Download Notes on circular motion - University of Miami Physics Department

Circular Motion
By Prof. Massimiliano Galeazzi, University of Miami

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If you spin a bucket full of water in a vertical circle fast enough, the water will stay in the bucket. How fast do you
need to spin it to make sure you don’t get showered?
What is the force that makes your car turn when you turn the steering wheel? How big is that force?
These are some of the questions we will answer in this chapter. The general topic is that of circular motion, that is,
objects moving on a circular trajectory. Along the way we will learn about ferris wheels, cars turning without going off
the road, etc. But first, we need to improve our math understanding by reviewing the product between vectors.
--------------------------------------------------------------------------------------------------------------------------------------------------MATH INSERT
In chapter 1 we have introduced the concept of vectors and found how vectors can be added and or subtracted. We
also mentioned the fact that vectors can be multiplied with each other in two different ways, what are called the scalar (or
dot) product and the vector (or cross) product. In this chapter we will make use, for the first time, of the dot-product, so it
is appropriate to take a look at it before we proceed.
1. The scalar (dot) product between vectors
FIG. 1
�⃗ = 𝐵𝑥 𝚤̂ + 𝐵𝑦 𝚥̂ + 𝐵𝑧 𝑘�. The scalar product between 𝐴⃗ and 𝐵
�⃗,
Let’s consider two vectors, 𝐴⃗ = 𝐴𝑥 𝚤̂ + 𝐴𝑦 𝚥̂ + 𝐴𝑧 𝑘� and 𝐵
�⃗ (“A dot B”, hence the term “dot product”), is a scalar quantity equal to the product between the magnitudes
denoted 𝐴⃗ ∙ 𝐵
of the two vectors, times the cosine of the angle 𝜙 between the two:
�⃗ = �𝐴⃗��𝐵
�⃗� cos 𝜙,
𝐴⃗ ∙ 𝐵
[1]
�⃗ = 𝐵
�⃗ ∙ 𝐴⃗
𝐴⃗ ∙ 𝐵
[2]
�⃗ = �𝐴⃗���𝐵
�⃗� cos 𝜙�,
𝐴⃗ ∙ 𝐵
[3]
where the angle 𝜙 is measured by drawing the two vectors with the tails in the same point (see Fig. 1a-b). We notice that,
from Eq. 1, the dot product, like the “regular” product between scalars is commutative, i.e.,
Also, we can write Eq. [1] as:
�⃗ is equal to the magnitude of 𝐴⃗ times the projection of 𝐵
�⃗ in the direction of 𝐴⃗,
That is, the scalar product between 𝐴⃗ and 𝐵
�⃗� cos 𝜙 (see Fig. 1c). Similarly, we can write:
i.e., �𝐵
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�⃗ = �𝐵
�⃗���𝐴⃗� cos 𝜙�,
𝐴⃗ ∙ 𝐵
[4]
�⃗ is equal to the magnitude of 𝐵
�⃗ times the projection of 𝐴⃗ in the direction of 𝐵
�⃗
That is, the scalar product between 𝐴⃗ and 𝐵
(see Fig. 1d).
From the definition of the scalar product it is possible to derive a few simple properties:
•
•
•
•
•
The scalar product between two vectors can be positive, negative, or zero. In particular, if 0 ≤ 𝜙 < 90𝑜 then
the product is positive, while if 90𝑜 < 𝜙 ≤ 180𝑜 then the product is negative.
The scalar product of two parallel vectors is equal to the product of the magnitudes of the two (𝜙 = 0).
The scalar product of two vectors that are anti-parallel is equal to the negative of the product of the two
magnitudes (𝜙 = 180𝑜 ).
The scalar product of two perpendicular vectors is equal to zero (𝜙 = 90𝑜 ).
From the previous result it follows that, if the scalar product between two vectors is zero and their magnitudes
are different from zero, then the two vectors must be perpendicular. This provides a simple, yet very powerful
way to determine whether two vectors are perpendicular.
While the definition of scalar product in Eq. [1] is very useful, sometimes it is easier to work with vector
components, rather than magnitude and direction. It is also possible to calculate the scalar product of two vectors in terms
of their components:
�⃗ = �𝐴𝑥 𝚤̂ + 𝐴𝑦 𝚥̂ + 𝐴𝑧 𝑘� � ∙ �𝐵𝑥 𝚤̂ + 𝐵𝑦 𝚥̂ + 𝐵𝑧 𝑘��.
