Download Rotary Motion

Center of Gravity
The center of gravity is the point at which all of
an objects weight can be considered to be
concentrated.
The center of gravity can be considered to be
the average position of the weight distribution.
Any object will balance if allowed to pivot at its
center of gravity.
 The weight of an object, Fw, behaves as if it
is concentrated at its cog.
 Fw can be thought of as the resultant of an
infinite number of points contributing to an
objects weight.
For a uniform object, the cog is located at its
geometric center.
 The cog may be either inside or outside of
an object.
The location of the center of gravity determines
the stability of an object.
 If a line drawn from the center of gravity
straight down falls inside the base, the
object will be stable.
 Stable in the sense of not toppling over or
rotating about a point.
 In football, lineman line up in a three point
stance to become more difficult to tip over.
Whenever the base of an object is made
larger with its cog closer to the base, the
object becomes more rotationally stable.
If a net force is applied at the cog, the object will
not rotate but accelerate in a straight line.
Torque is Cheap
In order to make an object rotate, a force must
be applied but more specifically a torque must
be applied.
Torque is the twisting effect of a force and is
given by:
T=F×
d
Torque is a vector quantity.
where F is the force in N,
d is the
perpendicular distance from the axis of rotation,
and T has units of N•m.
When it comes to objects rotating, think of the
hands of a clock.
 Tcw
Tccw
 The hands of a clock, a toy top, or a
gyroscope are examples of rotary motion.
Torque Problem
A bar 6.0 m long has its center of gravity 1.8 m
from the heavy end. If it is placed on the edge
of a block 1.8 m from the light end and a weight
of 650 N is added to the light end, the bar is in
rotational equilibrium. What is the weight of the
bar?
l = 6.0 m
F1 = 650 N
COG 1.5 m from heavy end
.
COG
Fup
1.8 m
Δ
Fw
AOR
1.8 m
6.0 m
ΣF = 0
Fup = Fw + F1
ΣT = 0
Tcw = Tccw
Translational
Equilibrium
Rotational
Equilibrium
F1
AOR at the fulcrum.
ΣT = 0
Tcw = Tccw
F1 × d1 = Fw × dw
650 N × 1.8 m = Fw × 2.4 m
Fw = 490 N
Talking About Torque
The AOR (axis of rotation) is an arbitrary point
that can be placed on or off the object.
 All perpendicular distances are measured
from this point.
The COG (center of gravity) is the point on or
off the object that Fw is considered to be acting.
 An example of the COG not found on the
object is a hallow ring.
Remember, the distances measured from the
AOR must be perpendicular distances.
 For example, if F1 had not been
perpendicular to the bar, you would have
to resolve F1 to find the component that is
perpendicular.
You know the object is not symmetrical
because the problem torques about a heavy
end.
To maintain translational equilibrium (motion in
a straight line) the ΣF must equal zero.
To maintain rotational equilibrium (rotary motion
about a point), the ΣT must equal zero.
If a force is applied at the COG and the object is
free to move, the object will move in a straight
line.
Another Torque Problem
A uniform board weighing 360 N and 8.0 m long
rests on a dock with 2.4 m of the board
extending out over the water. How far from the
dock could a 460 N boy walk out on the board
without getting wet?
Fwb = 360 N
Fwboy = 460 N
l = 8.0 m
Fup
.
1.6 m
Fwb
Δ
2.4 m
Fwboy
8.0 m
ΣF = 0
Fup = Fwb + Fwboy = 360 N + 460 N = 820 N
ΣT = 0
Tcw = Tccw
AOR at the fulcrum.
ΣT = 0
Tcw = Tccw
Fwboy × dwboy = Fwb × dwb
460 N × d = 360 N × 1.6 m
dwboy = 1.3 m
Beyond 1.3 m the Tcw > 0 and both the boy and the
board will fall into the water.
Rotary Motion
Rotary motion is an object spinning on its axis.
 Examples include the hands of a clock or
a bicycle tire spinning on its axis.
 Rotary motion is not as intuitive as
circular, projectile, or rectilinear motion.
 Rotary motion uses the same terms as
rectilinear motion but are defined
differently.
Analogous Terminology
Angular Velocity (ω)
The magnitude of angular velocity is given by
Δθ
ω=
Δt
ω is the angular velocity and Δθ is the angular
displacement.
The angular displacement, Δθ, is usually
measured in radians.
A radian is defined such that length of the
intercepted arc of a circle is equal to the radius
of the circle.
r
s
When the length of s equals the
length of the radius (s = r), θ is
equal to 57.3° or 1 radian.
Why 57.3°?
 1 revolution = 360° = 2π radians
1 radian = 360°/2/π ≈ 57.3°
Keep in mind that a radian is a dimensionless
quantity.
The direction of the angular velocity is not
intuitive either as shown on the following slide.
