Newtonian mechanics

concern about the relation between force and acceleration

first studied by Newton (1642-1727)

limitations:

1. for small scale, quantum mechanics should be used instead.

2. When the speed is close to speed of light, special relativity is needed (in fact, Newtonian mechanics is simply a special case where v<<c.)

3. If gravity is strong, general relativity is applied.

Force and motion (chap 5 & 6)

Newton’s Laws

First law: if the net force ( 淨力 ) is zero, there is no change of velocity, i.e. acceleration = 0.

• an object remains at rest if it is originally at rest

• a moving object keeps moving without change of velocity. It seems to contradict our daily experience, in which frictional force ( 摩擦力 ) is present.

(a)1,4,3,2

(b) 1,4,3,2

Second law: F=ma, the net force = product of mass and acceleration.

• this defines the force F

• Special case: W=mg the weight of mass m

2

2

R

MGg

mgR

MmGW

Third law: when two objects interact, the forces acting on each other is equal. But the directions are opposite.

• e.g. Normal force

Newton’s law is not true in all reference frames ( 參考坐標系 ). For those accelerating frames, Newton’s law is invalid.

Standing on the ground, we are actually in an accelerating frame because the Earth is rotating. An object moving from the North pole towards the equator experiences a deflection to the west in the accelerating frame, even though there is no net force acting on the object. This kind of reference frame is called non-inertial reference frame.

Inertial reference frame ( 慣性參考坐標系 ) refers to those that Newton’s law is valid. In other words, it refers to those frames that moves in constant velocity.

Applications of Newton’s law:

• each object of a system can be consider independently

• the motion of each object must obey Newton’s law

Tension T and acceleration a can be found by solving two equations of motion

Other examples:

Free body diagram

sin6

1

sin6

mg

T

mgT(a) (b) If the legs is straightened, θ increases. Therefore T reduces.

T1

T2

Mg

a

)sin(

sin

21

21

gaMTT

MaMgTT

Try problems of Chap 5: 26, 50, 54, 88

In a car race, it is crucial to understand the role of frictional force 摩擦力 (when the car accelerates), drag force 阻力 (air resistance), and centripetal force 向心力 (for the car to turn).

Friction is very important in our daily life. Without friction, we cars and motor bike cannot move. On the other hand, 20f% of gasoline is used to overcome friction in the engine.

fs is the static frictional force

fk is the kinetic frictional force

Origin of friction:

1. in practice, no surface is perfectly flat. However, the roughness can only explain part of the frictional force.

2. The contact area of two surfaces is about 104 times smaller the total area.

3. Since there is no air and space between the surfaces in the contact area, atoms of one surface make contact to the atoms of the other surface and the surfaces cold-weld ( 冷熔接 forming a piece of material).

1. The force needs to break the welds is the static friction

2. Note that the larger the normal force, the larger the contact area and hence the larger frictional force.

The max. static friction is given by:

Normal force

coefficient of static friction

The kinetic friction is given by:

coefficient of kinetic friction

g

m

mg

m

fa k

kk

xgv

axv

k2

2020

20

2

The acceleration is due to the frictional force:

(a) Fx decreases

(b) fs = Fx decreases

(c) FN = Fy+Mg increases

(d) fs,max = FNs increases

(e) fk= FNk increases

Terminal velocity 終端速度 • When a body moves inside a

fluid (gas or liquid), it experiences a drag force D opposed its motion.

• The magnitude of D is given by:

D

Drag coefficientDensity of

fluid

Cross section area 截面積

velocity

• For a free falling body, v increases together with the drag force until the drag force equals to the gravity.

• Then the net force=0 and so v remains to be a constant, vt, the terminal velocity.

Note: here we ignore the shape of the body and assume it is like a ball.

Cross section area:

2R Volume:3

3

4R

smC

gR

RC

Rgv

a

w

a

w

t /4.73

834

2

2

3

hkmsmv

ghv

/550/3.153

22

Very fast!

Uniform circular motionThe magnitude of the acceleration is given

by:

while the direction always point to the center of the circle.

By Newton’s second law, there must be a centripetal force that always acting towards the center.

In this example, it is the tension T that provides the centripetal force.

Example:• For a car turning around a corner,

the frictional force provides the centripetal force.

• If the friction fs is not large enough, then the acceleration is less than the required value of v2/R. The car will slide away from the center until it reaches a R that fs=mv2/

R.• the same thing happens to the p

assengers, they will slide to the wall inside the car until the normal force fN from the wall provided fN+fs=mv2/R.

• For a spacecraft orbiting the earth, the gravity provides the centripetal force.

• For an orbital radius R, there is only one possible velocity, independent of the mass.

• Note that the astronaut inside the spacecraft is now weightless.

R

GMv

R

vm

R

mGMF

E

EG

2

2

2

• What provides the centripetal force?

• The weight and the normal force.

FN

Fg

Note that the speed v is not a constant. It changes throughout the whole loop. What causes it change?

The minimum speed v at the top of the loop can be determined as follows:

R

vmmgF

maFF

N

gN

2

The min. speed v is obtained when FN = 0

gRv

gRv

mgR

vm

mgF

Ff

N

gs

2

What is force holding the rider?

Frictional force

(a) fs remains the same

(b)FN increases

(c) Fs,max increases

R

vmF

mgF

N

N

2

sin

cos

Horizontal:

Vertical:

Rg

v2tan

mgFR

vmF

L

L

cos

sin2

gR

v2tan

The same equation for the banked track.

Try problems of chap 6: 9, 23, 49