𝐴⃗ ∙ 𝐵
[5]
To find the value of the scalar product in terms of components, we simply expand the product in Eq. [5]:
�⃗
𝐴⃗ ∙ 𝐵
= 𝐴𝑥 𝚤̂ ∙ 𝐵𝑥 𝚤̂ + 𝐴𝑥 𝚤̂ ∙ 𝐵𝑦 𝚥̂ + 𝐴𝑥 𝚤̂ ∙ 𝐵𝑧 𝑘� + 𝐴𝑦 𝚥̂ ∙ 𝐵𝑥 𝚤̂ +
+𝐴𝑦 𝚥̂ ∙ 𝐵𝑦 𝚥̂ + 𝐴𝑦 𝚥̂ ∙ 𝐵𝑧 𝑘� +
[6]
+𝐴𝑧 𝑘� ∙ 𝐵𝑥 𝚤̂ + 𝐴𝑧 𝑘� ∙ 𝐵𝑦 𝚥̂ + 𝐴𝑧 𝑘� ∙ 𝐵𝑧 𝑘�
And taking into account that, using the properties of the scalar product and of unit vectors 𝚤̂ ∙ 𝚤̂ = 𝚥̂ ∙ 𝚥̂ = 𝑘� ∙ 𝑘� = 1,
and 𝚤̂ ∙ 𝚥̂ = 𝚤̂ ∙ 𝑘� = 𝚥̂ ∙ 𝑘� = 0, we get:
�⃗ = 𝐴𝑥 𝐵𝑥 + 𝐴𝑦 𝐵𝑦 + 𝐴𝑧 𝐵𝑧 .
𝐴⃗ ∙ 𝐵
[7]
END OF MATH INSERT
--------------------------------------------------------------------------------------------------------------------------------------------------2. Motion along a circular path
Consider a particle with mass m that is constrained to move on a circular path, as shown in Fig. 2. To describe the
motion of the particle we can set up a coordinate system with the origin coincident with the center of the circle and then
use the x and y coordinate of the particle. We could, however, also use the angle 𝜃 that the particle position makes with
the positive x-axis, or the distance 𝑙 of the particle from the positive x-axis along the circular path. Note that in our choice
of coordinates the particle is moving around the z-axis which is perpendicular to the page and passes through the center of
the circle. A general convention is to consider the angle 𝜃 and the distance l as positive if measured counterclockwise. The
Cartesian coordinates x and y of the particle and the vector position 𝑟⃗ can then be easily calculated as:
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FIG. 2
𝑥 = 𝑅 cos 𝜃
[8]
𝑦 = 𝑅 sin 𝜃
[9]
𝑟⃗ = 𝑥𝚤̂ + 𝑦𝚥̂ = (𝑅 cos 𝜃)𝚤̂ + (𝑅 sin 𝜃) 𝚥̂,
[10]
𝑙 = 𝑅𝜃.
[11]
where R is the radius of the circle. Moreover, since the distance l traveled by the particle is nothing more than the arc
subtended by the angle 𝜃, we can also write:
From Eq. [10] we can also easily find an expression for the particle velocity. By using the chain rule for derivatives
and the fact that in circular motion R is constant, i.e., the distance from the origin does not change, we get:
𝑣⃗ =
𝑑𝑟⃗
𝑑𝑡
= (−𝑅 sin 𝜃)
𝑑𝜃
𝑑𝑡
𝚤̂ + (𝑅 cos 𝜃)
𝑑𝜃
𝑑𝑡
𝚥̂ ,
[12]
where we made use of the fact that, indeed, the angle 𝜃 is the only quantity in Eq. [10] that changes with time.
The first derivative of the angle versus time is called the angular velocity of the particle and is indicated with the
symbol 𝜔𝑧 (the Greek letter omega):
𝜔𝑧 =
𝑑𝜃
𝑑𝑡
,
[13]
Where the index “z” is used to indicate that, as mentioned before, the rotation is around the z-axis. We note that, in
general, the angular velocity is a vector, with each component quantifying the rotation around a different axis. However,
in this chapter (and in most of the book) we will only consider rotations around the z-axis, and thus only use the zcomponent 𝜔𝑧 of the vector. We also point out that the angular velocity measures angles per time, or radians per second.
