Chapter 9 Rotational Motion - Texas A&M...

● To study angular velocity and angular acceleration.● To examine rotation with constant angular acceleration.● To understand the relationship between linear and angular quantities.● To determine the kinetic energy of rotation and the moment of inertia.● To study rotation about a moving axis.

Chapter 9 Rotational Motion

9.2 Rotation with Constant Angular Acceleration

𝑣𝑥 𝑡 = 𝑣0𝑥 + 𝑎𝑥𝑡 ……...(2.6)

𝑥 𝑡 = 𝑥0 + 𝑣0𝑥𝑡 +1

2𝑎𝑥𝑡

2…(2.10)

𝑣𝑥2 = 𝑣0𝑥

2 + 2𝑎𝑥 𝑥 − 𝑥0 …(2.11)

𝑣𝑎𝑣,𝑥 =1

2[𝑣𝑥 𝑡 + 𝑣0𝑥]……(2.7)

Kinematic Equations for

Rotational Motion

Kinematic Equations for

Linear Motion

𝜔 𝑡 = 𝜔0 + 𝛼𝑡 ………....(9.7)

𝜃 𝑡 = 𝜃0 +𝜔0𝑡 +1

2𝛼𝑡2…(2.11)

𝜔2 = 𝜔02 + 2𝛼 𝜃 − 𝜃0 …(9.12)

𝜔𝑎𝑣 =1

2[𝜔 𝑡 + 𝜔0]……(9.8)

compare

9.3 Relationship Between Linear

and Angular Quantities

Length of Arc: 𝑠 = 𝑟𝜃

Average Speed: 𝑣𝑎𝑣 =∆𝑠

∆𝑡=

∆(𝑟𝜃)

∆𝑡= 𝑟

∆𝜃

∆𝑡= 𝑟𝜔𝑎𝑣

Instantaneous Tangential Velocity:

𝑣 = lim∆𝑡→0

∆𝑠

∆𝑡= 𝑟 lim

∆𝑡→0

∆𝜃

∆𝑡= 𝑟𝜔

Direction: tangent to the circle.

Average Tangential Acceleration:

𝑎𝑡𝑎𝑛,𝑎𝑣 =∆𝑣

∆𝑡=

∆(𝑟𝜔)

∆𝑡= 𝑟

∆𝜔

∆𝑡= 𝑟𝛼𝑎𝑣

Instantaneous Tangential Acceleration:

𝑎𝑡𝑎𝑛 = lim∆𝑡→0

∆𝑣

∆𝑡= lim

∆𝑡→0

∆(𝑟𝜔)

∆𝑡= 𝑟 lim

∆𝑡→0

∆𝜔

∆𝑡= 𝑟𝛼

Radial Acceleration: 𝑎𝑟𝑎𝑑 =𝑣2

𝑟=

(𝑟𝜔)2

𝑟= 𝜔2𝑟

Magnitude of Acceleration: 𝑎 = 𝑎𝑟𝑎𝑑2 + 𝑎𝑡𝑎𝑛

2

∆𝜃

9.4 Kinetic Energy of Rotation and Moment of Inertia

How to calculate the kinetic energy of rotating rigid body?

Cut the rigid body into many small pieces, A, B, C, …

Kinetic Energy for piece A:

𝐾𝐴 =1

2𝑚𝐴𝑣𝐴

2 =1

2𝑚𝐴(𝑟𝜔)𝐴

2=1

2𝑚𝐴𝑟𝐴

2𝜔2

Total Kinetic Energy:

𝐾 =1

2𝑚𝐴𝑟𝐴

2𝜔2 +1

2𝑚𝐵𝑟𝐵

2𝜔2 +1

2𝑚𝐶𝑟𝐶

2𝜔2 +⋯

=1

2(𝑚𝐴𝑟𝐴

2 +𝑚𝐵𝑟𝐵2 +𝑚𝐶𝑟𝐶

2 +⋯)𝜔2

=1

2𝐼𝜔2

Define the Moment of Inertia:

𝐼 = 𝑚𝐴𝑟𝐴2 +𝑚𝐵𝑟𝐵

2 +𝑚𝐶𝑟𝐶2 +⋯

Moment of Inertia for Objects of Various Shapes

9.4 Rotation about a

Moving Axis

How to calculate the kinetic energy of a rigid body that is rotating

and also having a linear motion?

