Rotational Motion – Part I
AP Physics C
The radian
 There are 2 types of pure unmixed motion:
 Translational - linear motion
 Rotational - motion involving a rotation or revolution around a fixed
chosen axis( an axis which does not move).
 We need a system that defines BOTH types of motion working together on a
system. Rotational quantities are usually defined with units involving a radian
measure.
If we take the radius of a circle and
LAY IT DOWN on the circumference, it
will create an angle whose arc length
is equal to R.
In other words, one radian angle is
subtends an arc length s equal to
the radius of the circle (R)
The radian
Half a radian would subtend an arc length equal to half the radius and 2
radians would subtend an arc length equal to two times the radius.
A general Radian Angle (q) subtends an arc length (s) equal to R. The
theta in this case represents ANGULAR DISPLACEMENT.
∆𝑠 = 𝑅∆𝜃
∆𝑠
∆𝜃 =
𝑅
Angular Velocity
Since velocity is defined as the rate of change of displacement.
ANGULAR VELOCITY is defined as the rate of change of
ANGULAR DISPLACEMENT.
∆𝑥
→ 𝑡𝑟𝑎𝑛𝑠𝑙𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
∆𝑡
∆𝜃
𝜔=
→ 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
∆𝑡
𝑑𝑥
𝑣=
𝑑𝑡
𝑑𝜃
𝜔=
𝑑𝑡
𝑣=
Angular Acceleration
∆𝑣
𝑎=
→ 𝑡𝑟𝑎𝑛𝑠𝑙𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛
∆𝑡
∆𝜔
𝛼=
→ 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛
∆𝑡
𝑑𝑣
𝑎=
𝑑𝑡
𝑑𝜔
𝛼=
𝑑𝑡
Once again, following the same lines
of logic. Since acceleration is defined
as the rate of change of velocity. We
can say the ANGULAR ACCELERATION
is defined as the rate of change of the
angular
velocity.
Also, we can say that the ANGULAR
ACCELERATION is the TIME
DERIVATIVE OF THE ANGULAR
VELOCITY.
All the rules for integration apply as
well.
𝑥=
𝑣𝑑𝑡 , 𝑣 =
𝜃=
𝜔𝑑𝑡 , 𝜔 =
𝑎𝑑𝑡
𝛼𝑑𝑡
Linear Angular Relationships
 We often want to be able to compare the linear quantity
of something to its angular counterpart.
 We are going to use our arc length-angle relationship to
derive a relationship between linear and angular velocity.
∆𝑠 = 𝑟∆𝜃
∆𝑠
∆𝜃
=𝑟
∆𝑡
∆𝑡
𝑣 = 𝑟𝜔
 This relationship shows that on a rotating rigid body, points
farther away from the rotational axis move faster.
 We can also use this idea to compare linear and angular
acceleration.
𝑣 = 𝑟𝜔
∆𝑣
∆𝜔
=𝑟
∆𝑡
∆𝑡
𝑎 = 𝑟𝛼
Rotational Kinematics
 Using these new linear-angular relationships, we can take our linear
kinematic equations and translate them into angular kinematics.
𝑣 = 𝑣𝑜 + 𝑎𝑡 → 𝜔 = 𝜔𝑜 + 𝛼𝑡
𝑥 = 𝑥𝑜 + 𝑣𝑜 𝑡 + 1 2 𝑎𝑡 2 → 𝜃 = 𝜃𝑜 + 𝜔𝑜 𝑡 + 1 2 𝛼𝑡 2
𝑣 2 = 𝑣𝑜2 + 2𝑎∆𝑥 → 𝜔2 = 𝜔𝑜2 + 2𝑎∆𝜃
 These kinematics are used in the same way we used our linear
equations. They allow us to describe the rotational motion of an
object using our basic definitions of motion and time.
Example
A car engine is idling at 500 RPM. When the light turns green,
the crankshaft rotation speeds up at a constant rate to 2500
RPM over and interval of 3.0 s. How many revolutions does the
crankshaft go through during this interval?
