𝑣𝑓 = 𝑣𝑖 + 𝑎𝑡;

𝑠 =1

2𝑣𝑖 + 𝑣𝑓 𝑡;

𝑠 = 𝑣𝑖𝑡 +1

2𝑎𝑡2; 𝑣𝑓

2 = 𝑣𝑖2 + 2𝑎𝑠

Constant Acceleration

Displacement 𝑦 Downwards

Initial velocity 𝑣𝑖 Upwards

Acceleration, a Downwards throughout

It is important to set convention, either Upwards as +ve or Downwards as +ve and apply it consistently to displacement, velocity and acceleration when using equations.

𝑔 = 9.8 𝑚/𝑠2

Example: Tut 1 Q1 For

PH

10

12

, 20

15

H

o S

Y, S

PM

S N

TU

Constant Speed / Velocity; acel = 0

𝑠𝑝𝑒𝑒𝑑 =𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑡𝑖𝑚𝑒

Any Acceleration (constant or non-constant)

𝑥 = 𝑘𝑡𝑛 ⇒ 𝑑𝑥

𝑑𝑡= 𝑘 𝑛𝑡𝑛−1

Eg. 𝑥 = 4𝑡2 + 3 ⇒𝑑𝑥

𝑑𝑡= 8𝑡 + 0

𝑥𝑚 𝑑𝑥 =1

𝑚 + 1𝑥𝑚+1 (𝑤ℎ𝑒𝑟𝑒 𝑚 ≠ −1)

Eg. ∫ 8𝑡 𝑑𝑡 = 8 ×1

2𝑡2 + 𝐶 = 4𝑡2 + 𝐶

where C is an arbitrary constant

Basic Calculus

𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 𝑣 =𝑥𝑓 − 𝑥𝑖

𝑡𝑓 − 𝑡𝑖=Δx

Δ𝑡

𝐼𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦,

𝑣 = limΔ𝑡→0

Δx

Δ𝑡=𝑑𝑥

𝑑𝑡

𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛,

𝑎 =𝑣𝑓 − 𝑣𝑖

𝑡𝑓 − 𝑡𝑖=Δv

Δ𝑡

𝐼𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛,

𝑎 = limΔ𝑡→0

Δv

Δ𝑡=𝑑𝑣

𝑑𝑡=𝑑

𝑑𝑡

𝑑𝑥

𝑑𝑡=𝑑2𝑥

𝑑𝑡2

Speed constant 𝑣 (magnitude of arrows fixed) [tangential acceleration=0; angular acceleration=0] However, velocity is changing, (direction of arrows changing). Hence, change in velocity and centripetal acceleration are not zero:

𝑎𝑟 =𝑣2

𝑟= 𝑟𝜔2

𝜃

In time Δ𝑡, angular displacement is 𝜃;

Hence angular velocity 𝜔 =𝜃

Δ𝑡.

Constant Speed (𝜶 = 𝟎; 𝒂𝒕𝒂𝒏 = 𝟎)

Constant Angular Acceleration (𝜶 ≠ 𝟎 = 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕;𝒂𝒕𝒂𝒏 = 𝒓𝜶)

r

𝑠 = 𝑟𝜃

𝑣 = 𝑟𝜔 = 𝑟2𝜋

𝑇

T – time to complete one round

𝜃

𝑠 = 𝑟𝜃

𝑣 = 𝑟𝜔 = 𝑟2𝜋

𝑇

𝑎𝑡𝑎𝑛 = 𝑟𝛼

𝜔𝑓 = 𝜔𝑖 + 𝛼𝑡

𝜃 =1

2𝜔𝑖 +𝜔𝑓 𝑡

𝜃 = 𝜔𝑖𝑡 +1

2𝛼𝑡; 𝜔𝑓

2 = 𝜔𝑖2 + 2𝛼𝜃

𝑎𝑟 =𝑣2

𝑟= 𝑟𝜔2

But note that 𝜔 is changing so the magnitude of 𝑎𝑟 is changing.