However, radians are usually treated as if they are not a real unit, and the SI unit for angular velocity is simply 𝑠 −1 . Using
Eq. [13] we can rewrite Eq. [12] as:
𝑣⃗ = −𝑅𝜔𝑧 sin 𝜃 𝚤̂ + 𝑅𝜔𝑧 cos 𝜃 𝚥̂.
[14]
We can also use Eq. [12] to find the speed of the particle 𝑣 = |𝑣⃗| = �𝑣𝑥2 + 𝑣𝑦2 . However, it is easier to calculate it
using the fact that the speed is the first derivative of the distance l traveled in Eq. [11]:
𝑣=
𝑑𝑙
𝑑𝑡
=
𝑑(𝑅𝜃)
𝑑𝑡
=𝑅
𝑑𝜃
𝑑𝑡
𝑣
= 𝑅𝜔𝑧 , or 𝜔𝑧 = ,
𝑅
[15]
where we used the fact that the radius R is constant. Equation [15] represents a very simple and straightforward relation
between the particle speed and its angular velocity.
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3. Position vs. Velocity in circular motion
Before continuing with our analysis, let’s take a quick look at the properties of the vectors position and velocity
associated with circular motion. The vector position 𝑟⃗ is the vector going from the origin of the coordinate system to the
particle position. However, since in our choice the origin of the coordinate system coincides with the center of the circular
path, the magnitude of the vector position will always be constant and equal to the radius R of the circle and will always
have a “radial” direction, i.e., it will always be oriented with the radius of the circle at the position of the particle. This
property can be expressed as
𝑟⃗ = 𝑅 𝑟̂ ,
[16]
where 𝑟̂ is the generic unit vector (magnitude 1) aligned with the radius of the circle at the position of the particle.
Using Eqs. [10] and [14] we can also demonstrate that, in circular motion, the velocity 𝑣⃗ is perpendicular to the
position 𝑟⃗. The simplest way to demonstrate it is by calculating the dot-product between the two vectors, using the
definition of dot-product with components derived in Eq. [7] and the fact that the two vectors only have x and y
components:
𝑟⃗ ∙ 𝑣⃗ = (𝑥𝚤̂ + 𝑦𝚥̂) ∙ �𝑣𝑥 𝚤̂ + 𝑣𝑦 𝚥̂� = 𝑥𝑣𝑥 + 𝑦𝑣𝑦 = (𝑅 cos 𝜃)(−𝑅𝜔𝑧 sin 𝜃) + (𝑅 sin 𝜃)(𝑅𝜔𝑧 cos 𝜃) ,
[17]
𝑟⃗ ∙ 𝑣⃗ = −𝑅2 𝜔𝑧 cos 𝜃 sin 𝜃 + 𝑅2 𝜔𝑧 sin 𝜃 cos 𝜃 = 0.
[18]
𝑣⃗ = 𝑣 𝜃� = 𝑅𝜔𝑧 𝜃�,
[19]
which simplifies to:
Equation [18] is valid in the most general case of circular motion, and implies that, if a particle can only move on a
circular path, then its velocity is always perpendicular to its vector position. Moreover, since the vector position is always
aligned with a radius of the circular path, this means that the velocity is always perpendicular to the radius, i.e., it is
always tangent to the circular trajectory. This property is sometimes expressed as
where 𝜃� is the generic unit vector (magnitude 1) tangent to the circle at the position of the particle (see Fig. 3), and we
made use of Eq. [15].
FIG. 3
4. Uniform circular motion
Before moving to the more general description of circular motion, let’s review the simpler case of uniform circular
motion, that is, the motion of an object moving at constant angular velocity 𝜔𝑧 , or equivalently, at constant speed v. First
of all, if the particle moves at constant speed, we can easily calculate the time it takes for the particle to complete a full
circle, called the period of the motion. We can do that in two different ways, using either the speed or the angular
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velocity. To complete a full circle, the particle must travel a distance equal to the circumference of the circle 2𝜋𝑅. The
period T is therefore given by
𝑇=
2𝜋𝑅
𝑣
.
[20]
Similarly, we can say that to complete a full circle the particle must cover an angle, expressed in radians, equal to
2𝜋, traveling at constant angular velocity 𝜔𝑧 . The period is therefore given by the angle divided by the angular velocity:
𝑇=
2𝜋
𝜔𝑧
.
