NASA-Threads
Work and Mechanics
Lesson 39: Circular Motion
Uniform Circular Motion
Uniform circular motion refers to an object moving at constant speed along a circular path;
examples include a ball twirled overhead while attached to a string and a satellite orbiting
the earth. If a force from a string or gravity constantly pulls an object toward the center of
rotation, as shown below, the object must continually accelerate inward toward the center
in response to the applied force. The force that pulls the object toward the center is called
the centripetal force, and the resulting acceleration is the centripetal acceleration. The
centripetal force and centripetal acceleration act perpendicular (or normal) to the direction
of motion of the body (the “n” subscript on an means “normal”). If the centripetal force is
suddenly released(i.e., the string is cut), the object will begin to travel in a straight line.
𝑣 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑎𝑛 = centripetal
𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑝𝑎𝑡ℎ
acceleration
𝐹 = 𝑐𝑒𝑛𝑡𝑟𝑖𝑝𝑒𝑡𝑎𝑙 𝑓𝑜𝑟𝑐𝑒
𝑐𝑒𝑛𝑡𝑒𝑟 𝑜𝑓 𝑚𝑜𝑡𝑖𝑜𝑛
𝜔 = 𝑎𝑛𝑔𝑙𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑟 = 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑝𝑎𝑡ℎ
NASA-Threads
Work and Mechanics
Lesson 39: Circular Motion
Radians and Angular Velocity
A radian is a unit of angular measure; radians and degrees are related as follows:
𝑑𝑒𝑔 = 𝑟𝑎𝑑 ∙
180°
𝜋
A radian is the length of an arc divided by the radius. An angle measured in radians is
commonly used to compute the arc length of circular segment.
𝑺=𝒓∙𝜽
where
 S = arc length (m or in)
 r = radius, which is constant (m or in)

 = angle in radians (rad)
S
r

r
The angular velocity 𝜔 quantifies how fast an object is rotating about the center of motion
and is specified in radians per second. The centripetal force can be computed from the
velocity of the particle and the radius of the curved path, as shown below.
CLASS PROBLEM: An object traveling in a circular path at constant speed moves through an
angle of 𝜋 radians in 1 second.
(a) If the radius of the path is 1 m, then how far does the object travel (the arc length)?
(b) What is the angular velocity of the object (𝜔)?
(c) How many revolutions does the object make per minute (RPM)?
NASA-Threads
Work and Mechanics
Lesson 39: Circular Motion
S
v1
1

2
v1
v
v2
Consider a particle moving along a circular path of radius r from
position 1 to position 2 over at time period of t. If the velocity (v) is
constant, then the arc length (S) traced by the particle will be . . .
𝑆 = 𝑣 ∙ Δ𝑡
r
r

(1)
We can also see that the arc length (S) is equal to . . .
𝑆 =𝑟∙𝜃
(2)
Equating S from (1) and (2) and solving for  yields . . .
𝜃=
𝑣 ∙ Δ𝑡
𝑟
(3)
Moving the velocity vector from position 1 over to position 2 and
using vector addition shows that v1 + v = v2. The value of v in this
vector triangle can be approximated as v = v1 ∙  = v ∙  when the
angle  is very small:
Δ𝑣 = 𝑣 ∙ 𝜃
𝑜𝑟
𝜃=
Δ𝑣
𝑣
(4)
Equating (3) and (4) and taking the centripetal acceleration (an) as
v/t yields . . .
0
𝑣 ∙ ∆𝑡 ∆𝑣
=
𝑟
𝑣
𝑜𝑟
𝑎𝑛 =
∆𝑣 𝑣 2
=
∆𝑡
𝑟
NASA-Threads
Work and Mechanics
Lesson 39: Circular Motion
CLASS PROBLEM: Several NASA Threads physics students ride the jet-O-spin at the state fair for
academic purposes. Riders stand with their backs against a ring that is 6 m in diameter and that
spins at a rate of 1 revolution per second (2𝜋
𝑟𝑎𝑑
𝑠
).
(a) What is the velocity of a rider?
(b) What is the centripetal acceleration of a rider?
(c) What is the centripetal force of the rotating ring
on the back of a 55 kg rider?
𝜔 = 2𝜋
𝑟𝑎𝑑
𝑠
𝑟 =3𝑚
Universal Circular Motion Applied to Orbits
A satellite traveling around the Earth in a geosynchronous orbit will roughly pass above the
same position on earth at the same time each day. Geosynchronous orbits are circular and
have an orbital radius of approximately 42,164 km and make one revolution every sidereal
day (23.93447 hours).
NOTE: The orbital altitude is 35,786 km (the orbital radius minus the radius of the earth
which is 6,378 km).
CLASS PROBLEM: What is the velocity of a satellite in geosynchronous orbit?
NASA-Threads
Work and Mechanics
Lesson 39: Circular Motion
CLASS PROBLEM: A 500 kg satellite follows a geosynchronous orbit.
(a) Compute the centripetal force acting on the satellite.
(b) Using Newton’s Universal Law of Gravitation, compute the pull of the Earth on
the satellite.
𝑚1 𝑚2
𝑟2
r = 42,164 km
𝐹=𝐺
G = 6.674x(10)-11
𝑁∙ 𝑚2
𝑘𝑔2
mass of Earth = 5.9736 × 1024 kg
(c) Compare your answers from part (a) and (b).
NASA-Threads
Work and Mechanics
Lesson 39: Circular Motion
NOTE: Although the orbital distance was given above as 42,164 km, this distance can be
computed by equating the centripetal force to the force determined from Newton’s
Universal Law of Gravitation.
𝐺
𝑚𝑒𝑎𝑟𝑡ℎ 𝑚𝑜𝑏𝑗𝑒𝑐𝑡
𝑣2
=
𝑚
𝑜𝑏𝑗𝑒𝑐𝑡
𝑟2
𝑟
𝑤ℎ𝑒𝑟𝑒 𝑣 =
2𝜋𝑟
𝑡𝑑𝑎𝑦
The mass of the object can be cancelled from each side:
𝐺𝑚𝑒𝑎𝑟𝑡ℎ
=
𝑟2
2𝜋𝑟 2
(𝑡 )
𝑑𝑎𝑦
𝑟
Solving for the radius r yields . . .
𝑟3 =
𝐺 𝑚𝑒𝑎𝑟𝑡ℎ 𝑡𝑑𝑎𝑦 2
(2𝜋)2
Inserting the constants and solving yields . . .
6.674 ∙ (10)−11
𝑟=
2
1
2 3
𝑁∙ 𝑚
3600 𝑠
∙ 5.9736(10)24 𝑘𝑔 ∙ (23.93447 ℎ𝑟 ∙
)
ℎ𝑟
𝑘𝑔2
(2𝜋)2
[
]
= 42,160,000 𝑚
This sequence of calculations shows that the orbital radius for geosynchronous orbit does
not depend on the mass of the object in orbit.