Resultant acceleration 𝑎 = a𝑡𝑎𝑛 + 𝑎 𝑟

For extended objects in equilibrium: ∑𝜏 = 0 ∑𝐹 = 0

𝜏 = 𝐹𝑅 sin 𝜃

Torque

𝐹 𝐴𝐵 - Force pointing from A to B 𝑣 𝑊𝑆 - Water with respect to Shore

𝑣 𝐵𝑆 = 𝑣 𝐵𝑊 + 𝑣 𝑊𝑆

Relative Velocity

𝐹𝐴𝐵

Impt – duration of flight, t Horizontal: 𝑥 = 𝑢𝑥𝑡

Vertical: 𝑦 = 𝑢𝑦𝑡 +1

2𝑎𝑡2

Horizontal velocity: 𝑣𝑥 = 𝑢𝑥 Vertical Velocity: 𝑣𝑦 = 𝑢𝑦 + 𝑎𝑡

Projectile Motion

𝑣𝑓

𝑣𝑖

𝑣𝑖

𝑣𝑓 Δ𝑣

𝑣 𝑓 = 𝑣 𝑖 + Δ𝑣

Δ𝑣 = 𝑣 𝑓 − 𝑣 𝑖

Velocity change

Note that the meaning of the subscripts changes with the context of the problem.

Vector addition

v 𝜃

𝑣𝑥 = 𝑣 cos𝜃

𝑣𝑦 = 𝑣 sin 𝜃

tan 𝜃 =𝑣𝑦𝑣𝑥

𝑣2 = 𝑣𝑥2 + 𝑣𝑦

2

Components of vectors

Types of Forces

Weight, mg (due to gravitational attraction)

Contact Forces Normal Force, 𝑭𝑵 (90∘ to surface) Frictional Force,

𝐹𝑓 = 𝜇𝐹𝑁

𝜇𝑠 - static 𝜇𝑘- kinetic, moving

Tension, 𝑭𝑻 (Pulling force, in strings etc, always pointing away from objects.)

Upthrust, 𝑼 (= weight of fluid displaced by obj.)

Viscous Force , 𝑭𝒗 = 𝒌𝒗 or 𝑭𝒗 = 𝒌𝒗

𝟐 (depends on velocity of object moving through fluid.)

Spring Force , 𝑭𝒔 (Spring-mass system, depends on displacement of mass from equilibrium position.

Other Forces Gravitational Force Electrostatic Force Magnetic Force

∑𝐹 = 𝑚𝑎

Force (thrust)

𝐹 = 𝑣𝑑𝑚

𝑑𝑡

Related to Newton’s 2nd/3rd Law, conservation of momentum

𝐹𝑁 Friction, Fk

q

mg

𝑚𝑔𝑠𝑖𝑛 𝜃 𝑚𝑔𝑐𝑜𝑠 𝜃

𝐹𝑁 −𝑚𝑔𝑐𝑜𝑠𝜃 = 0 𝑚𝑔𝑠𝑖𝑛𝜃 − 𝐹𝑘 = 𝑚𝑎

𝐹𝑘 = 𝜇𝐾𝐹𝑁 = 𝜇𝐾𝑚𝑔𝑐𝑜𝑠𝜃 (sliding down slope)

+

m1

m2

T T

m2g

m1g

𝑇 −𝑚1𝑔 = 𝑚1𝑎 𝑚2𝑔 − 𝑇 = 𝑚2𝑎

1. Same tension 𝑇 and 𝑎 for frictionless pulleys

mg

𝐹𝑣 = 𝑘𝑣2

𝑚𝑔 − 𝑘𝑣2 = 𝑚𝑎 1. Upthrust negligible in air here. 2. Set 𝑎 = 0 to obtain terminal velocity, 𝑣𝑇. 3. Moving in viscous fluid use 𝐹𝑣 = 𝑘’𝑣. Moving

thru’ air

Spring Force 𝐹 = −𝑘𝑥

𝑥 = 𝐴𝑐𝑜𝑠 𝜔𝑡 + 𝜙 or 𝑥 = 𝐴𝑠𝑖𝑛(𝜔𝑡 + 𝜙)

Stationary helicopter

𝑚𝑔 − 𝑣𝑑𝑚

𝑑𝑡= 0 Fo

r P

H1

01

2, 2

01

5

Ho

SY,

SP

MS

NTU

Basics : Momentum: 𝑝 = 𝑚𝑣

Force: ∑𝐹 = 𝑚𝑎 =𝑚(𝑣−𝑢)