[21]
Notice how Eqs. [20] and [21] are related, and can be derived from each other using Eq. [15]. Another quantity used
to describe uniform circular motion is the frequency f of the motion, that is, the number of full circles (revolutions)
completed by the particle per unit time. It is easy to see that the frequency is just the inverse of the period, e.g., if a
particle takes ½ s to compete a full circle, then it will make 2 revolutions per second.
1
𝑓= .
[22]
𝜔𝑧 = 2𝜋𝑓.
[23]
𝑇
Another way of looking at the frequency is through the angular velocity 𝜔𝑧 . In fact, the angular velocity represents
the number of radians completed by the particle per time, while the frequency represents the number of revolutions per
time, and since there are 2𝜋 𝑟𝑎𝑑𝑖𝑎𝑛𝑠 per revolution, it follows that
Notice that, indeed, Eq. [23] can be derived easily by combining Eqs. [21] and [22]. Finally, it is important to
remember that Eqs. [20] through [23] are not general, but ARE ONLY VALID FOR UNIFORM CIRCULAR MOTION,
which we defined as motion at constant speed (or constant angular velocity).
Having discussed the properties of uniform circular motion, let’s now review how it can be generated. Even if the
speed of the particle is constant, the direction of the velocity, and thus the velocity itself, change constantly as the particle
moves around the circle. This implies that the particle is subject to a non-zero acceleration. To find the value of such
acceleration we can differentiate Eq. [14], keeping into account that, in this specific case, the angular velocity 𝜔𝑧 is
constant:
𝑎⃗ =
�⃗
𝑑𝑣
𝑑𝑡
= −𝑅𝜔𝑧 cos 𝜃
𝑑𝜃
𝑑𝑡
𝚤̂ − 𝑅𝜔𝑧 sin 𝜃
𝑑𝜃
𝑑𝑡
𝚥̂ = −𝑅𝜔𝑧2 cos 𝜃 𝚤̂ − 𝑅𝜔𝑧2 sin 𝜃 𝚥̂.
[24]
By comparing Eq. [24] with Eq. [10] we obtain:
𝑎⃗ = −𝜔𝑧2 𝑟⃗ = −𝜔𝑧2 𝑅𝑟̂ ,
[25]
that is, the acceleration responsible for the circular trajectory of the particle is aligned with the radius of the circle at the
particle position (unit vector 𝑟̂ ), is directed toward the center of the circle (the negative sign), and has constant magnitude
equal to:
𝑣 2
𝑎 = 𝜔𝑧2 𝑅 = � � 𝑅 =
𝑅
𝑣2
𝑅
,
[22]
where we made use of Eq. [15]. Due to its direction toward the center of the circular trajectory, such acceleration is often
called centripetal acceleration.
Using Newton’s second law, we can move one step further and say that, to move on a circular trajectory at constant
speed, a particle requires a net force directed toward the center of the circle with constant magnitude
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𝐹 = 𝑚𝑎 = 𝑚𝜔𝑧2 𝑅 =
𝑚𝑣 2
𝑅
.
[23]
Let’s now review how we can apply the concepts just learned to real problems, with a few examples.
Example 1
A car with mass m moves at constant speed v on a vertical circular track of radius R (Fig. 4a). Find the normal force
exerted by the track on the car at [a] the highest and [b] the lowest point, and [c] the minimum speed needed by the car not
to fall from the track.
FIG. 4
[a] To find the normal force we simply use Newton’s second law. The only two forces acting on the car at the top are the
weight and the normal force. To stay on the track, the acceleration, for uniform circular motion, must be equal to the
centripetal acceleration or 𝑣 2 /𝑅. The free body diagram of the car at the highest point is show in Fig. 4b. Using a
reference frame with the y-axis vertical, only the y-component of the forces is non-zero, thus we can write:
−𝑚𝑔 − 𝑛 = 𝑚𝑎𝑦 = −𝑚
𝑣2
𝑅
.
[24]
Solving for the magnitude of the normal force n we get
𝑛=𝑚
𝑣2
𝐿
− 𝑚𝑔 = 𝑚 �
𝑣2
𝐿
− 𝑔� therefore 𝑛�⃗ = −𝑚 �
𝑣2
𝐿
− 𝑔� 𝚥̂.
