Chapter 8- Rotational Motion
Assignment 8
Textbook (Giancoli, 6th edition), Chapter 7-8:
Due on Thursday, November 19
1.
On page 190 of Giancoli, problem 40.
2.
On page 193 of Giancoli, problem 79.
3.
On page 219 of Giancoli, problem 20.
4.
The flywheel of a steam engine runs with a constant angular speed of
156 rev/min. When steam is shut off, the friction of the bearings and of the air
brings the wheel to rest in 2.20 h.
(a) What is the constant angular acceleration of the wheel, in rev/min2 ?
(b) How many rotations will the wheel make before coming to rest?
(c) What is the tangential linear acceleration of a point 52.4 cm from the
axis of rotation when the flywheel is turning at 72.5 rev/min
(d) What is the magnitude of the total linear acceleration of the point in part
(c)?
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
All marks, including assignments, have been
posted on the web.
http://ilc2.phys.uregina.ca/~barbi/academic/phys109/2009/grades-web.pdf
You have until Friday, Nov. 13, to verify that
all your marks have been entered and are
correct in the list.
The link will be removed after 1pm on Friday.
Recalling Last Lecture
Angular Quantities
In purely rotational motion, all points on the object move in circles around the axis
of rotation (“O”) which is perpendicular to this slide.
The radius of the circle is r.
All points on a straight line drawn through the axis
move through the same angle in the same time.
The angle θ in radians is defined:
(8-1)
Where
r = radius of the circle
l = arc length covered by the angle θ
The angular displacement is what characterizes
the rotational motion.
Angular Quantities
One radian is defined such that it corresponds
to an arc of circle equal to the radius of the circle.
Or, if we use eq. 8.1:
Radians are dimensionless.
(8-2)
Angular Quantities
(8-3)
Angular displacement:
(8-4)
Angular Quantities
Average angular velocity and instantaneous angular velocity:
(8-5)
(8-6)
Average angular acceleration and instantaneous
angular acceleration:
(8-7)
Both the velocity and acceleration are
the same for any point in the object.
(8-8)
Angular Quantities
∆
At each angular position, point P will have a
linear velocity whose directions are tangent to
its circular path.
∆
(8-9)
Tangential linear acceleration:
(8-10)
Centripetal acceleration (radial direction towards
the center of rotation):
(8-11)
Total acceleration:
(8-12)
(8-13)
Angular Quantities
Angular Quantities
The frequency is the number of complete revolutions per second:
(8-14)
Frequencies are measured in hertz.
The time required for a complete revolution is called period, or in other words: period
is the time one revolution takes:
(8-15)
Constant Angular Acceleration
The equations of motion for constant angular acceleration are the same as those for
linear motion, with the substitution of the angular quantities for the linear ones.
Today
Linear Momentum
Problem 8-8 (textbook) A rotating merry-go-round makes one complete revolution
in 4.0 s (Fig. 8–38).
(a) What is the linear speed of a child seated 1.2 m from the center?
(b) What is her acceleration (give components)?
Linear Momentum
Problem 8-8
The angular speed of the merry-go-round is
ω = 2π rad 4.0 s = 1.57 rad s
(a)
v = ω r = (1 .5 7 r a d s e c ) (1 .2 m
)=
1 .9 m s
(b) Ignoring air our other resistance, there is no tangential acceleration (no tangential
forces are applied).Therefore, the acceleration is purely radial.
a R = ω 2 r = (1.57 rad sec ) (1.2 m ) = 3.0 m s 2 towards the center
2
and
atan = 0
Linear Momentum
Problem 8-13 (textbook) A turntable of radius R1 is turned by a circular rubber roller
of radius R2 in contact with it at their outer edges. What is the ratio of their angular
velocities, ω1/ ω2.
Linear Momentum
Problem 8-13
The tangential speed of the turntable must be equal to the tangential speed of the
roller, if there is no slippage.
v1 = v 2
→
ω 1 R1 = ω 2 R 2
→
ω 1 ω 2 = R 2 R1
Linear Momentum
Problem 8-19 (textbook) A cooling fan is turned off when it is running at 850 rev/min.
It turns 1500 revolutions before it comes to a stop.
(a) What was the fan’s angular acceleration, assumed constant?
(b) How long did it take the fan to come to a complete stop?
Linear Momentum
Problem 8-19
(a)
The angular acceleration can be found from
ω
2
= ω o2 + 2 α θ
ω 2 − ω o2 0 − ( 850 rev min )
rev   2π rad   1 min 
rad

α =
=
=  −241
=
−
0.42



2θ
2 (1500 rev )
min 2   1 rev   60 s 
s2

2
(b)
The time to come to a stop can be found from
θ =
1
2
(ω o
+ω
)t
2 (1 5 0 0 re v )  6 0 s 
t =
=

