S
Extension of Circular Motion & Newton’s Laws
Chapter 6
Mrs. Warren
Kings High School
Review from Chapter 4
S Uniform Circular Motion
S Centripetal Acceleration
Uniform Circular Motion,
Force
• A force is associated with the
centripetal acceleration.
• The force is also directed
toward the center of the circle.
• Applying Newton’s Second
Law along the radial direction
gives
rF
2
c
vF ma m
r= =å
Section 6.1
Uniform Circular Motion, cont.
SA force causing a centripetal
acceleration acts toward the center of
the circle.
SIt causes a change in the direction of
the velocity vector.
SIf the force vanishes, the object
would move in a straight-line path
tangent to the circle.
S See various release points in
the active figure
Section 6.1
S A small ball of mass, m, is suspended
from a string of length, L. The ball
revolves with constant speed, v, in a
horizontal circle of radius, r. Find an
expression for v in terms of the geometry.
The Conical
Pendulum
sin tanv Lg q q=
The Conical Pendulum
S A puck of mass 0.500 kg is attached to the end of a cord. The puck moves in a horizontal
circle of radius 1.50 m. If the cord can withstand a maximum tension of 50.0 N, what is
the maximum speed at which the puck can move before the cord breaks? Assume the
string remains horizontal during the motion.
Trv
m=
The Flat
Curve… • Model the car as a particle in
uniform circular motion in the horizontal direction.
• Model the car as a particle in equilibrium in the vertical direction.
• The force of static friction supplies the centripetal force.
• The maximum speed at which the car can negotiate the curve is:
Note, this does not depend on the mass of the car.
sv grm=
The Flat Curve…
S A 1500 kg car moving on a flat, horizontal road negotiates a curve as shown on the
previous slide. If the radius of the curve is 35.0 m and the coefficient of static friction
between the tires and dry pavement is 0.523, find the maximum speed the car can
have and still make the turn successfully.
The Banked Roadway…
SThese are designed with friction equaling
zero.
SModel the car as a particle in equilibrium in
the vertical direction.
SModel the car as a particle in uniform circular
motion in the horizontal direction.
SThere is a component of the normal force that
supplies the centripetal force.
SThe angle of bank is found from
The Banked Curve…
S A civil engineer wishes to redesign the curved roadway in such a way that a car will not
have to rely on friction to round the curve without skidding. In other words, a car moving
at the designated speed can negotiate the curve even when the road is covered with ice.
Such a road is usually banked, which means that the roadway is tilted toward the inside
of the curve. Suppose the designated speed for the road is to be 30.0 mi/h and the radius
of the curve is 35.0 m. At what angle should the curve be banked?
Ferris Wheel S A child of mass m rides
on a Ferris wheel as
shown. The child moves
in a vertical circle of
radius 10.0 m at a
constant speed of 3.00
m/s. Determine the force
exerted by the seat on the
child at the bottom of the
ride. Express your answer
in terms of the weight of
the child, mg. Do the
same when the child is at
the top of the ride.
Non-Uniform Circular Motion
SThe acceleration and force have tangential
components.
S produces the centripetal acceleration
S produces the tangential acceleration
SThe total force is
S r t= +å å åF F F
tF
rF
Section 6.2
Vertical Circle with Non-
Uniform Speed
SA small sphere of mass, m, is attached
to the end of a cord of length R and set
into motion in a vertical circle about a
fixed point O as illustrated. Determine the
tangential acceleration of the sphere and
the tension in the cord at any instant when
the speed of the sphere is v and the cord
makes an angle with the vertical.
2
cosv
T mgRg
qæ ö
= +ç ÷è ø
Section 6.2
Motion in Accelerated Frames
SA fictitious force results from an accelerated frame of reference.
S The fictitious force is due to observations made in an accelerated
frame.
S A fictitious force appears to act on an object in the same way as a real
force, but you cannot identify a second object for the fictitious force.
S Remember that real forces are always interactions between two objects.
S Simple fictitious forces appear to act in the direction opposite that of
the acceleration of the non-inertial frame.
Section 6.3
“Centrifugal” Force
“Coriolis Force”
Motion with Resistive Forces
SThe medium exerts a resistive force, , on an object moving through the medium.
SThe magnitude of. ….
SThe direction of is…
SThe magnitude of can depend on the speed in complex ways.
SWe will discuss only two:
S is proportional to v
S Good approximation for…
S is proportional to v2
S Good approximation for…
R
R
R
Section 6.4
R
R
R
Resistive Force Proportional
To Speed
SThe resistive force can be expressed as
Sb depends on the property of the medium, and on the shape
and dimensions of the object.
SThe negative sign indicates is in the opposite direction to .
= -bR v
R v
Section 6.4
Resistive Force Proportional
To Speed, Example
SAssume a small sphere of mass m is
released from rest in a liquid.
SForces acting on it are:
S Resistive force
S Gravitational force
SAnalyzing the motion results in
- = =
= = -
dvmg bv ma m
dt
dv ba g v
dt m
Section 6.4
Resistive Force Proportional
To Speed, Example, cont.
SInitially, v = 0 and dv/dt = g
SAs t increases, R increases and a decreases
SThe acceleration approaches 0 when R mg
SAt this point, v approaches the terminal speed of the object.
Terminal Speed
STo find the terminal speed, let a = 0
SSolving the differential equation gives
St is the time constant and
t = m/b
Section 6.4
Terminal Speed
S A small sphere of mass 2.00 g is released from rest in a large vessel filled with
oil, where it experiences a resistive force proportional to its speed. The sphere
reaches a terminal speed of 5.00 cm/s. Determine the time constant and the
time at which the sphere reaches 90.0% of its terminal speed.
Resistive Force Proportional
To v2
S For objects moving at high speeds through air, the resistive force is approximately equal to the square of the speed.
S R = ½ DrAv2
S D is a dimensionless empirical quantity called the drag coefficient.
S r is the density of air.
S A is the cross-sectional area of the object.
S v is the speed of the object.
Section 6.4
Resistive Force Proportional
To v2, example
SAnalysis of an object falling
through air accounting for air
resistance.
r
r
= - =
æ ö= - ç ÷
è ø
å 2
2
1
2
2
F mg D Av ma
D Aa g v
m
Resistive Force Proportional
To v2, Terminal Speed
SThe terminal speed will occur
when the acceleration goes to
zero.
SSolving the previous equation
gives
r=
2T
mgv
D A
Example: Skysurfer
SStep from plane
S Initial velocity is 0
S Gravity causes downward
acceleration
S Downward speed increases, but so
does upward resistive force
SEventually, downward force of gravity
equals upward resistive force
S Traveling at terminal speed
Section 6.4
Skysurfer, cont.
SOpen parachute
S Some time after reaching terminal speed, the parachute is
opened.
S Produces a drastic increase in the upward resistive force
S Net force, and acceleration, are now upward
S The downward velocity decreases.
S Eventually a new, smaller, terminal speed is reached.
Section 6.4
HW
S Objective Questions: 1, 4, 5, 6, 7
S Conceptual Questions: 4, 5, 6, 8
S Problems: (Section 6.1): 1, 2, 6, 7, 8, 9/ (Section 6.2): 13,
14, 15, 16, 18/ (Section 6.3): 21, 22/ (Section 6.4): 27, 28,
30, 31