frames
• Recall, Ch. 5: Newton’s Laws technically apply only in
inertial (non-accelerated) reference frames. Consider a
train moving with respect to the ground at constant velocity vtg to the
right. A person is walking in the train at a constant velocity vpt with
respect to the train.
vtg 
vpt 
Newton’s Laws hold
for the person walking
in the train as long as
the train velocity vtg with respect to the ground is constant (no
acceleration). But, if the train accelerates with respect to the
ground (vtg ≠ constant) & the person does an experiment, the
result will appear to violate Newton’s Laws because objects
will undergo an acceleration in the opposite direction as the train’s
acceleration. This acceleration will (seem to be) unexplained
because it occurs in the absence of forces!
More Discussion
• Inertial Reference Frame:
– Any frame in which Newton’s Laws are valid!
– Any reference frame moving with uniform (nonaccelerated) motion with respect to an “absolute”
frame “fixed” with respect to the stars.
• By definition, Newton’s Laws are only valid in
inertial frames!!
∑F = ma
Is not valid in a non-inertial frame!
• By definition, a “Force”, as defined in Ch. 5 (&
by Newton), is that it comes about from an
interaction between objects.
Fictitious “Forces”
• Suppose a person on a train places a flat object on a flat,
horizontal, frictionless surface. There are no horizontal forces
on the object. The train accelerates forward. The person will
see the flat object accelerate horizontally backward at the
same time. This
Appears to violate Newton’s Laws
because there is no horizontal force on the object
(accelerations are caused by forces).
This only appears to violate Newton’s Laws, but
actually doesn’t,
because the accelerating train is
NOT AN Inertial Reference Frame.
• Based on this experiment, the person will say that there
is a Fictitious (or Apparent) Force on the object.
• Fictitious (Apparent) Forces:
– If we insist that Newton’s 2nd Law holds in a noninertial reference frame. To make the force equations
look as if the reference frame is an inertial one, it’s
necessary to introduce Fictitious Forces.
– Technically, it’s the coordinate transformation
from the inertial frame to the non-inertial one
introduces terms on the “ma” side of ∑F = mainertial.
If we want eqtns in the non-inertial frame to look like
Newton’s Laws, these terms moved to the “F” side &
we have ∑“F” = manoninertial. where “F” = F + terms
from coordinate transformation
 “Fictitious (Apparent) Forces”.
• Simple Example of a Fictitious Force
You ride in a car going around a curve at constant 
constant velocity v.
 There is a centripetal acceleration on the car:
ac = (v2/r), towards the center of the curve
(r = curve radius).
The car takes the curve fast enough that, in the passenger
seat, you slide to the right & bump against the door. The
force between door & your body keeps you from being
Your reference frame, the car, is a
NON-INERTIAL frame.
You feel an apparent force moving you towards the door (away from
the center of the curve). A popular & WRONG explanation is that
you feel a “Centrifugal Force” away from the center of the curve.
“Centrifugal Force” ” is an example of a fictitious force.
• Simple Example Continued
From your viewpoint, in the non-inertial
reference frame of the car, there is an
apparent force, pointed away from
the center of the curve & pushing you to the door.
Correct Explanation
Before the car starts to round the curve, it (& you)
are going in a straight line at constant velocity. As
the car enters the curve, from N’s 1st Law, you will
first tend to keep moving straight. In the inertial
frame of the Earth, You feel a real force
(friction), acting inward towards the curve center
between you & the car seat that makes you stay in
your seat. N’s Laws say that you will tend to
continue to move in your original straight line until
you bump against the door.
Example 6.7