[25]
[b] The procedure is identical to what we did for part [a], however, this time the normal force and the acceleration are
directed toward the positive y-axis. The free body diagram is shown in Fig. 4c, and Newton’s second law for the ydirection is:
𝑛 − 𝑚𝑔 = 𝑚
𝑣2
𝐿
∴ 𝑛 = 𝑚�
𝑣2
𝐿
+ 𝑔� ∴ 𝑛�⃗ = 𝑚 �
𝑣2
𝐿
+ 𝑔� 𝚥̂.
[26]
Comparing Eqs. [25] and [26] it is clear that not only the direction of the normal force changes from the highest to the
lowest point, but also its magnitude is bigger at the bottom than at the top.
[c] If the car is not fast enough, then it will fall from the track without completing a full circle. A practical experiment to
show this is by filling a bucket with water and spinning it in a vertical circle with your arm. If the speed of the bucket is
high enough, then the water stays in the bucket, otherwise… well, I will let you do the experiment 
The reason for it is evident in Eq. [24]. If the speed is high enough, then there is a normal force that, combined with the
weight, generates the right acceleration for circular motion. However, if the speed is too small, then we would need a
negative normal force to maintain circular motion. Unless the car is strapped to the track, it will fall before reaching the
maximum height. To find the minimum speed necessary to avoid it, all we need to do is set the normal force equal to zero
in Eq. [25]. If the normal force is zero, then gravity alone is responsible for the centripetal acceleration. Any speed smaller
than that, gravity is too big and thus the acceleration of the car will be bigger than the centripetal acceleration making it
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fall. For any speed bigger than that, then the normal force will be able to balance the gravity to provide the proper
centripetal acceleration. Setting 𝑛 = 0 in Eq. [25] we get:
𝑣2
𝐿
− 𝑔 = 0 ∴ 𝑣 = �𝐿𝑔.
[27]
Example 2
A car with mass m, traveling at constant speed v, moves around a flat curve with radius R. How much force must be
exerted on the car to stay on the road and what provides it?
FIG. 5
Solution: For the car to move of uniform circular motion, there must be a force directed toward the center of the circular
trajectory, capable of generating the proper acceleration. Figure 5 shows the car with the free body diagram. Notice that in
the figure the car is moving on a plane perpendicular to the page, and the instant velocity is straight out of the page. In the
figure we also show the center of the circle and the vector position. Gravity is perpendicular to direction of the
acceleration, so it cannot be responsible for it, and it is simply “balanced” by the normal force of the road. We thus added
the “unknown” force f. Using a reference frame with the x-axis horizontal pointing to the right, and the y-axis vertical,
pointing up, Newton’s second law in the x and y directions gives:
𝑓=𝑚
𝑣2
𝑅
𝑛 − 𝑚𝑔 = 0 ∴ 𝑛 = 𝑚𝑔
[28]
The x-component of Newton’s law gives the magnitude of the force responsible for the circular motion, while the ycomponent gives the magnitude of the normal force.
To answer to question of what is responsible for the force f, we note that the only thing, other than gravity, acting on the
car is the surface of the road and that the force is parallel to such surface. We have already seen in previous chapters that
the parallel force between an object and the surface over which it is placed is the friction between the two. The force
responsible for the circular path of the car is, indeed, the static friction, i.e., 𝑓 = 𝑓𝑠 .
We should also remember that there is a maximum value the static friction may have, proportional to the magnitude of the
normal force:
𝑓𝑠 ≤ 𝜇𝑠 𝑛,
Where 𝜇𝑠 is the coefficient of static friction. Using Eqs. [28] and [29] we get:
𝑚
𝑣2
𝑅
≤ 𝜇𝑠 𝑛 = 𝜇𝑠 𝑚𝑔 ∴ 𝑣 ≤ �𝜇𝑠 𝑅𝑔.
[29]
[30]
That is, there is a maximum value for the speed of the car, which depends on the coefficient of stating friction and the
radius of the curve (but not the mass of the car), and if the speed of the car exceeds this value we all know what happens
to the car. To understand the meaning of Eq. [30] we notice that, if the coefficient of static friction is small, then the
maximum speed at which the car can travel to stay on the road is small as well. For example, if the road is wet, or icy,
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then the speed at which the car may slip off the road is much smaller. Also, the maximum speed depends on the radius,
that is, the maximum speed at which the car can travel staying on the road is smaller for sharp curves, as we well know
from everyday experience. However, the maximum speed does not depend on the mass of the car, so a light car can travel
at the same maximum speed of a heavy one.