75 Revolutions
Rotational Kinetic Energy
 A rotating object has kinetic
energy because all particles in
the object are in motion.
 The kinetic energy due to
rotation is called rotational
kinetic energy.
 We add up the individual kinetic
energies of each particle to find
the rotational kinetic energy of
the object.
𝐾𝐸𝑟𝑜𝑡 = 1 2 𝑚1 𝑣12 + 1 2 𝑚2 𝑣22 + ⋯
𝑣 = 𝑟𝜔
𝐾𝐸𝑟𝑜𝑡 = 1 2 𝑚1 𝑟12 𝜔2 + 1 2 𝑚2 𝑟22 𝜔2 + ⋯
𝐾𝐸𝑟𝑜𝑡 = 1 2 𝜔2
𝑚𝑖 𝑟𝑖2
𝑖
Rotational Kinetic Energy and
Moment of Inertia
 Looking at our formula for rotational kinetic energy, there is an interesting term:
𝑚𝑖 𝑟𝑖2
𝑖
 We define this term as our moment of inertia, which is the rotational analog to
inertia.
 We use the letter I for moment of inertia, so we will substitute that into our
formula for rotational kinetic energy.
𝑚𝑖 𝑟𝑖2
𝐼=
𝑖
𝐾𝐸𝑟𝑜𝑡 = 1 2 𝐼𝜔2
 This is not a new form of energy, merely the familiar kinetic energy of
motion written in a new way.
Moment of Inertia
 Moment of Inertia represents an object’s resistance to change in rotational
motion. This is a fancy way of saying it represents and object’s resistance to
angular acceleration.
 The units of moment of inertia are kg m2.
 A system consisting of discrete point masses has a moment of inertia defined
using the summation we discussed earlier:
𝐼=
𝑚𝑟 2
 Moment of inertia depends
on the axis of rotation.
 Mass farther from the
rotation axis contributes
more to the moment of
inertia than mass nearer the
axis.
Moment of Inertia
The summation equation for moment of inertia worked fine for a
collection of point masses, but what about more continuous masses like
disks, rods, or sphere where the mass is extended over a volume or area?
In this case, calculus is needed.
𝐼=
𝑚𝑟 2 → 𝐼 =
This suggests that we will take small
discrete amounts of mass and add
them up over a set of limits. Indeed,
that is what we will do. Let’s look at
a few example we “MIGHT”
encounter. Consider a solid rod
rotating about its CM.
𝑟 2 𝑑𝑚
Example – Thin rod rotated about
the middle
We begin by using the same technique used to
derive the center of mass of a continuous body.
𝑀
𝐿
𝑑𝑚
𝑀𝑖𝑐𝑟𝑜 → 𝜆 =
𝑑𝑟
𝑑𝑚 = 𝜆𝑑𝑟
𝑀𝑎𝑐𝑟𝑜 → 𝜆 =
𝑟 2 𝑑𝑚
𝐼=
𝐿/2
𝐿/2
𝑟 2 (𝜆𝑑𝑟)
𝐼=
−𝐿/2
𝑟 2 𝑑𝑟
=𝜆
−𝐿/2
𝑀
= ∗
𝐿
dr
𝐿 3
𝐿
−
2 −
2
3
3
3
𝑀 𝐿3 𝑀𝐿2
= ∗
=
𝐿 12
12
Example – Thin rod rotated about
one end
What if the rod were rotating on one of its
ENDS?
𝑀
𝑀𝑎𝑐𝑟𝑜 → 𝜆 =
𝐿
𝐼 = 𝑟 2 𝑑𝑚
𝑑𝑚
𝑀𝑖𝑐𝑟𝑜 → 𝜆 =
𝑑𝑟
𝑑𝑚 = 𝜆𝑑𝑟
𝐿
𝐿
𝑟 2 (𝜆𝑑𝑟)
𝐼=
0
𝑟 2 𝑑𝑟
=𝜆
0
𝑀 𝐿3
𝑀𝐿2
= ∗
−0 =
𝐿
3
3
dr
Well that was fun, but…
 Will you be asked to derive the moment of inertia of an object? Maybe.
