# Slide 1

```Ch 7 - Circular Motion
• Circular motion: Objects moving in a
circular path.
Measuring Rotational Motion
• Rotational Motion – when an object turns
– Ex. Earth’s is every 24 hrs.
• Axis of rotation – the line about which the
rotation occurs
• Arc length – the distance (s) measured
along the circumference of the circle
• Radian- an angle whose arc length is
equal to its radius, which is approximately
equal to 57.3&deg;
• When the arc length “s” is equal to the
length of the radius, “r”, the angle θ swept
by r is equal to one rad.
• Any angle θ is radians if defined by:
• θ = s/r
• When a point moves 360&deg;,
θ=s/r = 2πr/r = 2π rad
Therefore, to convert from rads to degrees
180&deg;
For angular displacement,
Δθ=Δs/r
Angular displacement (in radians)= change in
arc length/distance from axis
Example Problem
• While riding on a carousel that is rotating
clockwise, a child travels through an arc
length of 11.5 m. If the child’s angular
displacement is 165&deg;, what is the radius of
the carousel?
• Ans. 3.98 m
Angular Substitutes for Linear
Quantities
• Linear (Straight Line)
– Displacement = x
– Velocity = v
– Acceleration = a
• Rotational
– Displacement = θ
– Velocity = ω
– Acceleration = α
• Angular speed – the rate
at which a body rotates
• Symbol = ω (omega) Unit
• ω(avg) = Δθ/Δt
• ω can also be in rev/s
• To convert:
Example Problem
• A child at an ice cream parlor spins on a
stool. The child turns counter-clockwise
with an average angular speed of 4.0
rad/s. In what time interval will the child’s
feet have an angular displacement of 8.0
• Ans. 6.3 s
• Angular Acceleration –
the time rate of change of
angular speed, expressed
second
•  avg = ω2-ω1/t2-t1 =Δω/Δt
• Average angular
acceleration = change in
angular speed / time
interval
Example Problem
• A car’s tire rotates at an initial angular
speed of 21.5 rad/s. The driver
accelerates, and after 3.5 s the tire’s
angular speed is 28.0 rad/s. What is the
tire’s average angular acceleration during
the 3.5 s time interval?
Frequency vs. Period
• Frequency – # of revolutions per unit of
time. Unit: revolutions/second (rev/s).
• Period – time for one revolution. Unit =
second (s).
• Inversely related:
t = 1/f and f = 1/t
Tangential Velocity
• Speed that moves
along a circular path.
• Right angles to the
• Direction of motion is
always tangent to the
circle.
Rotational Speed
• The number of rotations per unit of time.
• All parts of the object rotate about their
axis in the same amount of time.
• Units: RPM (revolutions per minute).
Tangential vs. Rotational
• If an object is rotating:
– All points on the object have the same rotational
(angular) velocity.
– All points on the object do not have the same linear
(tangential) velocity.
• Tangential speed is greater on the outer edge than
closer to the axis. A point on the outer edge moves a
greater distance than a point at the center.
• Tangential speed = radial distance x rotational speed
v  r
Centripetal Acceleration
• The acceleration of
an object moving in a
circle points toward
the center of the
circle.
• Means “center
seeking” or “toward
the center”.
ac  r
2
v
ac 
r
2
7.3 Forces that maintain
circular motion
• Consider a ball swinging on a string. Inertia
tends to make the ball stay in a straight-line
path, but the string counteracts this by exerting a
force on the ball that makes the ball follow a
circular path.
• This force is directed along the length of the
string toward the center of the circle.
The force that maintains circular motion
(formerly known as centripetal force)
• Fc = (mvt2)/r
•
Force that maintains circular motion = mass x (tangential speed)2 &divide; distance to axis of motion
• Fc = mrω2
•
Force that maintains circular motion = mass x distance to axis x (angular speed)2
• Because this is a Force, the SI unit is the Newton (N)
Practice Problem
• A pilot is flying a small plane at 30.0 m/s in
a circular path with a radius of 100.0 m. If
a force of 635N is needed to maintain the
pilot’s circular motion, what is the pilot’s
mass?
• Answer: m = 70.6 kg
Common Misconceptions
• Inertia is often misinterpreted as a force
• Think of this example: How does a
washing machine remove excess water
from clothes during the spin cycle?
Newton’s Law of Universal
Gravitation
• Gravitational force: a field force that
always exists between two masses,
regardless of the medium that separates
them; the mutual force of attraction
between particles of matter
• Gravitational force depends on the
distance between two masses
Newton’s Law of Universal
Gravitation
Gm1m2
F
2
r
Where F = Force
M1 and m2 are the masses of the two objects
R is the distance between the objects
And G = 6.673 x 10-11 Nm2/kg2 (constant of universal gravitation)
Practice Problem
• Find the distance between a 0.300 kg
billiard ball and a 0.400 kg billiard ball if
the magnitude of the gravitational force is
8.92 x 10-11 N.
• Answer: r = 3.00 x 10-1 m
```