∆𝑡

Kinetic energy: 𝐾. 𝐸.=1

2𝑚𝑣2 =

𝑝2

2𝑚

Work done: W = FDx cos q

F

t

F

t 𝐼𝑚𝑝𝑢𝑙𝑠𝑒 = 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑐ℎ𝑎𝑛𝑔𝑒; 𝐹 = Δ𝑝/Δ𝑡

1. Stop gently (prolong ∆𝑡). 2. Deliver large pf (maximize F∆𝑡).

Conservation of linear momentum (two body collision): 𝑚𝑢1 +𝑚𝑢2 = 𝑚𝑣1 +𝑚𝑣2

1. Elastic collision (KE conserved): use 𝑢1 − 𝑢2 = −(𝑣1 − 𝑣2 ) also 2. 2-D collision: resolve 𝑝 and apply COLM in both directions 3. When appropriate, use sine rule or cosine rule; sin2 𝜃 + cos2 𝜃 = 1; tan 𝜃 = sin 𝜃/ cos 𝜃

Area -> impulse -> Momentum change

Work-Energy theorem: 1

2𝑚𝑣𝑓

2 −1

2𝑚𝑣𝑖

2 = 𝐹∆𝑥

Change in 𝐾. 𝐸. = 𝑊𝑜𝑟𝑘 𝐷𝑜𝑛𝑒

Conservation of Energy: 𝐾. 𝐸. 𝑖 + 𝑃. 𝐸. 𝑖 = 𝐾. 𝐸. 𝑓 + 𝑃. 𝐸. 𝑓 1. Zero of G.P.E. is arbitrary 2. + Fx if frictional / dissipative forces present

F

x e.g. E.P.E. = ½ Fx = ½ kx2

F

x e.g. G.P.E. = mgh , Work done against friction

𝑃𝑜𝑤𝑒𝑟: 𝑃 =𝐸𝑛𝑒𝑟𝑔𝑦

∆𝑡= 𝐹𝑣

Area -> work done -> K.E. change

For PH1012, 2015 Ho SY, SPMS NTU

Concepts Related to Forces, Momentum and Energy

Newton’s laws of Motion 1st law: Inertia, reluctance to change state of motion 2nd law: Force proportional to rate of change of momentum;

∑𝑭 = 𝒎𝒂 3rd law: (Two objects) Action-reaction pair. Equal and opposite forces (of same nature) acting on two objects

Applying force 𝑭 over period of time 𝚫𝐭 results in change in momentum

𝐼 = 𝐹Δ𝑡 = 𝑝𝑓 − 𝑝𝑖

(Impulse-momentum relation)

Conservation of Momentum (No external forces acting) 𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑏𝑒𝑓𝑜𝑟𝑒= 𝑡𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑎𝑓𝑡𝑒𝑟

Applying force 𝑭 on object along diaplacement 𝚫𝐱 results in change in Kinetic energy

𝑊 = 𝐹Δ𝑥 = 𝐾𝐸𝑓 − 𝐾𝐸𝑖

(Work energy Theorem)

Conservation of Energy 𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑎𝑡 𝑝𝑡 𝐴= 𝑡𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑎𝑡 𝑝𝑡 𝐵

q

T

mg

𝑇𝑐𝑜𝑠 𝜃 − 𝑚𝑔 = 0

𝑇𝑠𝑖𝑛 𝜃 =𝑚𝑣2

𝑟

= 𝑚𝑟𝜔2 q

𝑟

Conical pendulum (Uniform, constant horizontal circle) vtop

vmid

vbot

mg

mg

mg

Ttop

Tbot

Tmid

𝑇𝑡𝑜𝑝 +𝑚𝑔 =𝑚𝑣𝑡𝑜𝑝

2

𝑟

𝑇𝑏𝑜𝑡 −𝑚𝑔 =𝑚𝑣𝑏𝑜𝑡

2

𝑟

𝑇𝑚𝑖𝑑 =𝑚𝑣𝑚𝑖𝑑

2

𝑟

inwards +ve

Vertical circle (Non-uniform) (consider conservation of energy)