Problem 1
The conical pendulum: An amusement park ride consists of small cars of mass m attached to steel cables of length L. The
cables have negligible mass and are attached to a spinning post (see Fig. 6). When at maximum speed, an empty car forms
an angle 𝛼 with the vertical. What is the angular velocity 𝜔𝑧 of the ride at maximum speed?
FIG. 7
FIG. 6
Strategy
At maximum speed the cars move of uniform circular motion around the spinning post and the forces acting on them are
the weight and the tension in the steel cables. We can therefore use Newton’s second law to find the value of the
acceleration and solve for the angular velocity.
Setup
Figure 6 already shows us one car at an instant in which it is in the x-y plane. We can thus simply choose a coordinate
system with the x-axis horizontal and pointing to the right of the page, and the y-axis vertical, pointing to the top of the
page. While the choice of the origin is not important in this problem, we can set it at the fixed point at the top of the steel
cable. Notice that, with this choice, the car motion is actually around the y-axis, in a plane parallel to the x-z plane
(horizontal and perpendicular to the page). We will therefore change the index for the angular velocity accordingly, but
the physics stays the same.
Solution
Figure 7 the free body diagram of the car. In the figure we have also marked the radius of the circular motion which, with
simple trigonometry, can be written as 𝑅 = 𝐿 sin 𝛼. Based on the free body diagram, we can write Newton’s second law
for the x and y direction as:
𝑇 sin 𝛼 = 𝑚𝑎𝑥 = 𝑚𝜔𝑦2 𝑅 = 𝑚𝜔𝑦2 𝐿 sin 𝛼 ∴ 𝜔𝑦 = �
𝑇 cos 𝛼 − 𝑚𝑔 = 0 ∴ 𝑇 =
𝑚𝑔
cos 𝛼
𝑇
𝑚𝐿
[31]
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Combining the two equations we get:
𝜔𝑦 = �
𝑚𝑔
1
cos 𝛼 𝑚𝐿
𝑔
=�
𝐿 cos 𝛼
.
[31]
Check dimensions: The units of angular velocity are simply 𝑠
−1
, therefore 𝑠
−1
𝑚
2
= � 𝑠 = √𝑠 −2 = 𝑠 −1 OK
𝑚
Notice that, if instead of the angular velocity we are interested in the speed of the particle, we can simply use Eq. [15]:
𝑣 = 𝜔𝑦 𝑅 = �
𝐿𝑔 sin2 𝛼
𝑔
𝐿 sin 𝛼 = �
𝐿 cos 𝛼
5. Non-uniform circular motion
cos 𝛼
.
[32]
We will briefly mention here the case of non-uniform circular motion, that is, when the speed of the particle is not
constant. The simplest way to calculate the acceleration in non-uniform circular motion is by starting from Eq. [19]:
𝑣⃗ = 𝑣𝜃� ,
and since the acceleration is the first derivative of the velocity, we get
𝑎⃗ =
�⃗
𝑑𝑣
𝑑𝑡
=
𝑑𝑣
𝑑𝑡
�
𝑑𝜃
𝜃� + 𝑣 .
[33]
𝑑𝑡
To find an expression for the second term of the Eq. [33] we can simply use what we have already done for uniform
circular motion. In fact, for uniform circular motion
𝑑𝑣
𝑣 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ∴
𝑑𝑡
= 0 ∴ 𝑎⃗ = 𝑣
�
𝑑𝜃
𝑑𝑡
,
[34]
But we have already calculated the value of the acceleration for uniform circular motion is Eq. [25]:
𝑎⃗ = −𝜔𝑧2 𝑅𝑟̂ ∴ 𝑣
�
𝑑𝜃
= −𝜔𝑧2 𝑅𝑟̂
[35]
𝜃� = 𝑎⃗𝑟𝑎𝑑 + 𝑎⃗𝑡𝑎𝑛 ,
[36]
𝑑𝑡
We can therefore write:
𝑎⃗ = −𝜔𝑧2 𝑅𝑟̂ +
𝑑𝑣
𝑑𝑡
with |𝑎⃗𝑟𝑎𝑑 | = 𝜔𝑧2 𝑅 and |𝑎⃗𝑡𝑎𝑛 | =
𝑑𝑣
𝑑𝑡
.