𝐾 = 𝐾𝑙𝑖𝑛𝑒𝑎𝑟 + 𝐾𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 =1

2𝑀𝑣𝑐𝑚

2 +1

2𝐼𝑐𝑚𝜔

2

Example 9.9 on page 272

Given: Given M, R, and h

Find: Velocity of the center of mass vcm

Solution:

Ui = Mgh, Ki = 0, Uf = 0

𝐾𝑓 = 𝐾𝑐𝑚 + 𝐾𝑟𝑜𝑡 =1

2𝑀𝑣𝑐𝑚

2 +1

2𝐼𝑐𝑚𝜔

2

=1

2𝑀𝑣𝑐𝑚

2 +1

2(1

2𝑀𝑅2)(

𝑣𝑐𝑚

𝑅)2=

3

4𝑀𝑣𝑐𝑚

2

Apply the conservation of energy: 𝑀𝑔ℎ + 0 = 0 +3

4𝑀𝑣𝑐𝑚

2

𝑣𝑐𝑚 = 4𝑔ℎ/3

y

o

● To understand the concept of torque.● To relate angular acceleration and torque.● To work and power in rotational motion.● To understand angular momentum.● To understand the conservation of angular momentum.● To study how torques add a new variable to equilibrium.● To see the vector nature of angular quantities.

Chapter 10 Dynamics of Rotational Motion

More on the Magnitude of Torque

Three ways to calculate torque:

= Fl = Frsin

= F(rsin) = Fr

= (Fsin)r = Fr

10.2 Torque and Angular Acceleration

Again, cut the rigid body into many small pieces, A,

B, C,…. The force acting on piece A is റ𝐹𝐴.

Consider the motion of piece (particle) A. According

to Newton’s Second Law,

𝐹𝐴,𝑡𝑎𝑛 = 𝑚𝐴𝑎𝐴,𝑡𝑎𝑛 = 𝑚𝐴𝑟𝐴𝛼

𝜏𝐴 = 𝑟𝐴𝐹𝐴,𝑡𝑎𝑛 = 𝑚𝐴𝑟𝐴2𝛼

Sum over the torques for all the pieces:

𝜏𝐴 + 𝜏𝐵 + 𝜏𝐶 … = 𝑚𝐴𝑟𝐴2𝛼 +𝑚𝐵𝑟𝐵

2𝛼 +𝑚𝐶𝑟𝐶2𝛼 +⋯

= (𝑚𝐴𝑟𝐴2 +𝑚𝐵𝑟𝐵

2 +𝑚𝐶𝑟𝐶2 +⋯)𝛼

orσ𝜏 = 𝐼𝛼

This is also known as Newton’s Second Law for

rotational motion.

10.4 Angular Momentum

In Chapter 8 we defined the momentum of a particle as റ𝑝 = 𝑚 റ𝑣,

we could state Newton’s Second Law as റ𝐹 = lim∆𝑡→0

∆ റ𝑝

∆𝑡.

Here we define the angular momentum of a rigid body:

𝐿 = 𝐼𝜔Notes: It is also a vector, same as 𝜔.

Units: kg∙m2/s

Angular momentum of a point particle:

𝐿 = 𝑚𝑣𝑙 (kg∙m2/s)

Notes: (a) 𝑙 is effective the “moment arm”

(b) a particle moving along a straight line can still have

an angular momentum about a pivot or rotational axis

The Relationship Between Torque and Angular Momentum

Since σ𝜏 = 𝐼𝛼 = 𝐼 lim∆𝑡→0

∆𝜔

∆𝑡= lim

∆𝑡→0

𝐼∆𝜔

∆𝑡= lim

∆𝑡→0

∆(𝐼𝜔)

∆𝑡= lim

∆𝑡→0

∆𝐿

∆𝑡

We can state Newton’s Second Law for rotational motion as

σ𝜏 = lim∆𝑡→0

∆𝐿

∆𝑡

10.5 Conservation of Angular Momentum

Conservation of Angular Momentum

If σ𝜏 = 0,

Then 𝐿 = constant

10.6 Equilibrium of a Rigid Body

The equilibrium of a point particle is determined by the conditions of

σ𝐹𝑥 = 0 and σ𝐹𝑦 = 0.

The Equilibrium of a Rigid Body

For the equilibrium of a rigid body, both its linear motion and rotational

motion must be considered. Therefore, in addition to

σ𝐹𝑥 = 0 and σ𝐹𝑦 = 0,

We must add another condition regarding its rotational motion

σ𝜏 = 0.

This third condition can be set up about any chosen axis.

Strategy for Solving Rigid Body Equilibrium Problems

General principle: The net force must be zero.

The net torque about any axis must be zero.

● Draw a diagram according to the physical situation.

●Analyze all the forces acting on each part of a rigid body.

● Sketch all the relevant forces acting on the rigid body.

● Based on the force analysis, set up a most convenient x-y coordinate system.

● Break each force into components using this coordinates.

● Sum up all the x-components of the forces to an equation: σ𝐹𝑥 = 0.

● Sum up all the y-components of the forces to an equation: σ𝐹𝑦 = 0.

