Chapter 8
Rotational Motion and Equilibrium
Prior to chapter 7, we only considered translational motion, in which
objects could always be considered to have no spatial extent.
Rotational motion was irrelevant at the time, since such a “point
object” can not physically rotate (its radius is zero).
In chapter 7, we introduced circular motion.
Although some examples involved extended
objects such as turntables, we really only
considered circular motion of point masses (such
as the moon’s orbit about the Earth, or the circular
motion of a boy on a merry-go-round). Here, we
consider the rotation of extended objects … for
example, that of the turntable itself, rather than
the circular motion of an ant riding on top of it.
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 1
8.1 Rigid Bodies, Translations, and Rotations
In PC141, rotational motion can only be analyzed for a rigid
body:
a rigid body is an object or system of particles in
which the distances between particles are fixed
For instance, a snooker ball is a rigid body. If you consider any
two atoms in the ball, the distance between them will be
constant even as the ball is rolling, sliding, colliding, etc. On the
other hand, a ball of silly putty is not a rigid object, as it can be
easily deformed.
Realistically, no object is perfectly rigid, since its atoms are
constantly vibrating. However, many solid objects can be treated
as rigid bodies for the sake of PC141. Even people!
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 2
8.1 Rigid Bodies, Translations, and Rotations
Rigid bodies may be subject to either or both of two types of
motion – translational and rotational. For example, a chest pass
of a basketball gives it almost entirely translational motion.
Spinning the ball on your finger gives it entirely rotational
motion. Rolling the ball down the court gives it both types (the
ball rotates about an axis of symmetry while this axis translates
down the court).
In pure translational motion, every particle of an object has the
same instantaneous velocity.
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 3
8.1 Rigid Bodies, Translations, and Rotations
On the other hand, for pure rotational motion (“pure” means
that the rotation is about an axis that is fixed in space…this axis is
not translating or rotating), every particle in the body has the
same instantaneous angular velocity, each traveling in circles
about the axis of rotation.
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 4
8.1 Rigid Bodies, Translations, and Rotations
When an object undergoes rolling without slipping, the motion
can be described as a combination of pure translation and pure
rotation.
We’ll discuss this in class.
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 5
8.1 Rigid Bodies, Translations, and Rotations
Consider an object that undergoes rolling without slipping. In a
given time t, it rotates through an angle 𝜃. A point on the object
that was originally in contact with the surface moves through an
arc length s. We know from chapter 7 that 𝑠 = 𝑟𝜃. We also
know from chapter 6 that the CM of the object is directly
overtop of the point of contact, and that it moves a distance s
(if it did not, this would imply that
slipping has occurred). Therefore,
𝑠 𝑟𝜃
𝑣𝐶𝑀 = =
= 𝑟𝜔
𝑡
𝑡
By the same argument, it can be
shown (p. 269) that
𝑣𝐶𝑀 𝑟𝜔
𝑎𝐶𝑀 =
=
= 𝑟𝛼
𝑡
𝑡
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 6
Problem #1: Pure Rotational Motion
WBL LP 8.1
In pure rotational motion of a rigid body…
A
…all the particles of the body have the same angular velocity
B
…all the particles of the body have the same tangential velocity
C
…acceleration is always zero
D
…there are always two simultaneous axes of rotation
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 7
Problem #2: Rolling Ball
WBL EX 8.5
A ball with a radius of 15 cm rolls on a level surface, and the translational speed
of the center of mass is 0.25 m/s. What is the angular speed about the center of
mass if the ball rolls without slipping?
Solution: In class
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 8
8.2 Torque, Equilibrium, and Stability
We have known since chapter 4 that a force is necessary to
produce translational acceleration of an object. To produce
rotational acceleration of an object, a force is also necessary –
however, the situation is a bit more complex.
This can be illustrated by an easy experiment. Find a door with a
simple hinge mechanism that opens away from you. Using one
finger, push the door near the handle. It should rotate quite
easily. Then, close the door and push again, this time at a point
much closer to the hinge. You will find that it is much more
difficult to rotate the door…a much greater force is required in
order to achieve the desired angular acceleration of the door
about the hinges. Next, try to grab the door handle and pull in a
direction away from the hinges. The door won’t rotate at all.
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 9
8.2 Torque, Equilibrium, and Stability
What this experiment tells us is that a force’s ability to cause
rotational acceleration depends not only on its magnitude, but
also on (i) its point of application relative to the axis of rotation,
and (ii) the angle between the force vector and a vector directed
from the axis of rotation to the application point. These two
concepts are combined into the moment arm, 𝑟⫠ = 𝑟 sin 𝜃,
where 𝑟 is the distance between the axis and the force
application point, and 𝜃 is the angle between r and F.