 = 210 s
ωo + ω
8 5 0 re v m in  1 m in 
2θ
2
Centripetal Forces (Section 55-2, textbook)
We can now go back to chapter 5 and discuss the concept of centripetal forces.
For an object to be in uniform circular motion, there must be a net force acting on it.
In fact, we have discussed that there is a radial (centripetal) force associated to a
particle (point) moving in circular motion.
According to Newton’s second law, there should exist a net force Frnet such that:
(8-16)
FRnet is known as centripetal force.
The centripetal force points in the direction of the
radial acceleration (inwards).
We can use eq. 8-11 (
) and rewrite 8-16 as:
(8-17)
Centripetal Forces (Section 55-2, textbook)
A misconception is to believe that there exist a force (centrifugal force) pointing
outwards acting on an object when it moves in a circular path.
An example is the case of a person swinging a ball on the end of a string as shown
in the figure.
The force on the person’s hand is the reaction of the string to the inward pull the
person exerts on it to produce the circular motion.
When the string is released, the ball flies away
tangentially as depicted in the figure b below.
So, there was no forces in the outward direction,
otherwise we would have motion in this direction
after the string is released (figure a).
Centripetal Forces (Section 55-2, textbook)
Note: If a particle is developing a circular motion at constant angular velocity ω, its
angular acceleration α is zero (eq. 8-8), and the tangential acceleration atan is
consequently zero (eq. 8-10).
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 α , then there is a
tangential acceleration atan
The net force will NOT be pointing inward.
(We will come back to it when we discuss torque)
Centripetal Forces (Section 55-2, textbook)
Problem 5-7 (textbook) A ball on the end of a string is revolved at a uniform rate in
a vertical circle of radius 72.0 cm, as shown in Fig. 5–33. If its speed is 4.00 m/s and
its mass is 0.300 kg, calculate the tension in the string when the ball is
(a) at the top of its path, and
(b) at the bottom of its path.
Centripetal Forces (Section 55-2, textbook)
Problem 5-7
A free-body diagram is shown in the figure. Since the object is moving in a circle with
a constant speed, the net force on the object at any point must point to the center of
the circle.
(a) Take positive to be downward. Write Newton’s 2nd law in the downward direction.
∑F
R
= mg + FT1 = ma R = m v 2 r
→
2


4.00
m
s
(
)
2
2
FT1 = m ( v r − g ) = ( 0.300 kg ) 
− 9.80 m s  = 3.73 N
 0.720 m



This is a downward force, as expected.
Centripetal Forces (Section 55-2, textbook)
Problem 5-7
A free-body diagram is shown in the figure. Since the object is moving in a circle with
a constant speed, the net force on the object at any point must point to the center of
the circle.
(b) Take positive to be upward. Write Newton’s 2nd law in the upward direction.
∑F
R
= FT 2 − m g = m a = m v 2 r
(
FT 1 = m v 2
→
 ( 4 .0 0 m s ) 2

2
r + g ) = ( 0 .3 0 0 k g ) 
+ 9 .8 0 m s  = 9 .6 1 N
 0 .7 2 0 m



This is a upward force, as expected.
Rolling Motion (Without Slipping)
When a wheel or radius R rolls without slipping along a flat straight path, the points of
the wheel in contact with the surface are instantaneously at rest and the wheel
rotates about a rotation axis through the contact point.
In fact, the wheel is experiencing both a translational and
rotational motion.
P
The center C of the wheel does not rotate relative to the
contact point. It undergoes only translational motion with
velocity
.
r
R
ω
A point P at a distance r from the center of rotation (point of
contact) will undergo a rotation with angular velocity given by eq. 8-9:
(nonslip condition for the speed of any point at a distance r).
The center of the wheel moves with a linear speed given by:
(nonslip condition for the speed of the center of the wheel)
Rolling Motion (Without Slipping)
The linear acceleration of the center of the wheel will be given by:
P
r
R
(nonslip condition for the acceleration of the
center of the wheel)
ω
θ
Rolling Motion (Without Slipping)
Let’s now look at this problem from a different perspective.
So far we have placed ourselves on a reference system
which is at rest relative to the surface. Let’s now assume
a reference system at rest with respect to the center of
the wheel.
In this system, the surface (and therefore the point of contact)
moves with a velocity
.
As the wheel rotates about its center through and 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
The center of the wheel remains directly over the
point of contact, therefore it also moves by the same
distance.
l
Rotational Kinetic Energy and Moment of Inertia
An object can be seen as made of many individual tiny particles located at different
positions. In the real word, this is actually the way things are made. The object can
then be interpreted as a system of particles.
We have already seen that the total kinetic energy of a system is the sum of the
kinetic energy of each of its constituents. So, the kinetic energy of this object can be
written as:
If the object is undergoing a rotational motion, we can use eq. 8-9 (noting that the
angular velocity is the same for all particles) to write:
Rotational Kinetic Energy and Moment of Inertia
Then
⇒
(8-18)
We can now define the following quantity:
(8-19)
Such that:
(8-20)
The quantity I is known as moment of inertia.
Rotational Kinetic Energy and Moment of Inertia
Note that in the case of the wheel discussed in the rolling motion without slipping few
slides ago, the wheel also undergoes translational motion.
The total kinetic energy will then be the sum of the kinetic
energy due to its linear and angular motions:
P
r
R
(8-21)
Where
ω
is the total mass of the object.
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.
These two objects have the same mass, but the one on the left has a greater
rotational inertia, as so much of its mass is far from the axis of rotation.
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
Compare (f) and (g), for example.
We will come back to moment of inertia
when we discuss about torque.
Linear Momentum
Problem 8-45 (textbook) A bowling ball of mass 7.3 kg and radius 9.0 cm rolls
without slipping down a lane at 3.3 m/s. Calculate its total kinetic energy.
Linear Momentum
Problem 8-45
The total kinetic energy is the sum of the translational and rotational kinetic energies.
Since the ball is rolling without slipping, the angular velocity is given by
ω=v R
The rotational inertia of a sphere about an axis through its center is
I =
2
5
mR2
(see table presented two slides ago)
KEtotal = KEtrans + KErot = 12 mv 2 + 12 I ω 2 = 12 mv 2 + 12 25 mR 2
= 0.7 ( 7.3 kg )( 3.3 m s ) = 56 J
2
v2
R2
= 107 mv 2