● Based on the force analysis, set up a most convenient rotational axis.

● Calculate the torque by each force about this rotational axis.

● Sum up all the torques by the forces to an equation: σ𝜏 = 0.

● Use the above three equations to solve for unknown quantities.

● To understand stress, strain, and elastic deformation.● To understand elasticity and plasticity.● To understand simple harmonic motion (SHM).● To solve equations of simple harmonic motion.● To understand the pendulum as a an example of SHM.

Chapter 11 Elasticity and Periodic Motion

Tensile and Compressive Stress and Strain

Tensile and compressive stress

Tensile stress = 𝐹⊥

𝐴

Units: N/m2 or pascal (Pa), in SI unit system

psi or pounds per square inches, in the British units

Tensile and compressive strain

Tensile strain = 𝑙−𝑙0

𝑙0=

∆𝑙

𝑙0

Units: none

Young’s modulus

𝑌 =𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑆𝑡𝑟𝑒𝑠𝑠

𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑆𝑡𝑟𝑎𝑖𝑛=

𝐹⊥/𝐴

∆𝑙/𝑙0=

𝑙0

𝐴

𝐹⊥

∆𝑙

Units: N/m2 or Pa

Example 11.1 on page 324

A stretching elevator cable

Given: m, 𝑙0, A, and ∆𝑙

Find: the cables stress, strain, and Young’s modulus

Solution:

Stress =𝐹⊥

𝐴=

𝑊

𝐴=

𝑚𝑔

𝐴=

(550 𝑘𝑔)(9.8 𝑚/𝑠2)

0.20×10−4 𝑚2 = 2.7 × 108 Pa

Strain = 𝑙−𝑙0

𝑙0=

∆𝑙

𝑙0=

0.40×10−2𝑚

3.0 𝑚= 0.00133

Young’s Modulus

𝑌 =𝑆𝑡𝑟𝑒𝑠𝑠

𝑆𝑡𝑟𝑎𝑖𝑛=2.7 × 108 Pa

0.00133= 2.0 × 1011 𝑃𝑎

11.2 Periodic Motion

Simple Harmonic Motion (SHM) illustrated by the oscillations

of a mass-loaded spring

Spring restoring force: 𝐹𝑥 = −𝑘𝑥

Acceleration of the mass: 𝑎𝑥 =𝐹𝑥

𝑚= −

𝑘

𝑚𝑥

Defining simple harmonic motion: motion driven by a

restoring force that is always opposite to the displacement and

directly proportional to the displacement in magnitude.

Note:

(a) The restoring force Fx is opposite to displacement;

(b) Fx is not a constant;

(b) As a result, the acceleration ax is not a constant;

(c) ax varies between (+ kA/m) and (– kA/m);

(d) The magnitude of ax has a maximum amax = kA/m.

11.3 Energy in Simple Harmonic Motion

Conservation of Energy in SHM

𝐸 =1

2𝑘𝐴2 =

1

2𝑚𝑣𝑥

2 +1

2𝑘𝑥2

Velocity of an object in SHM as a function of position

𝑣𝑥 = ±𝑘

𝑚(𝐴2 − 𝑥2) = ±

𝑘

𝑚(𝐴2 − 𝑥2)

Note:

(a) 𝑣𝑥 = 0 when 𝑥 = ±𝐴.

(b) Maximum speed 𝑣𝑥,𝑚𝑎𝑥 = 𝐴 𝑘/𝑚 when 𝑥 = 0.

A few notes about SHM:

● The angular frequency, period, and frequency are all independent of the amplitude.

𝜔 = 𝑘/𝑚; 𝑇 =1

𝑓= 2𝜋 𝑚/𝑘; 𝑓 =

𝜔

2𝜋=

1

2𝜋𝑘/𝑚

● If 𝜙𝑜 ≠ 0 at 𝑡 = 0, the previously derived expressions remain correct if the angular

position 𝜙 = 𝜔𝑡 is replaced by 𝜙 = 𝜙𝑜 +𝜔𝑡. For example, the position is

𝑥 = 𝐴𝑐𝑜𝑠 𝜙 = 𝐴𝑐𝑜𝑠(𝜙𝑜 + 𝜔𝑡)

● Since 𝑥 = 𝐴𝑐𝑜𝑠 𝜙 , we may re-write

𝑣𝑥 = −𝜔𝐴𝑠𝑖𝑛 𝜔𝑡 = ±𝜔𝐴 1 − cos 𝜔𝑡 2

= ±𝜔 𝐴2 − 𝐴 cos 𝜔𝑡 2 = ±𝜔 𝐴2 − 𝑥2

= ±𝑘

𝑚(𝐴2 − 𝑥2)