Chapter 8
Rotational Motion
Some definitions
• Rigid body: any object with a definite shape so
that all particles stay in a fixed position with
regards to each other.
• Pure rotational motion: all points in the body
move in a circle. The absence of translational
motion (the center of mass does not move).
• Axis of rotation: the line on which all the
centers of all the circles lie. This will be the
center of mass.
Angular Quantities
• When dealing with rotational motion, it is
often useful to work with angular quantities
such as angular velocity and angular
acceleration.
• These are analogous to linear velocity and
linear acceleration, which means we must
now define angular position.
Angular Position
• As a body rotates, a point on the circle moves
from point P to P’ along the circumference of
its circle.
• It will sweep out an angle, θ, and an arc
length, l.
• We usually measure θ in degrees, but for
rotational motion it is usually beneficial to use
radians.
Radians
• Radian(rad): one rad is defined as the angle
subtended by an arc whose arc length (l) equals
its radius (r)
• If r = l, then θ = 1 rad
• Typically, θ = l/r
• Because there are 360o in a circle and l would
equal the circumference, 2πr, θ = l/r =
2πr / r = 2π
• So 360o = 2π radians or 1 radian = 360/ 2π, which
is approximately 57.3o.
Example
• A particular bird’s eye can distinguish objects
that subtend an angle no smaller than
3E-4 rad.
• A) how many degrees is this?
• B) how small an object can the bird see from
100m in the air?
Solution
• A. (3E-4 rad)(360 degrees / 2π rad) = 0.017
degrees.
• B. from θ = l/r we get l = θr
note: for small θ (less than 15 degrees), θ is
very close to the cord length.
l = 100m(3E-4rad) = 3E-2m or 3cm
• Radians are unit less because it is the ratio of
two lengths (units cancel).
Angular Velocity
• Velocity is displacement over time this is true
in the angular sense as well.
• Displacement: if we begin at some initial θ0
and ending at θ, we can find our angular
displacement, Δθ, by Δθ = θ – θ0
• Angular velocity (ω, omega) = Δθ/Δt
Angular Acceleration
• Acceleration is velocity/time, so
• Angular acceleration (α, alpha) = Δω/Δt
Linear Velocity
• Imagine a line of points along a radius of a
rotation circle.
• All of the points would have the same ω, but
the points further from the center would have
more distance to cover in the same amount of
time, meaning its has a greater linear, or
tangential, velocity, v.
• v = rω
Linear Acceleration
• The tangential acceleration is given by
atan = Δv/Δt = r Δω/Δt = rα
• But wait! We already covered in chapter 5 that
objects in circular motion experience a
centripetal acceleration aR.
• So the total linear acceleration of a rotating
point is the vector sum of atan and aR, where
aR can be rewritten as aR = v2/r = (ωr)2/r = ω2r
Frequency
• f can be written in terms of ω easily,
f = ω/2π
• Or ω = 2πf
Example
• What is the linear velocity of a child sitting
1.2m from the center of a merry-go-round
that makes one complete revolution on 4.0s?
• What is the child’s acceleration?
Solution
•
•
•
•
•
•
First find f: f = 1/T = 1rev/4.0s = 0.25 rev/s
ω = 2πf = (2π rad/rev)(0.25 rev/s) = 1.6 rad/s
v = rω = 1.2m(1.6 rad/s) = 1.9m/s
As for the acceleration:
atan = 0 because the rotation is constant
aR = ω2r = (1.6 rad/s)2(1.2m) = 3.1m/s2
Rotational Kinematics
• The rotational kinematic equations are simply
the 4 linear kinematic equations from chapter
2 with the following substitutions:
• x becomes θ
• v becomes ω
• a becomes α
The “New” 4 Kinematic Equations
•
•
•
•
ω = ω0 + αt
θ = ω0t + ½αt2
ω2 = ω02 + 2αθ
__
  0
 
2
Rolling Motion
• Rolling without slipping combines rotational
and translational motion.
• 1 revolution of the tires corresponds to a
translational distance of 2πr
• Also, remember that v = ωr
Rotational Dynamics
• Obviously rotations require forces and the
stronger the force the higher the acceleration
just like in the linear case.
• BUT… there’s more to it than that.
• Imagine opening a swing door.
– You apply a force, F1, perpendicular to the door at
the far end vs.
– You apply a force, F2, that is equal in magnitude to
F1, you apply it at the middle of the door.