Non-uniform circular motion with constant angular acceleration 𝛼,

Same analysis for banking, planes making turns etc

𝐹 =𝐺𝑀𝐸𝑚

𝑟2=𝑚𝑣2

𝑟 𝑜𝑟 𝑚𝑟𝜔2; 𝜔 =

2𝜋

𝑇

𝑔 =𝐺𝑀𝐸𝑟2

𝐹 = 𝜇𝑠𝐹𝑛 =𝑚𝑣2

𝑟

Turning on fan 𝛼 > 0, 𝑎𝑡𝑎𝑛 > 0 At steady speed 𝛼 = 0, 𝑎𝑡𝑎𝑛 = 0 (uniform circular motion, only centripetal accn.) Turning off fan 𝛼 < 0, 𝑎𝑡𝑎𝑛 < 0

𝜔𝑓 = 𝜔𝑖 + 𝛼𝑡

𝑎 𝑅 = 𝑟𝜔𝑓2; 𝑎 ta𝑛 = 𝑟𝛼

𝑎 = 𝑎 𝑅 + 𝑎 𝑡𝑎𝑛

Tut 6 Q7 (Non-uniform)

∑𝐹𝑡𝑎𝑛 = 𝑚𝑔cos𝜃 = 𝑚𝑎𝑡𝑎𝑛

∑𝐹𝑟𝑎𝑑 = 𝐹𝑇 −𝑚𝑔 sin𝜃 = 𝑚𝑎𝑟𝑎𝑑 =𝑚𝑣2

𝑟

Circular Motion For PH1012, 2015 (Key concepts) Ho SY, SPMS NTU

Q

|𝐹| = 𝑘𝑄𝑞

𝑟2

𝑈(𝑟) = 𝑘𝑄𝑞

𝑟

𝐸 = 𝑘𝑄

𝑟2 𝑟

𝑉(𝑟) = 𝑘𝑄

𝑟

When a charge q is placed in the E-field of charge Q.

𝐹 = 𝑞𝐸

𝑈 = 𝑞𝑉

𝐸𝑟 = −𝜕𝑉

𝜕𝑟

𝑉𝑏 − 𝑉𝑎 = − 𝐸. 𝑑𝑟

𝑏

𝑎

𝐸 = 𝐸1+ 𝐸2+ 𝐸3+⋯

For a system of charges, the electric field and potential at any point due to charges q1, q2 etc is given by

𝑉 = 𝑉1 + 𝑉2 + 𝑉3 +⋯

To sketch E-field or equipotential lines for system of charges: 1. Check for charge magnitudes and

geometry of charge distribution; making use of any symmetry.

2. Check for possible any points/lines with E = 0 or components of E which vanishes.

3. More field lines entering / leaving larger charge

4. E-fields lines are perpendicular to equipotential lines.

1 2

Point Charges (Radial Fields)

Parallel Plates (Uniform Fields)

𝐹 = 𝑞𝐸 ; U = qV Since the E-field is uniform (except near the edges), the force acting on a charge q is the same anywhere in the uniform field

𝐸 =Δ𝑉

𝑑

Dipole Field and Equipotential lines

3 4

Solving for resultant E-fields

1. Sketch arrows to indicate the contributions from various charges at a point.

2. Identify charge to point distances for application of Coulomb’s law

3. Identify angles for resolving of E-fields. 4. Combine E fields vectorially.

Very often, conservation of energy is used when the force is not constant when a charge is moving through a non-uniform E field.

For PH1012, 2015 (Key concepts)

Ho SY, SPMS NTU

Electric Field

Electrostatic Equilibrium (no net motion of charges) Fo

r PH

10

12

20

15

(Lectu

re Sum

ma

ry an

d K

ey con

cepts)

Ho

SY, SPM

S NTU

Φ𝐸 = 𝐸 ∙ d𝐴 =𝑞

𝜖𝑜 Φ𝐸 = 𝐸 ∙ d𝐴 = 0

Infinite plane of uniform charge Infinite line of uniform charge

s = Q/A – charge per unit area

Φ𝐸 = 𝐸 ∙ d𝐴 = 2𝐸𝐴 =𝜎𝐴

𝜖𝑜

𝑬 =𝝈

𝟐𝝐𝒐

𝜆 =𝑄

𝑙– - charge per unit length

Φ𝐸 = 𝐸 ∙ d𝐴 = 𝐸 𝑑𝐴 =𝜆𝑙

𝜖𝑜

𝐸 2𝜋𝑟𝑙 =𝜆𝑙

𝜖𝑜⇒ 𝑬 =

𝝀

𝟐𝝅𝝐𝒐𝒓

𝐸 =𝜎

𝜖𝑜

𝑄 - total charge, 𝜌 – charge density

Φ𝐸 = 𝐸 ∙ d𝐴 = 𝐸 𝑑𝐴 =𝑄

𝜖𝑜

𝐸 4𝜋𝑟2 =𝑄

𝜖𝑜⇒ 𝑬 =

𝑸

𝟒𝝅𝝐𝒐𝒓𝟐

(or E =𝜌𝑎3

3𝜖𝑟2 where 𝜌 is charge density)

