Chapter 8- Rotational Motion
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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
Centripetal Forces (Section 55-2, textbook)
FRnet is known as centripetal force.
Circular motion at constant angular velocity ω
The net force will be exclusively due to the radial acceleration and will point
inward
On the other hand, if there is a non-zero angular acceleration
The net force will NOT be pointing inward.
Rolling Motion (Without Slipping)
(nonslip condition for the speed of any point at a distance r).
(nonslip condition for the speed of the center of the wheel)
(nonslip condition for the acceleration of the
center of the wheel)
P
r
R
ω
Rolling Motion (Without Slipping)
As the wheel rotates about its center through an angle θ
with angular velocity ω, the point of contact between the
wheel and the surface will have moved a distance l (arc of circle)
given by eq. 8-1:
R
ω
R θ
l
l
Rotational Kinetic Energy and Moment of Inertia
The quantity I is known as moment of inertia.
P
r
R
KE is the rotational kinetic energy
ω
Total kinetic energy:
θ
Rotational Kinetic Energy and Moment of Inertia
Moment of inertia depends not only on the mass of each particle, but also on their
distribution (position r) in the object.
We will come back to moment of inertia when we discuss about torque.
Rotational Kinetic Energy and Moment of Inertia
The rotational inertia of an object
depends not only on its mass distribution
but also the location of the axis of rotation
Today
Rotational Kinetic Energy and Moment of Inertia
Example 1: A bowling ball that has an 11-cm radius and a 7.2-kg mass is rolling
without slipping at 2.0 m/s on a horizontal ball return. It continues to roll without
slipping up a hill to a height h before momentarily coming to rest and then rolling back
down the hill. Model the bowling ball as a uniform sphere and find h.
v=0
ω=0
h
v
M R
ω
Rotational Kinetic Energy and Moment of Inertia
Example 1 (solution):
No slipping occurs, so no energy is dissipated by kinetic friction. No external forces
act on the ball-hill-Earth system, so no external forces do work on the system. The
total energy of the system is then conserved:
The total kinetic energy is the translation plus rotational kinetic energies. So, the
initial kinetic energy is:
v=0
ω=0
The final kinetic energy is zero:
h
Assume the center of the ball before
it starts rolling as the zero of the potential
energy, the initial potential energy is
0,
, and the final potential energy
will be (positive points upward):
v
M R
ω
Rotational Kinetic Energy and Moment of Inertia
v=0
Example 1 (solution):
ω=0
Therefore:
h
v
M R
ω
using the non-slipping condition for velocity
And using table fig. 8-21 from the text book to obtain:
Rotational Kinetic Energy and Moment of Inertia
Problem 8-31 (textbook): Calculate the moment of inertia of the array of point
objects shown in Fig. 8–43 about (a) the vertical axis, and (b) the horizontal axis.
Assume m = 1.8 Kg, M = 3.1 Kg, and the objects are wired together by very light,
rigid pieces of wire. The array is rectangular and is split through the middle by the
horizontal axis. (c) About which axis would it be harder to accelerate this array?
Rotational Kinetic Energy and Moment of Inertia
Problem 8-31
(a) To calculate the moment of inertia about the y-axis (vertical), use
I =
∑M R
i
2
ix
= m ( 0.50 m ) + M ( 0.50 m ) + m (1.00 m ) + M (1.00 m )
2
2
2
2
2
2
2
2



= ( m + M ) ( 0.50 m ) + (1.00 m ) = ( 4.9 kg ) ( 0.50 m ) + (1.00 m )  = 6.1 kg m2




(b) To calculate the moment of inertia about the x-axis (horizontal), use
I =
∑
M i R = (2m + 2M
2
iy
(c) Because of the larger I value, it
is harder to accelerate the array
about the vertical axis.