Assume as you look at the wheel above that it
is rotating counterclockwise in a horizontal
plane.
If you curl the fingers of your right hand, your
thumb points in the direction of the angular
velocity (pointing straight, up perpendicular to
the horizontal.
A rotating object can have a constant angular
velocity of zero or be rotating at a constant rate.
 Reminiscent of Newton’s 1st Law.
In accordance with Newton’s 2nd Law, a rotating
may accelerate by changing its rate of rotation,
the direction of the rotation, or both.
Angular acceleration (α) is defined by
α=
Δω
Δt
with units of rad/s2.
Uniformly Accelerated Rotation
Equations for uniformly accelerated rotary
motion.
 ωf = ωi + αΔt
 Δθ = ωiΔt + ½αΔt2
 ωf2 = ωi2 + 2αΔθ
Rotary Motion Problem
A circular saw blade completes
1400 revolutions in 60. s while coming to a stop.
(a) Assuming the deceleration is constant, what
is the angular deceleration in rad/s2?
ωave = Δθ
Δt
1400 rev ×
ωave =
60. s
2π rad
1 rev
= 150 rad/s
ωave = (ωi + ωf)/2
ωf = 0; ωi = 2 × ωave = 2 × 150 rad/s = 300 rad/s
ωf = ωi + αΔt
α =
0 – 300 rad/s
= - 5.0 rad/s2
60. s
Its like two people on a merry-go-round.
One is close to the axis of rotation and the
other on the outer edge but yet they both
sweep out 360° or make one revolution in
the same amount of time.
(d) Do all of the points on the circular saw blade
have the same linear speed (v)?
No because the further a point is away from
the center, the faster it must travel to “keep
up” with the other points.
(b) Assuming the deceleration is constant, what
was the initial angular speed?
As determined in Part (a), ωi = 300 rad/s.
(c) Do all of the points on the circular saw blade
have the same angular speed (ω)?
Yes, because every point on the blade
moves through the same angle in the same
time interval.
All circles consist of 360° but the larger the
circle the larger the circumference.
The relationship between linear velocity and
angular velocity is given by
v = ωr
and radians are dimensionless units so
m/s = 1/s × m = m/s
More Rotary Motion Problems
A ball rolls 5.7 m along a horizontal surface
before coming to a stop. If the ball makes
12 revolutions, what is the diameter of the ball?
2×π×r
= 75 r
5.7 m = 12 rev ×
1 rev
r = 7.6 x 10-2 m
D = 2 × r = 1.5 x 10-1 m
An electric saw blade is spinning at
1750 rev/min. If the blade has a diameter of
0.45 m:
(a) What is its angular velocity in rad/s?
rev
2πrad
1 min
×
ω = 1750 min ×
1 rev
60 s
ω = 180 rad/s
(b) What is the linear speed of a point on the
edge of the saw blade?
v = ωr = 180 rad/s × 0.45 m = 81 m/s
(c) What is the acceleration of a point on the
edge of a saw blade?
The tangential acceleration (aT) is given by
aT = rα.
The angular velocity is constant, therefore
the tangential acceleration (aT) is zero.
The centripetal (radial) acceleration is given by
aR =
v2
(wr)2
2r
=
ω
=
r
r
aR =
v2
(180 rad/s)2
4 m/s2
=
=
7.2
x
10
r
0.45 m
The total linear acceleration is equal to the
vector sum of a = aT + aR.
Moment of Inertia
The moment of inertia is an interesting concept
and a little more involved than just plain old
“inertia” found in Newton’s 1st Law.
As with Newton’s 1st Law, a non-rotating wheel
will remain at rest unless a torque is applied.
A wheel rotating at a constant angular velocity
will continue to do so until a net torque is
applied.
To explain the inertia of a rotating object not
only is the mass important but also the
distribution of mass with respect to the axis of
rotation.
A familiar example is that of a figure skater
spinning or rotating in place.
While a figure skater has her arms extended,
she rotates rather slowly but she rotates much
faster when she pulls her arms into her body.
 Her mass does not change but the
distribution of her mass changes when
she pulls her arms into her body.
The close the mass is to the axis of
rotation, the less rotational inertia an
object has.
Rotational inertia is called the moment of
inertia and is symbolized by I (capital eye).
The geometric shape and the location of the
axis of rotation determines the moment of
inertia.
Consider the following three objects all having
the same mass and their moments of inertia:
 A solid sphere, I = 2/5mr2, a solid cylinder,
I = ½mr2, and a hollow cylinder, I = mr2.
 What happens when they are released at
the same time on the same incline from
the same height?
The sphere reaches the bottom first,
followed by the solid cylinder, and lastly
the hollow cylinder.
 The hollow cylinder because of its largest
moment of inertia will be the hardest to
start and stop rotating.