Most of the time the moment of inertia is given within the free response
question.
 Moments of Inertia for various geometric objects can be found on tables
such as the one below.
Parallel Axis Theorem
This theorem will allow us to calculate the moment of inertia of any rotating body
around any axis, provided we know the moment of inertia about the center of
mass.
𝐼𝑃 = 𝐼𝐶𝑀 + 𝑀𝑑 2
It basically states that the Moment of Inertia (Ip) around any axis "P" is equal to the
known moment of inertia (Icm) about some center of mass plus M (the total mass
of the system) times the square of "d" (the distance between the two parallel axes)
Let’s use the thin rod example to prove this.
Parallel Axis Theorem
dr
dr
𝐼𝐶𝑀
𝑀𝐿2
=
12
𝑀𝐿2
𝐼𝑃 =
3
𝐼𝑃 = 𝐼𝐶𝑀 + 𝑀𝑑 2
𝑀𝐿2 𝑀𝐿2
𝐿 2
=
+𝑀∗
3
12
2
2
2
2
𝑀𝐿
𝑀𝐿
𝑀𝐿
𝑀𝐿2 𝑀𝐿2
=
+
→
=
3
12
4
3
3
Moment of Inertia of Compound Objects
 If you have two geometric objects attached to each other, you can find
the moment of inertia of the system by adding the moment of inertia of
the two objects (assuming they rotate about the same point).
 This means that moment of inertia obeys the principle of superposition.
 Notice that if the sphere is very small (L>>R), then the sphere can be
approximated as a point mass, making the moment of inertia of this system:
1
𝐼 = 𝑀𝑟𝑜𝑑 𝐿2 + 𝑀𝑠𝑝ℎ𝑒𝑟𝑒 𝐿2
3
Rolling Motion
 Rolling is a combination of rotation and translation. For an object that rolls
without slipping, the translation of the center of mass is related to the
angular velocity by:
2𝜋𝑟
𝑣𝐶𝑀 =
= 𝑟𝜔
𝑇
 We also find that this is the velocity at both the top and bottom of the
object.
Rolling Motion
 The figure below shows how the velocity vectors at the top, center, and bottom of a
rolling wheel are found.
 vtop = 2vcm.
 vbottom = 0.
 The point on the bottom of a rolling object is instantaneously at rest.
Rolling Motion and Energy
 To describe the kinetic energy of a rolling object, we must take into account
the linear velocity and the rotational velocity of the object.
 This means that the total kinetic energy of a rolling object can be calculated
by adding the kinetic energy of the center of mass and the rotational kinetic
energy about the center of mass.
2
𝐾𝐸𝑇𝑜𝑡𝑎𝑙 = 𝐾𝐸 + 𝐾𝐸𝑟𝑜𝑡 = 1 2 𝑚𝑣𝐶𝑀
+ 1 2 𝐼𝜔2
Example
A very common problem is to find the velocity of a ball
rolling down an inclined plane. It is important to realize
that you cannot work out this problem the way you used
to. In the past, everything was SLIDING. Now the object is
rolling and thus has MORE energy than normal. So let’s
assume the ball is like a thin spherical shell and was
released from a position 5 m above the ground. Calculate
the velocity at the bottom of the incline.
Ebefore  Eafter
Ebefore  Eafter
U g  KT  K R
U g  KT
mgh  1 mv 2  1 I 2
2
2
v  R I sphere@ cm  2 mR 2
3
mgh  1 mv2
2
mgh  1 mv2
2
gh  1 v 2
2
v  2 gh  2(9.8)(5)  9.90 m / s
v2
mgh  1 mv 2  1 ( 2 mR 2 )( 2 )
2
2 3
R
gh  1 v 2  1 v 2
2
3
v  6 gh  6 (9.8)(5) 
5
5
7.67 m/s
If you HAD NOT
included the
rotational
kinetic energy,
you see the
answer is very
much different.