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 10
8.2 Torque, Equilibrium, and Stability
The product of the force and the moment arm is called torque,𝛕
(Greek lower-case “tau”). The magnitude of the torque is
𝜏 = 𝑟⫠ 𝐹 = 𝑟𝐹 sin 𝜃
Torque has SI units of m·N. Many textbooks write this as N·m.
For once, our authors have made a wise choice – remember, N·m
is also the SI unit for work. However, work and torque are not
the same thing (for example, the latter is a vector while the
former is a scalar…furthermore, work’s units can be expressed as
Joules, but torque’s units can not).
The direction of 𝛕 is determined by the right-hand rule. If you
curl the fingers of your right hand from the 𝐫 direction to the 𝐅
direction, your right thumb points in the direction of 𝛕.
Combining this fact with the equation above indicates that
𝛕=𝐫×𝐅
PC141 Intersession 2013
Slide 11
Day 16 – June 13 – WBL 8.1-8.3
8.2 Torque, Equilibrium, and Stability
The role of torque in rotational motion is analogous to the role of
force in translational motion. Any non-zero net torque will result
in an angular acceleration, just as any non-zero net force results
in a linear acceleration.
Examples 8.2 and 8.3 of the text (pp. 270-272) discuss the
concept of torque as it applies to kinesiology.
You can see (and experience) this concept for
yourself at the gym. Most newer exercise
equipment is based on the variable cam, which
causes a constant mass to appear to get lighter and
heavier during a full range of motion. We’ll discuss
this in class.
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 12
8.2 Torque, Equilibrium, and Stability
Equilibrium
Since unbalanced forces cause translational acceleration and
unbalanced torques cause rotational acceleration, it stands to
reason that an object that experiences neither of these
imbalances will be at mechanical equilibrium; it will experience
no acceleration of either type. The defining equations are
F𝑛𝑒𝑡 = F𝑖 = 0 and 𝛕𝑛𝑒𝑡 = 𝛕𝑖 = 0. The situations depicted
in parts (a) and (b) of the figure have zero net force and zero net
torque (zero torque at all,
in fact). The situation in
part (c) has zero net force
but non-zero net torque.
It will experience a
rotational acceleration.
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 13
8.2 Torque, Equilibrium, and Stability
Stability and Center of Gravity
An object / particle / system which is at equilibrium can be either
stable or unstable in a gravitational field. The stability is
determined using the center of gravity (the point at which all the
weight of an object may be considered to be acting). For PC141,
we will assume that the center of gravity coincides with the
center of mass, which we know how to calculate. If you can’t
remember the difference between the CM and CG, take a few
minutes to review section 6.5 of the text.
If an object is in stable equilibrium, any small displacement from
equilibrium results in a restoring force or torque, which acts to
return the object to the original equilibrium position. In
unstable equilibrium, the displacement tends to move the
object farther away from the equilibrium position.
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 14
8.2 Torque, Equilibrium, and Stability
Stability and Center of Gravity cont’
Two examples of systems at stable equilibrium are shown here.
On the left, a ball sits at the bottom of a bowl. There are no
torques, and the only forces (the ball’s weight, and the normal
force) cancel each other out. If the ball is moved to the right, the
net force is no longer zero; there is a resultant restoring force
that returns the ball to the bottom. On the left, an object is
tipped up onto one edge. When tipped slightly, the object’s
weight (applied at the
CG) produces a restoring
torque that returns the
object to its original
position.
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 15
8.2 Torque, Equilibrium, and Stability
Stability and Center of Gravity cont’
On the other hand, if a ball is placed at the top of an overturned
bowl, it is still in equilibrium (no torques, and weight balances
the normal force). However, if the ball is displaced to the right,
the net force now acts to continue moving the ball to the right;
the ball will not return to the same equilibrium position.
Likewise, for the object on the right, a slight clockwise pivot
produces a torque about the rotation axis that acts to continue
rotating the object in the
same clockwise
direction. Again, the
object will not return to
the original equilibrium
position; it is unstable.
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 16
8.2 Torque, Equilibrium, and Stability
Stability and Center of Gravity cont’
Stability of translational equilibrium is usually easy to determine
by inspection. For rotational equilibrium, it can be a bit trickier.
In summary, we can say that
an object is in stable rotational equilibrium as long as
its center of gravity after a small displacement still lies
above and inside the object’s original base of support.
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 17
8.2 Torque, Equilibrium, and Stability
The counterweight on a
construction crane is used to
ensure rotational stability of
the crane under heavy loads
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 18
Problem #3: Unstable Equilibrium
WBL LP 8.7
If an object in unstable equilibrium is displaced slightly…
A
…its potential energy will decrease
B
…the center of gravity is directly above the axis of rotation
C
…no gravitational work is done
D
…stable equilibrium follows
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 19
Problem #4: Tipping a Crate
WBL EX 8.15
A worker applies a horizontal force to the top edge of a
crate to get it to tip forward. If the crate has a mass of 100
kg and is 1.6 m tall and 0.80 m in depth and width, what is
the minimum force needed to make the crate start tipping?
