Download Chapter 8- Rotational Motion

Chapter 8- Rotational Motion
Assignment 9
Textbook (Giancoli, 6th edition):
Due on Thursday, November 26.
1.
On page 131 of Giancoli, problem 18.
2.
On page 220 of Giancoli, problem 24.
3.
On page 221 of Giancoli, problem 39.
4. A cue stick strikes a cue ball horizontally at a point a distance d above the
centre of the ball as in the figure below. Find the value of d for which the cue ball
will roll without slipping from the beginning. Express your answer in terms of the
radius R of the ball. (Hint one: you will find the moment of inertia of a ball or
sphere on page 208 of Giancoli. Hint two: write the “no slip" condition as
aball = Rα where aball is the linear acceleration of a point at the centre of the ball.).
Assignments and midterm exam
Solutions are available on
the web
Chapter 8
• Angular Quantities
• Constant Angular Acceleration
• Rolling Motion (Without Slipping)
• Centripetal Forces
• Torque
• Rotational Dynamics; Torque and Rotational Inertia
• Rotational Kinetic Energy
• Angular Momentum and Its Conservation
Recalling Last Lecture
Rotational Dynamics; Torque
(8-22)
mi
ri
(8-23)
Or , using
τi is known as torque, or moment of the
force.
Note that F|| does not exert torque since its lever arm is zero.
ri
Rotational Dynamics; Torque
We can extend equation 8-23 to apply to the entire door. For this purpose, we
have repeat our calculations for each section “i” of the door and add the results
to find the net torque exerted on the door:
(8-24)
mi
ri
But, according to eq. 8-19:
ri
We can then write the net torque in terms of the
moment of inertia and angular acceleration
(8-25)
Rotational Dynamics; Torque
Equation 8-25 is the rotational equivalent of Newton’s 2nd law for linear motion.
(8-25)
Here, the moment of inertia I plays the same role as the object’s mass m in
F = ma. It tells us how difficult is to set an object in rotational motion.
Today
Problem 8-37 (textbook): A centrifuge rotor rotating at 10,300 rpm is shut off and is
eventually brought uniformly to rest by a frictional torque of 1.20 m.N . If the mass of
the rotor is 4.80 kg and it can be approximated as a solid cylinder of radius 0.0710 m,
through how many revolutions will the rotor turn before coming to rest, and how long
will it take?
Problem 8-37:
The torque on the rotor will cause an angular acceleration given by
α =τ I
The torque and angular acceleration will have the opposite sign of the initial angular
velocity because the rotor is being brought to rest.
The rotational inertia is that of a solid cylinder. Substitute the expressions for angular
acceleration and rotational inertia into the equation
ω 2 = ω o2 + 2 α θ
and solve for the angular displacement.
Problem 8-37:
ω 2 = ω o2 + 2αθ
−ωο2
− MR 2ωο2
ω 2 − ωο2 0 − ωο2
θ =
=
=
=
2
1
2α
2 (τ I ) 2 (τ 2 MR )
4τ
rev   2π rad   1 min  
2 
− ( 4.80 kg )( 0.0710 m )   10, 300



min   1 rev   60 s  
 1 rev 


=
= 5865 rad 

4 ( −1.20 N m )
 2π rad 
2
= 993 rev
The time can be found from
θ =
t =
1
2
(ω o + ω ) t
2θ
ωo + ω
=
2 (9 9 3 r e v )
 60 s 

