chapter 9 rotational motion - texas a&m...
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● 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)