Assume that the CM of the crate is at its center, and that the
static friction between the crate and ground is sufficient to
prevent sliding.
Solution: In class
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 20
Problem #5: Bicep Muscle
WBL EX 8.19
Determine the force exerted by the bicep muscle, assuming
that the hand is holding a ball with a mass of 5.00 kg.
Assume that the mass of the forearm is 8.50 kg with its center
of mass located 20.0 cm away from the elbow joint. Assume
also that the center of mass of the ball in the hand is 30.0 cm
away from the elbow joint. The muscle contact point is 4.00
cm from the elbow joint.
Solution: In class
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 21
8.3 Rotational Dynamics
Moment of Inertia
We have seen that net torque causes rotational acceleration in
the same way that net force causes translational acceleration.
For the latter, the defining equation is Newton’s 2nd law, 𝐹𝑛𝑒𝑡 =
𝑚𝑎. We would like to derive a similar equation that relates τ to
α.
Consider a constant net force F that acts on a particle of mass m
as shown in the figure. The particle is a distance r from the axis
of rotation. We already know that 𝜏𝑛𝑒𝑡 = 𝑟⫠ 𝐹 = (𝑟 sin 𝜃) 𝐹.
This can be rearranged as 𝜏𝑛𝑒𝑡 = 𝑟⫠ 𝐹 =
𝑟 𝐹 sin 𝜃 = 𝑟𝐹⫠ . But by Newton’s 2nd
law, 𝐹⫠ = 𝑚𝑎⫠ . Finally, since 𝑎⫠ = 𝑟𝛼,
we can write 𝜏𝑛𝑒𝑡 = 𝑚𝑟 2 𝛼.
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 22
8.3 Rotational Dynamics
Moment of Inertia cont’
The preceding equation is valid for a single particle of mass m, a
distance r from the axis of rotation. For an object of finite size,
we simply have to add these net torques for every particle in the
object:
𝜏𝑛𝑒𝑡 = 𝜏1 + 𝜏2 + ⋯ + 𝜏𝑛
= 𝑚1 𝑟12 𝛼 + 𝑚2 𝑟22 𝛼 + ⋯ + 𝑚𝑛 𝑟𝑛2 𝛼
= 𝑚1 𝑟12 + 𝑚2 𝑟22 + ⋯ + 𝑚𝑛 𝑟𝑛2 𝛼
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 23
8.3 Rotational Dynamics
Moment of Inertia cont’
This constant is called the moment of inertia, and it is given the
symbol I (that’s a capital “i”). It has SI units of kg·m2. All
together, we can write
𝜏𝑛𝑒𝑡 = 𝐼𝛼
This is the rotational form of Newton’s 2nd law. In vector form,
the law is
𝝉𝑛𝑒𝑡 = 𝐼𝜶
The moment of inertia is a property of an object’s geometry and
the axis about which it is rotating.
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 24
8.3 Rotational Dynamics
Moment of Inertia cont’
This figure shows the
moments of inertia
for various common
shapes.
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 25
8.3 Rotational Dynamics
Moment of Inertia cont’
Comparing parts (j) and (k) of the previous figure, or parts (b)
and (c), we see that the moment of inertia changes when the
axis of rotation is changed. The parallel axis theorem (which
requires calculus to prove) tells us that if the moment of inertia
of an object of mass M about an axis passing through the
object’s center of mass is 𝐼𝐶𝑀 , then the moment of inertia for the
same object about a different axis that is parallel to the first one
but displaced by a distance d is
𝐼 = 𝐼𝐶𝑀 + 𝑀𝑑 2
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 26
Problem #6: Moment of Inertia
WBL LP 8.9
The moment of inertia of a rigid body…
A
…depends on the axis of rotation
B
…cannot be zero
C
…depends on mass distribution
D
all of the preceding
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 27
Problem #7: Angular Acceleration of a Ball
WBL EX 8.35
What net torque is required to give a uniform 20-kg solid ball with a radius of
0.20 m an angular acceleration of 20 rad/s2?
Solution: In class
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 28
Problem #8: Atwood’s Machine with Massive Pulley
WBL EX 8.39
Two masses are suspended from a pulley as shown in the figure.
The pulley itself has a mass of 0.20 kg, a radius of 0.15 m, and a
constant torque of 0.35 m·N due to the friction between the rotating
pulley and its axle. What is the magnitude of the acceleration of the
suspended masses if m1 = 0.40 kg and m2 = 0.80 kg ?
Solution: In class
PC141 Intersession 2013
Day 16 – June 13 – WBL 8.1-8.3
Slide 29