 = 1 0 .9 s
1 0 , 3 0 0 re v m in  1 m in 
Problem 8-40 (textbook): A helicopter rotor blade can be considered a long thin
rod, as shown in Fig. 8–46. (a) If each of the three rotor helicopter blades is 3.75 m
long and has a mass of 160 kg, calculate the moment of inertia of the three rotor
blades about the axis of rotation. (b) How much torque must the motor apply to bring
the blades up to a speed of 5.0 rev/s in 8.0 s?
Problem 8-40:
(a) The moment of inertia of a thin rod, rotating about its end, is given in Figure
8-21(g). There are three blades to add.
I total = 3
(
1
3
)
2.3×x10
1023kg
kg.m
kg.m
ML2 = ML2 = (160 kg )( 3.75 m ) = 2250 kg
m22 ≈ 2.3
m 22
2
(b) The torque required is the rotational inertia times the angular acceleration,
assumed constant.
τ = I totalα = I total
ω − ω0
t
(
= 2250kg m
2
)
( 5.0rev/sec )( 2π rad
8.0 s
rev )
= 8.8 ×103 m N
Angular Momentum and Its Conservation
We have defined several angular quantities in analogy to linear motion.
Type
Linear
Rotational
Relation
Displacement
x
θ
x=rθ
Velocity
v
ω
v=rω
Acceleration
a
α
a=rα
Dynamics
F (force)
τ (torque)
τ = F d sin(θ)
Inertia
m (mass)
I (moment of inertia)
Kinetic Energy
Newton’s 2nd Law
Equations of motion
Angular Momentum and Its Conservation
To complete the analogy to linear motion, few quantities as well as an important
law are missing from the previous table:
1. Angular momentum (equivalent to linear momentum in linear motion)
2. Work in rotational motion (equivalent to work done by a force in translational
motion)
3. Rotational Impulse (equivalent to linear impulse done by a force in linear
motion)
4. Generalized form of Newton’s 2nd law (equivalent to the generalized form of
Newton’s 2nd law in terms of momentum in linear motion)
5. Conservation of total angular momentum (equivalent to conservation of total
linear momentum in linear motion)
Let’s now find each of these items.
Angular Momentum and Its Conservation
1) Angular momentum
In linear motion, linear momentum is written as:
The equivalents for mass and velocity in rotational motion are moment of
inertia, I, and angular velocity, ω.
We can then define the following equivalent quantity in rotational motion:
(8-26)
L is called angular momentum.
Angular Momentum and Its Conservation
2) Work in angular momentum
In linear motion, the work done by a force is defined as:
(only the parallel component of the force to the direction of
motion matters)
The equivalents for force and displacement in rotational motion are torque
exerted by F, τ = Fd, and angular displacement θ.
We can then define the following equivalent quantity in rotational motion:
(8-27)
Eq. 8-27 represents the work done by the torque τ when rotating an object
through an angle ∆θ.
Using the above equation, we can obtain the power P (rate with it the work is done):
(8-28)
Angular Momentum and Its Conservation
3) Impulse in angular momentum
In linear motion, the impulse done by a force is defined as:
Similarly, we can define impulse in rotational motion as:
(8-29)
Angular Momentum and Its Conservation
4) Generalized form of Newton’s 2nd Law
In linear motion, the general form of Newton’s 2nd law can be expressed as:
This form includes general cases where the mass of an object can change
remember the mass of a rocket changing as it burns fuel to accelerate.
Similarly, we can write the generalized form of the rotational equivalent of Newton’s
2nd law in terms of the angular momentum L:
(8-30)
Here, changes in the momentum of inertia ( L = Iω ) are taken into account in the
formalism of Newton’s 2nd law.
Angular Momentum and Its Conservation
5) Conservation of Angular Momentum
In linear motion, the total linear momentum is conserved if there is no net external
force acting on the system:
In this case, the total linear momentum does NOT change with time it is
conserved.
Similarly, if there is no net torque acting on a system, we have:
(8-31)
And the total angular momentum does NOT change with time:
The total angular momentum is conserved if there is NO external net torque
acting on the system.
Angular Momentum and Its Conservation
There are clear applications of angular momentum conservation in your day-today life.
Let’s recall the angular momentum definition (eq. 8-26):
Assume now that no external torque is exerted on the skater depicted below
and ignore friction.
As she brings her arms closer to her body, her moment of inertia I will change (will
decrease the distance between the “particles” in her body will be closer to her
axis of rotation).
Since NO external torque is applied on her, her
total angular momentum should be conserved:
Therefore, if her momentum of inertia decreases,
her angular velocity has to increase to maintain
the total angular momentum constant.
Problem 8-53 (textbook): A person stands, hands at his side, on a platform that is
rotating at a rate of 1.30 rev/s. If he raises his arms to a horizontal position,
Fig. 8-48, the speed of rotation decreases to 0.80 rev/s.
(a) Why?
(b) By what factor has his moment of inertia changed?
Problem 8-53 (textbook):
(a)
Consider the person and platform a system for angular momentum analysis. Since
the force and torque to raise and/or lower the arms is internal to the system, the
raising or lowering of the arms will cause no change in the total angular momentum of
the system. However, the rotational inertia increases when the arms are raised.
Since angular momentum is conserved, an increase in rotational inertia must be
accompanied by a decrease in angular velocity.
(b)
Li = L f
→ I i ωi = I f ω f
ωi
1.30 rev s
→ I f = Ii
= Ii
= 1.625 I i ≈ 1.6 I i
0.80 rev s
ωf
The rotational inertia has increased by a factor of 1.6
Problem 8-57 (textbook):
(a) What is the angular momentum of a figure skater spinning at 3.5 rev/s with arms
in close to her body, assuming her to be a uniform cylinder with a height of 1.5 m,
a radius of 15 cm, and a mass of 55 kg?
(b) How much torque is required to slow her to a stop in 5.0 s, assuming she does not
move her arms?
Problem 8-57 (textbook):
(a)
L = I ω = 12 MR 2ω =
1
2
( 55 kg )( 0.15 m )2  3.5
rev   2π rad 


2
=
14
kg
m
s

s   1 rev 
(b)
If the rotational inertia does not change, then the change in angular momentum is
strictly due to a change in angular velocity.
τ =
∆L
∆t
=
0 − 14 kg m 2 s
5 .0 s
= − 2 .7 m N
The negative sign indicates that the torque is in the opposite direction as the initial
angular momentum.
Problem 8-62 (textbook): A 4.2-m-diameter merry-go-round is rotating freely with
an angular velocity of 0.80 rad/s. Its total moment of inertia is 1760 Kg.m2. Four
people standing on the ground, each of mass 65 kg, suddenly step onto the edge of
the merry-go-round. What is the angular velocity of the merry-go-round now?
What if the people were on it initially and then jumped off in a radial direction (relative
to the merry-go-round)?
.
Problem 8-62 (textbook):
The angular momentum of the merry-go-round and people combination will be
conserved because there are no external torques on the combination. This situation
is a totally inelastic collision, in which the final angular velocity is the same for both
the merry-go-round and the people.
Subscript 1 represents before the collision, and subscript 2 represents after the
collision. The people have no initial angular momentum.
L1 = L 2
ω 2 = ω1
→
I1
I2
I 1ω 1 = I 2 ω 2
= ω1
I m-g-r
I m-g-r + I people
→


I m-g-r
= ω1 
2 
I
+
M
R
4
person
 m-g-r



1760 kg m 2
= ( 0.80 rad s ) 
= 0.48 rad s
2 
2
 1760 kg m + 4 ( 65 kg )( 2.1 m ) 
If the people jump off the merry-go-round radially, then they exert no torque on the
merry-go-round, and thus cannot change the angular momentum of the merry-goround. The merry-go-round would continue to rotate at .0.80 rad/s.