Download pdf - Extension of Circular Motion & Newton’s Laws of Circular Motion & Newton’s ... Kings High School . Review from Chapter 4 SUniform Circular Motion ... 14, 15, 16, 18/ (Section 6.3):

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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


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


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.


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


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


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