1. The electric field is zero everywhere inside the conductor (hollow or solid).

2. If conductor is isolated (from external E-fields), charges reside on the surface.

3. The electric field near the surface of the conductor points perpendicular to the surface.

4. For irregularly shaped conductor, the surface density is greatest at locations where the radius of curvature of the surface is smallest.

∆𝑉 = − 𝐸. 𝑑𝑟 𝑏

𝑎

= 𝐸𝑑 =𝜎𝑑

𝜖𝑜=𝑄𝑑

𝐴𝜖𝑜

𝐶 =𝑄

∆𝑉=𝜖𝑜𝐴

𝑑

∆𝑉 = − 𝐸. 𝑑𝑙 𝑏

𝑎

= − 𝜆

2𝜋𝜖𝑜𝑟𝑑𝑟

𝑏

𝑎

= −𝑄

2𝜋𝜖𝑜𝑙ln(𝑏

𝑎)

𝐶 =𝑄

∆𝑉=2𝜋𝜖𝑜𝑙

ln(𝑎𝑏)

∆𝑉 = − 𝐸. 𝑑𝑟 𝑏

𝑎

= − 𝑄

4𝜋𝜖𝑜𝑟2𝑑𝑟

𝑏

𝑎

= −𝑄

4𝜋𝜖𝑜(1

𝑏−1

𝑎)

𝐶 =𝑄

∆𝑉=4𝜋𝜖𝑜𝑎𝑏

(𝑎 − 𝑏)

Gauss’s Law:

Capacitance:

Potential Difference:

Sphere of uniform charge

5 6

convention − d𝐴 𝑝𝑜𝑖𝑛𝑡𝑖𝑛𝑔 𝑜𝑢𝑡𝑤𝑎𝑟𝑑𝑠 𝑖𝑠 + 𝑣𝑒

7

a b

b

a

a

b

See footnote pg 631 of Giancoli, concerning –ve signs

On Capacitance Included for completeness, not required for PH1012

Moment of inertia / Rotational Dynamics

Discrete mass: 𝐼 = 𝑚1𝑟1

2 +𝑚2𝑟22 +⋯+𝑚𝑛 𝑟𝑛

2 = ∑ 𝑚𝑖𝑟𝑖2𝑛

𝑖=1 Continuous mass distribution: 𝐼 = ∫𝑅2𝑑𝑚

If the masses are the same, which has larger MI?

𝑇𝑟𝑎𝑛𝑠𝑙𝑎𝑡𝑖𝑜𝑛𝑎𝑙 → 𝑅𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑚𝑎𝑠𝑠 𝑚 → 𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝐼; 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑣 → 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝜔;

𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑎 → 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝛼 For a fixed axis in inertia reference frame,

𝐹 = 𝑚𝑎 → 𝜏 = 𝐼𝛼 ; 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 =1

2mv2 →

1

2I𝜔2

Pure rotational motion (Involving height change) Conservation of energy:

𝐺𝑃𝐸 + 𝐾𝐸𝑟𝑜𝑡 𝑖 = 𝐺𝑃𝐸 + 𝐾𝐸𝑟𝑜𝑡 𝑓

Newton’s 2nd law (rotation) ∑𝜏 = 𝐼𝛼 (about pivot)

For PH1012, 2015 (Key concepts)

Ho SY, SPMS NTU

For forces acting on extended objects in general:

∑𝜏 = 𝐼𝛼 ∑𝐹 = 𝑚𝑎

Cannot assume that 𝑎 and 𝛼 are related.

e.g. ∑𝐹 = 0 But ∑𝜏 ≠ 0 = 𝐹𝑅

For forces acting on extended objects (rolling without slipping):