That is, the acceleration for non-uniform circular motion can be divided into two terms, one directed along the
radius of the circle, equal to the centripetal acceleration, the other tangent to the circle, with magnitude equal to the first
derivative of the speed versus time. As in the case of uniform circular motion, the centripetal acceleration is responsible
for the circular trajectory of the particle, however, if the speed is not uniform, there is an additional term of the
acceleration, tangent to the particle trajectory, responsible for the change is speed.
We can go one step further by using Eq. [15] or 𝑣 = 𝑅𝜔𝑧 in Eq. [36] and considering the fact that the radius 𝑅 is
constant in circular motion. The tangent component of the acceleration then becomes:
|𝑎⃗𝑡𝑎𝑛 | =
𝑑𝑣
𝑑𝑡
=
𝑑(𝑅𝜔𝑧 )
𝑑𝑡
=𝑅
𝑑𝜔𝑧
𝑑𝑡
= 𝑅𝛼𝑧
[37]
9
Massimiliano Galeazzi: Circular Motion – Any Reproduction or distribution without the author’s consent is forbidden.
where 𝛼𝑧 =
𝑑𝜔𝑧
𝑑𝑡
, is called the angular acceleration.
In terms of problem solving, we can solve most problems using the same strategy shown for uniform circular
motion even when the speed is not constant. For example, in examples 1 and 2 and problem 1 the solution would be the
same even if the speed is not constant, as long as we use the instant speed to calculate the centripetal acceleration. In
general, however, there may be additional forces, tangent to the particle trajectory, responsible for the change in speed.
Example 3
FIG. 8
Let’s consider the same situation of example #1, but this time we will consider the normal force when the car is 90
degrees from the top of the circle. Figure 8 shows the free body diagram associated with the particle. Using Newton’s
second law in the x and y directions we get:
𝑛 = 𝑚𝑎𝑟𝑎𝑑 = 𝑚
𝑣2
𝑅
−𝑚𝑔 = 𝑚𝑎𝑡𝑎𝑛 = 𝑚
𝑑𝑣
𝑑𝑡
∴
𝑑𝑣
𝑑𝑡
[38]
= −𝑔
Equation [38] shows that, in the horizontal position only the normal force is responsible for the circular trajectory of the
particle, while the weight, which is tangent to the particle position at that point, is going to increase the particle speed on
the downward leg, and decrease it in the upward one. For the car to move of uniform circular motion there must therefore
be an additional force (given by the car’s engine) that makes 𝑎𝑡𝑎𝑛 = 0 (note that, in example #1, since he weight is radial
at the top and bottom, then the additional force would be zero). However, even if that is not the case, as long as we know
the instant speed of the car, then we can solve the problem. For example, we will see later this semester that in many
situations the speed of the particle at any point can be “easily” calculated using conservation of energy, even if it changes
with time and position. In those problems, we can still calculate the forces acting on the particle using the same procedure
used here.
6. Equation Sheet
𝜔𝑧 =
𝛼𝑧 =
𝑑𝜃
𝑑𝑡
𝑑𝜔𝑧
𝑑𝑡
𝑣 = 𝜔𝑧 𝑅
𝑟⃗ = 𝑅(𝑐𝑜𝑠𝜃𝚤̂ + 𝑠𝑖𝑛𝜃𝚥̂) = 𝑅𝑟̂
𝑣⃗ = 𝜔𝑧 𝑅(−𝑠𝑖𝑛𝜃𝚤̂ + 𝑐𝑜𝑠𝜃𝚥̂) = 𝜔𝑧 𝑅𝜃�
𝑎⃗ = 𝑎⃗𝑟𝑎𝑑 + 𝑎⃗𝑡𝑎𝑛 ; |𝑎⃗𝑟𝑎𝑑 | = 𝜔𝑧2 𝑅; |𝑎⃗𝑡𝑎𝑛 | =
𝑑𝑣
𝑑𝑡
= 𝑅𝛼𝑧
10
Massimiliano Galeazzi: Circular Motion – Any Reproduction or distribution without the author’s consent is forbidden.