Lever arm
• What we see is the magnitude of the force
alone is not enough.
• It also matters where the force is applied.
• The angular acceleration is proportional to the
magnitude of the force and the perpendicular
distance from the axis of rotation to the line
along which the force acts.
• This distance is called the lever arm or
moment arm, which is given the variable r.
Torque
• The acceleration is proportional to the
product of the force times the lever arm, rF.
• This product is called torque, τ (tau: t + ”ow”)
• There is a word we have been glossing over
thus far, perpendicular.
• Let’s go back to our door example:
– Picture another force, F3, that is equal in
magnitude to F1, acts on the same spot as F1, but
acts at an angle to the door.
Torque continued
• So just like in the linear case we only care
about the force component that causes the
motion.
• HOWEVER, unlike the linear case, we want the
perpendicular component. (the linear case
wanted the parallel component)
• Therefore, τ = rFperpendicular = rFsinθ
Units of Torque
• Torque is a distance (the lever arm) times a
force (the perpendicular component);
therefore, the units of torque are mN
• NOTE: we write this as mN and not Nm to help
you distinguish torque from energy. Energy is
a scalar quantity and torque is a vector
quantity.
• Nm is called joules, mN is NEVER referred to
as joules.
Example
• The biceps muscle exerts a vertical force on
the lower arm. The point of attachment for
the biceps muscle is 5.0cm from the elbow
(the axis of rotation) and can generate a force
of 700N. Calculate the torque when the arm is
at A.) a 90 degrees angle to the upper arm and
B.) an angle 30 degrees below perpendicular.
Solution
• A. F = 700N and the Fperp = 0.050m, so
• τ = (0.050m)(700N) = 35mN
• B. The angle of the arm shortens the lever arm
down to 0.050sin60, so
• τ = (0.050m)(sin60)(700N) = 30mN
It’s how you use it
• Adult chimpanzees only have about 1/3 the
muscle mass of an adult human male;
however, they have been observed to be twice
as strong as human for certain movements.
How is this possible?
Return of the sigma
• If more than one torque is acting on a body,
the acceleration α is proportional to the net
torque.
• Remember the vector nature of torque:
clockwise subtracts from counterclockwise
and vice versa.
• It is accepted by the scientific community that
counterclockwise is positive and clockwise is
negative.
Similar to chapter 4
• Let us begin with F = ma,
• Remember, a = rα so F = mrα
• We no longer have force in chapter 8, we have
torque (Fr).
• So, let’s multiply both sides by r so that we
can morph F into τ:
Fr = mr2α or
τ = mr2α
Rotational Inertia
• So we finally know exactly how τ relates to α,
but it depends on this quantity mr2.
• We give a name to this quantity, we call it the
moment of inertia.
I
• Now picture a rotating rigid body like a wheel.
We can think of the wheel as having many
particles located at various distances from the
axis of rotation.
• The sum of the various torques is given by
  (mr 2 )
• Σmr2 = m1r12 + m2r22 +…+mnrn2 = I = the
moment of inertia of the body.
Torque and Inertia
• Στ = Iα
• This is the rotational equivalent of Newton’s
second law.
• Rotational inertia plays the same roll in
rotations as regular inertia does in
translational motion: a resistance to changing
velocity.
Location Location Location
• I depends not only on mass but on how that
mass is distributed about the axis.
• UNLIKE for gravity, the mass CANNOT be
considered to be concentrated at its center of
mass here.
• Consider two round objects of equal mass.
One is a solid plate and the other is a thin ring.
They will rotate differently.
Example
• A 5.0kg mass and a 7.0kg mass are mounted
4.0m apart on a light rod (considered
massless). Calculate the moment of inertia of
the system when:
• A.) it is rotated about an axis halfway between
the masses.
B.) it is rotated about an axis 0.50m to the left
of the 5.0kg mass.
Solution
• A.) I = m1r12 + m2r22 =
(5.0kg)(2.0m)2 + (7.0kg)(2.0m)2 =
20kgm2 + 28kgm2 = 48kgm2
• B.) I = m1r12 + m2r22 =
(5.0kg)(0.50m)2 + (7.0kg)(3.5m)2 =
1.3kgm2 + 142kgm2 = 143kgm2
Analysis
• Obviously, the moment of inertia is different
for different axes of rotation.
• Secondly, masses very close to the axis of
rotation contribute very little to the inertia.