For advanced learners

1. Circular and
Rotational
Motion

2.  Rotational motion- motion of a body that
spins about an axis
 Axis of rotation-line about which the rotation
occurs
 A point on a rotating object undergoes
circular motion, because a single point always
travels in a circle.
 Arc Length The distance traveled by a point
in a time interval (∆t) moves a distance s.

3. r
s
When working with
rotational or circular
motion, angles are
measured in radians
(rad).
where:
θrad =
θ = Angular PositionDisplacement
s = arc length
r = radius
If a point makes a complete
revolution, then arc length, s, equals the
circumference of the circle, 2πr
Counterclockwise rotation
(+)
Clockwise rotation
(-)

4.  The angle for this piece of pie is equal to 1
radian (about 57.3 ). Thus, all three of its
sides are of equal length.

5. Describing Angular Motion
 Previously we used the degree () to measure
angles. In most scientific calculations it is the
radian (rad) that is used
 The length of circular arc = radius
 1 rev = 360 = 2π rad
 Conversion
θrad= θdeg(π/180)
θdeg= θrad(180/ π)

6. While riding on a carousel that is rotating
clockwise, a child travels through an arc length of
11.5m. If the child’s angular displacement is 165
degrees, what is the radius of the carousel?
Example Angular Position/Displacement
Θdeg =-165⁰ s=-11.5m
r
s
θrad =
θrad =s/r
r=s/ θrad
r= -11.5m/-2.88rad
r=3.99m

7. Average Angular speed- rate at which a
body rotates about an axis



t
avg =
SI units: rad/s or revolutions/s (1
revolution = 2 rad), so

8. Example: Average Angular Velocity
Θrad = avg = 4.0rad/s t=?



t
avg =
t=6.28s
How many revolutions?
4.0 revolutions

9. 


t
Angular acceleration- time rate of change of
angular speed; the units are rad per second per
second or rad/s2
avg = average angular acceleration. If the
object that is rotating is rigid, all points on that
object have the same angular speed and
acceleration.
1
2
1
2
t
t 


avg=
t


=

10. 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?
Ex: Angular Acceleration
1
2
1
2
t
t 


avg= =28.0 rad/s -21.5 rad/s
3.5s
= 1.86rad/s2

11. • If angular acceleration and angular
velocity has the same sign the object
is speeding up
• If angular acceleration and angular
velocity have opposite signs the
object is slowing down
Meaning of the Sign…

12. vf = vi + at f = i + t
∆x = vit + 1/2at2 ∆ = it + 1/2(t)2
vf
2 = vi
2 + 2a∆x f
2 = i
2 + 2∆
∆x = ½ `(vi + vf)t ∆ = ½ (i + f)t
Angular Kinematics

13. The wheel on an upside down bicycle
moves through 11.0 rad in 2.0 s. What is the
wheel’s angular acceleration if its initial
angular speed is 2.0 rad?
∆= 11.0 rad t=2.0s i= 2.0 rad/s =?
∆ = it + 1/2(t)2
 = 2(∆ - it )
t2
 = 2(11.0rad-(2.0rad)(2s) = 3.5rad/s2
(2.0s)2
Angular Kinematics Example

14. Tangential speed- the instantaneous linear
speed of an object directed along the tangent
to the object’s circular path; also called
instantaneous linear speed
Tangential speeds differ if the points are
different distances from the center of
rotation (different radii).
Vt=r  or Vt= 2Πr
T
T=t for one rotation

15. The radius of a CD in a computer is 0.0600 m. If a microbe
riding on the disc’s rim has a tangential speed of 1.88 m/s,
what is the disc’s angular speed?
r=0.0600m Vt= 1.88m/s =?
Vt=r  =Vt/r
= 1.88m/s = 31.3 rad/s
0.0600m
Tangential Speed Example:

16. Tangential acceleration: instantaneous linear
acceleration of an object directed along the
tangent to the object’s circular path;
acceleration is change in tangential velocity
over time.
at= r 
SI units = m/s2

17. A spinning ride at a carnival has an angular
acceleration of 0.50 rad/s2. How far from the
center is a rider who has a tangential
acceleration of 3.3 m/s?
=0.500rad/s2 at= 3.30m/s r =?
at= r  r = at/ 
r=3.3m/s2=6.60m
0.50rad/s
Tangential Acceleration Example

18. Centripetal Acceleration
 Centripetal means center seeking
 Centripetal acceleration – acceleration
towards the center of circle
ac = v2 / r ac = 4π2r / T2

19. Centripetal Acceleration
 ac = v2 / r ac = 4π2r / T2
 A rubber stopper is attached to a 0.93 m string. The
stopper is swung in a horizontal circle, making one
revolution in 1.18 s. What is the centripetal
acceleration of the stopper?
 A runner moving at a speed of 8.8 m/s rounds a bend
with a radius of 25 m. What is the centripetal
acceleration of the runner?

20. Centripetal Force
 Centripetal Force – Force that is directed
towards the center of circle that allows an
object to follow a circular path
Fc = m ac
 An athlete whirls a 7 kg hammer tied to the end of a
1.3 m chain. The hammer makes one revolution in
1.0 s. What is the centripetal acceleration of the
hammer? What is the tension in the chain?

21. Key Scientists and Developments
 1543 Copernicus publishes his theory that the solar system is heliocentric (sun centered).
He believes the orbits are circular.
 1576 Brahe begins meticulously observing and recording planetary motions. He believes
the solar system is geocentric (earth centered).
 1600 Kepler begins working for Tycho Brahe. After Tycho’s death, Kepler uses the Tycho’s
data to develop his three laws.
 ~1600 Galileo proves gravity is uniform (a constant).
 ~1687 Newton develops the Universal Law of Gravitation.
 1798 Cavendish proves the value of G in Newton’s Law of Gravitation
 ~1830 Michael Faraday develops the concept of a “field force”

22. Newton’s Universal Law of
Gravitation
 States that the attractive
force between two objects is
directly proportional to the
product of the masses and
inversely proportional to the
square of the distance
between the objects centers.

23. Law of Gravitation
 F = gravitational force
 G = gravitational constant (6.67 X 10 -11)
 m = mass of objects
 d = distance between the two objects
F = G (m1) (m2) / d2

24. Example Problem
 What is the gravitational attractive force between Mr.
Wilson (90 kg) and Mrs. Wilson (62 kg) when they are
standing 2 m apart?
 What is the gravitational force between two spherical 8
kg masses that are 5 m apart?

25. Henry Cavendish found the value of G
(Universal Gravitational Constant)
G = 6.67 x 10 -11 Nm2/kg2

26. Kepler’s Laws of Planetary
Motion
1. The paths of the planets are ellipses with the
center of the sun at one focus.
2. An imaginary line from the sun sweeps out
equal areas in equal time intervals. Thus, the
planets move fastest when closest to the sun.

27. Kepler’s Third Law
Tb
Ta
=
2
ra
rb
3
Ta = period of Planet A
Tb = period of Planet B
ra = Planet A’s average distance
from the sun
rb = Planet B’s average distance
from the sun
3. Ratio of the squares of the periods of any
two planets revolving about the sun is equal to
the ratio of the cubes of their average distance
from the sun.

28. Example Problem
 Mars has an orbital radius of 2.28 X 1011 m and Earth
has an orbital radius of 1.50 X 1011 m. How long does
it take Mars to orbit the sun (in earth days)?
 An asteroid revolves around the sun with an average
orbital radius twice that of Earth’s. What is the period
of the asteroid in Earth years?