)( 0.25 m )
2
= 0.61 kg m 2
Rotational Dynamics; Torque
We have discussed rotational motion in terms of angular velocity, angular
acceleration and angular displacement. We have also learnt the definition of
centripetal force. However, we are yet to see what causes rotational motion.
It is clear that to make an object start rotating, a force is needed; Unlike in linear
motion, the position and direction of the applied force matter as well.
In the figure, the applied force
will be more
effective in opening the door than force
assuming they both have the same magnitude.
The angular acceleration given to the door
by
will be greater that that given by
.
In fact, the angular acceleration is not only proportional
to the magnitude of the force, but also to the perpendicular
distance from the axis of rotation to the line along which the force acts.
This distance is called lever arm (or moment arm).
The line along which the force acts is called line of action.
Rotational Dynamics; Torque
Let me repeat the definition of lever arm:
It is the perpendicular distance from the axis of rotation to the line along which
the force acts.
The figure shows three forces acting on the
door, all of them with different directions.
It is clear from figure (a) that the lever
arm for force
is zero, therefore it plays
no role in rotating the door.
On the other hand, force
has a smaller
lever arm than force
(figs. a and b),
therefore it is less effective than
in
opening the door.
Rotational Dynamics; Torque
Let’s now find how we can express the effectiveness of a force to produce
rotational motion.
Consider fig. (a). A force
is applied on small
section “i” of a rigid door with mass mi .
The total mass M of the door will be the sum over
all sections “i” :
mi
ri
The line of action makes an angle θ with the
line connecting the point of application of the
force to the center of rotation.
Let’s concentrated on the rotation subjected by the section “i”
ri
Rotational Dynamics; Torque
The force can be written in term of its components in the perpendicular and
parallel directions to the line connecting the
point of application of the force to the center
of rotation.
is the tangential force to the direction
rotation.
mi
ri
We can use the Newton’s 2nd law to write:
(8-22)
Where
is the tangential acceleration.
We wish to obtain an equation involving angular quantities.
Using equation 8-10:
ri
Rotational Dynamics; Torque
We can then rewrite eq. 8-22 as:
Where ri is the lever arm and α is the
angular acceleration of the section “i”.
Multiplying both sides of the above equation
by ri:
mi
ri
Let’s now define:
(8-23)
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
Equation 8-23 can be extended to the entire door. For this purpose, we
repeat our calculations for each section “i” of the door and sum the results up
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.
In the figure below, the two cylinders have the same masses. However, the
cylinder with larger diameter will have greater moment of inertia (larger lever
arm) than the one with smaller cylinder.
It will be harder to start rotating as well to stop it than the other cylinder in the
very same way it is harder to start moving a heavy object compared to a lighter
one in linear motion.
Rotational Dynamics; Torque
Now you know why using a wrench
makes your work much easier.
A longer lever arm is very helpful in
rotating objects.
Solving Problems in Rotational Dynamics
1. Draw a diagram.
2. Decide what the system comprises.
3. Draw a free-body diagram for each object under consideration, including all
the forces acting on it and where they act.
4. Find the axis of rotation; calculate the torques around it.
5. Apply Newton’s second law for rotation. If the rotational inertia is not provided,
you need to find it before proceeding with this step.
6. Apply Newton’s second law for translation and other laws and principles as
needed.
7. Solve.
8. Check your answer for units and correct order of magnitude.
Problem 8-22 (textbook): A 55-kg person riding a bike puts all her weight on each
pedal when climbing a hill. The pedals rotate in a circle of radius 17 cm.
(a) What is the maximum torque she exerts?
Problem 8-22 (textbook):
(a) The maximum torque will be exerted by the force of her weight, pushing
tangential to the circle in which the pedal moves.
τ = r⊥ F = r⊥ mg = ( 0.17 m )( 55 kg ) ( 9.8 m s 2 ) = 92 m N
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
)
ML2 = ML2 = (160 kg )( 3.75 m ) = 2250 kg m 2 ≈ 2.3 × 102 kg m 2
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