MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 6: ROTATION (PART 1)
CLASS NOTES
CIRCULAR MOTION
Circular motion refers to the motion of a body about a fixed radius. One complete rotation about the
center (360o) is called a revolution.
RADIANS
The fundamental unit of circular motion is the radian, commonly abbreviated rad or rads:
One radian occurs when s = r.
RADIANS CONVERSION
The circumference of a circle is:
C = 2r
Therefore, for every complete turn along the circumference of a circle, we can trace out 2 radians.
This leads to the following conversion:
2 radians = 1 revolution
ANGULAR DISTANCE
 is the distance traveled in radians, it is an angular measurement.
ANGULAR VELOCITY
 is the angular velocity, measured in rads/sec.
TANGENTIAL VELOCITY
Tangential velocity is the linear speed traveled by an object along the arc of a circle:
v = r
Note: The direction of the velocity vector is a tangent to the circular path.
Example 1. A point on a wheel rotating at 5 rev/s and located 0.2 m from the axis has what tangential
velocity?
1A.
(1)  = (5 rev/s)(2 rad/rev) = 31.4 rad/s
(2) v = r
(3) v = (0.2 m)(31.4 rad/s)
(4) v = 6.3 m/s
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
ANGULAR ACCELERATION
 is the angular acceleration, measured in rads/s2.
TANGENTIAL ACCELERATION
Tangential acceleration is the linear acceleration along the arc of a circle:
at = r
CENTRIPETAL ACCELERATION
In circular motion, the centripetal acceleration is directed inward toward the center of the circle:
ac = v2/r
Example 2. A point on the rim of a 0.20 m radius rotating wheel has a centripetal acceleration of 4.0
m/s2. What is the angular velocity of the wheel?
2A.
(1) ac = v2/r
(2) v2 = rac
(3) v = (rac)1/2
(4) v = [(0.2 m)(4.0 m/s2)]1/2
(5) v = 0.89 m/s
(6) v = r
(7)  = v/r
(8)  = 0.89 m/s / 0.20 m
(9)  = 4.47 rad/s
Example 3. What angular velocity (in revs/sec) is needed for a centrifuge to produce an acceleration of
1000 g (1000 times the force of gravity) on a radius arm of 10 cm?
3A.
(1) ac = v2/r
(2) v2 = rac
(3) v = (rac)1/2
(4) v = [(0.10 m)(1000)(9.8 m/s2)]1/2
(5) v = 31.3 m/s
(6) v = r
(7)  = v/r
(8)  = (31.3 m/s)/(0.1 m)
(9)  = 313 rad/s
(10) (313 rad / 1 sec)(1 rev / 2 rad)
(11)  = 49.8 rev/s
CIRCULAR MOTION KINEMATICS
Circular motion equations derive from linear motion equations:
d (distance in meters)
~  (distance in radians)
v (velocity in m/s)
~  (velocity in rads/s)
2
a (acceleration in m/s )
~  (acceleration in rads/s2)
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
CIRCULAR MOTION EQUATIONS



Circular Equation
 = ot + ½ t2
 = o + t
2 = o2 + 2
Linear Equation
(d = vot + ½ at2)
(v = vo + at)
(v2 = vo2 + 2ad)
Example 4. A Ferris wheel, rotating initially at an angular velocity of 0.50 rad/s, accelerates over a 5 s
interval at a rate of 0.04 rad/s2. What angular displacement does the Ferris wheel undergo in this 5 s
interval?
4A.
(1)  = ot + ½ t2
(2)  = (0.50 rad/s)(5 s) + ½ (0.04 rad/s2)(5 s)2
(3)  = 3.0 rad
Example 5. A Ferris wheel, starting at rest, builds up to a final angular velocity of 0.71 rad/s while
rotating through an angular displacement of 2.5 rad. What is its average angular acceleration?
5A.
(1) 2 = o2 + 2
(2) 2 = 0 + 2
(3) 2 = 2
(4)  = 2 / 2
(5)  = (0.71 rad/s)2 / 2(2.5 rad)
(6)  = 0.10 rad/s2
CENTRIPETAL FORCE
All centripetal forces act toward the center of the circular path along which the object moves:
Fc = mv2/r
Example 6. A 0.3 kg mass, attached to the end of a 0.75 m string, is whirled around in a circular
horizontal path. If the maximum tension that the string can withstand is 250 N, then what maximum
velocity can the mass have if the string is not to break?
6A.
(1) Fc = mv2/r
(2) v2 = rFc/m
(3) v = (rFc/m)1/2
(4) v = [(0.75 m)(250 N)/(0.3 kg)]1/2
(5) v = 25.0 m/s
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
Example 7. A roller coaster, loaded with passengers, has a mass of 500 kg; the radius of curvature of
the track at the bottom point of the dip is 12 m. If the vehicle has a speed of 18 m/s at this point, what
force is exerted on the vehicle by the track?
7A.
(1) The track must support the weight of the coaster and the centripetal force of its motion:
(2) Track = weight + Fc
(3) Track = mg + mv2/r
(4) (500 kg)(9.8 m/s2) + [(500 kg)(18 m/s)2] / 12 m
(5) Track = 4900 N + 13,500 N
(6) Track = 1.84 x 104 N
Example 8. A car rounds a circular turn of radius r = 80 m in a horizontal (unbanked) road. A rear
view of the car is given below. The coefficient of static friction between tires and road is 0.65 (we
assume static friction because we don’t want the car to slide (that is skid)). The car’s weight mg is 1.2 x
104 N.
Example 8.
8a.
A.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
What is the magnitude of the static force of friction when the car’s speed v = 20 m/s?
Fc = mv2/r
w = mg
m = w/g
m = (1.2 x 104 N) / (9.8 m/s2)
m = 1224 kg
Fc = (1224 kg)(20 m/s)2 / (80 m)
Fc = 6120 N
8b. What is the maximum value of the static frictional force?
A.
(1) Fc (max): fs = n
(2) fs = w
(3) fs = (0.65)(1.2 x 104 N)
(4) fs = 7800 N
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PHYSICS
MR. SURRETTE
8c.
A.
(1)
(2)
(3)
(4)
(5)
(6)
VAN NUYS HIGH SCHOOL
Determine the maximum speed for the car to make it around the curve.
Fc = mv2/r
mv2 = rFc
v2 = (rFc) / m
v = (rFc /m)1/2
v = [(80 m)(7800 N) / (1224 kg)]1/2
v = 22.6 m/s
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PHYSICS