∑𝜏𝐶𝑀 = 𝐼𝐶𝑀𝛼𝐶𝑀 ∑𝐹 = 𝑚𝑎

𝑎 = 𝑟𝛼 or 𝑣 = 𝑟𝜔

𝐾. 𝐸.→ 𝐾. 𝐸.𝑡𝑟𝑎𝑛𝑠+𝐾. 𝐸.𝑟𝑜𝑡

𝑀𝑔ℎ =1

2𝑀𝑣𝐶𝑀

2 +1

2𝐼𝐶𝑀𝜔

2 +𝑀𝑔ℎ′

𝑎 = 𝑟𝛼 or 𝑣 = 𝑟𝜔

(Rolling without slipping):

Parallel Axis theorem IX = ICM +Mh

2

𝑣 = 𝑟𝜔

Rotational Dynamics

Non-uniform circular motion with constant angular acceleration 𝛼,

𝜔𝑓 = 𝜔𝑖 + 𝛼𝑡

𝑎 𝑅 = 𝑟𝜔𝑓2; 𝑎 ta𝑛 = 𝑟𝛼

𝑎 = 𝑎 𝑅 + 𝑎 𝑡𝑎𝑛

Moment of inertia / Rotational Dynamics

Discrete mass: 𝐼 = 𝑚1𝑟1

2 +𝑚2𝑟22 +⋯+𝑚𝑛 𝑟𝑛

2 = ∑ 𝑚𝑖𝑟𝑖2𝑛

𝑖=1 Continuous mass distribution: 𝐼 = ∫𝑅2𝑑𝑚

If the masses are the same, which has larger MI? Parallel Axis theorem IX = ICM +Mh

2

𝑇𝑟𝑎𝑛𝑠𝑙𝑎𝑡𝑖𝑜𝑛𝑎𝑙 → 𝑅𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑚𝑎𝑠𝑠 𝑚 → 𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝐼; 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑣 → 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝜔;

𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑎 → 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝛼 For a fixed axis in inertia reference frame,

𝐹 = 𝑚𝑎 → 𝜏 = 𝐼𝛼 ; 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 =1

2mv2 →

1

2I𝜔2

Perpendicular axis theorem 𝐼𝑧 = 𝐼𝑥 + 𝐼𝑦

𝐿 = 𝐼𝜔 = 𝐼𝜔𝑜; ∑𝐿𝑖 = ∑𝐿𝑓

E.g. 𝑚𝑟𝑣 = 𝐼𝜔 + 𝑚𝑅𝑜

2 𝜔

Conservation of Angular Momentum

Pure rotational motion (Involving height change) Conservation of energy:

𝐺𝑃𝐸 + 𝐾𝐸𝑟𝑜𝑡 𝑖 = 𝐺𝑃𝐸 + 𝐾𝐸𝑟𝑜𝑡 𝑓

Newton’s 2nd law (rotation) ∑𝜏 = 𝐼𝛼 (about pivot)

For PH1012, 2015 (Key concepts, Mechanics II)

Ho SY, SPMS NTU

For extended objects in equilibrium: ∑𝜏 = 0 ∑𝐹 = 0

For forces acting on extended objects in general:

∑𝜏 = 𝐼𝛼 ∑𝐹 = 𝑚𝑎

Cannot assume that 𝑎 and 𝛼 are related.

e.g. ∑𝐹 = 0 But ∑𝜏 ≠ 0 = 𝐹𝑅

For forces acting on extended objects (rolling without slipping):

∑𝜏𝐶𝑀 = 𝐼𝐶𝑀𝛼𝐶𝑀 ∑𝐹 = 𝑚𝑎

𝑎 = 𝑟𝛼 or 𝑣 = 𝑟𝜔

𝐾. 𝐸.→ 𝐾. 𝐸.𝑡𝑟𝑎𝑛𝑠+𝐾. 𝐸.𝑟𝑜𝑡

𝑀𝑔ℎ =1

2𝑀𝑣𝐶𝑀

2 +1

2𝐼𝐶𝑀𝜔

2 +𝑀𝑔ℎ′

𝑎 = 𝑟𝛼 or 𝑣 = 𝑟𝜔

(rolling without slipping):

- Mass of (non uniform) rod - mixing water; heat capacities - class averages - turning moments

10

Total Mass Center of Mass Moment of Inertia (about an axis)

Discrete

Continuous

For PH1012, 2015 (Key concepts)

Ho SY, SPMS NTU