 The units for moment of inertia are kg•m2.
Newton’s 2nd Law
Newton’s 2nd Law applied to rotating objects is
given by
T = Iα
where T is the torque discussed in the
beginning slides, I is the moment of inertia, and
α is the angular acceleration.
A Moment of Inertia Problem
A uniform cylinder has a mass of 0.680 kg and
a radius of 9.40 cm.
(a) Calculate its moment of inertia.
m = 0.680 kg
r = 9.40 cm
I = ½mr2
I = ½ × 0.680 kg × (9.40 cm × 1 m/102 cm)2
I = 3.00 x 10-3 kg•m2
(b) Calculate the torque needed to accelerate it
from rest to 1700 rpm in 8.00 s.
ωi = 0
ωf = 1700 rev/min
Δt = 8.00 s
ωf = ωi + αΔt
1700 rev × 2πrad × 1 min =
min
rev
60 s
0 + α × 8.00 s
α = 22 rad/s2
T = Iα = 3.00 x 10-3 kg•m2 × 22 rad/s2
T = 6.6 x 10-2 m•N
Angular Impulse and Momentum
Recalling impulse and the change of
momentum for rectilinear motion:
FnetΔt = Δmv = mΔv
an analogous equation can be written for rotary
motion:
TnetΔt = IΔω
TnetΔt = IΔω
where Tnet is the net torque in N•m, Δt is the
time in s that the Tnet is acting, I is the moment
of inertia in kg•m2, and ω is the angular velocity
in rad/s.
The quantity TnetΔt is the angular impulse
expressed in N•m•s and IΔω is the change in
angular momentum expressed in kg•m2/s.
Just as there is the conservation of linear
momentum given by:
Δp = 0
pi = p f
m1v1 + m2v2 = m1v1’ + m2v2’
there is a conservation of angular momentum
given by:
ΔL = 0
Li = L f
I1ω 1 = I 2ω 2
The example of a figure skater has already
been discussed.
Have you ever wondered why it is so much
easier to balance on a moving bike than a
stationary one?
 Stationary or not, the location of your
center of gravity does not change.
 Stationary or not, the size of your base
does not change unless your fall over.
An object spinning on its axis has a high degree
of stability.
A large torque is needed to change the angular
momentum.
 Remember, the direction of the angular
momentum is given by curling the fingers
of your right hand in the direction of
rotation.
Angular Momentum Problem
A merry-go-round has a mass of 200. kg and a
radius of 30. m. A boy whose mass is 50. kg
walks from the outer edge toward the center. If
the angular velocity of the merry-go-round is
4.0 rad/s when the boy is at the outer edge,
what will be the angular velocity of the boy
when he is 10. m from the center?
Assume I = ½mr2 for the merry-go-round and
I = mr2 for the boy.
mmgr = 200. kg mb = 20. kg
rb = 10. m
rmgr = 30. mω1 = 4.0 rad/s
ω2 = ?
ΔL = 0
Li = L f
I1ω 1 = I 2ω 2
((½mr2)mgr + (mr2)b) × ω12 =
((½mr2)mgr + (mr2)b) × ω22
(½ × 200. kg × (30. m)2 + 20. kg × (30. m)2) ×
4.0 rad/s =
(½ × 200. kg × (30. m)2 + 20. kg × (10. m)2) ×
ω2
ω2 = 4.7 rad/s
Counter Steering or
Counter Intuitive Physics
A rotating wheel will resist changes in its state
of rotation when a force is applied to it
• A rotating wheel is a gyroscope
• Newton’s 2nd Law applied to rotational
dynamics
• The resistance comes as a result of the I
(moment of inertia)
• The applied force produces an unbalanced
torque
Gyroscopic precession results when a force is
applied to a rotating mass and the force is
redirected 90° in the direction of the rotation
Gyroscopic precession is best illustrated when
going more than ≈ 20 mi/h on a motorcycle
• Not noticeable on a bicycle. Why?
Pushing the left handlebars causes the handle
bars to rotate to the right resulting in the
motorcycle turning to the left
This technique is used when taking turns (left
and right) and evasive maneuvers
A gyroscope does not precess in the absence
of gravity or in a state of free fall
Is this counter intuitive or what?
Let’s explore courtesy of Prentice Hall
When the wheel rotates,
the angular velocity (ωs)
combined with the torque
produced by the weight of
the wheel (Fw) causes
precession.
All vectors lie in the
same horizontal
plane. When Δωα is
small, vectors α, αΔt,
and Δωα are all
perpendicular to Δωs
and Δωs’.
Top View
ccw
Fw
Wrap Up Questions
The wheel shown below has a radius, R, and
rotates counterclockwise.
rA = 0
A•
R
•B
C
rB = R/2
rC = R
Do points A, B, and C have the same angular
velocity?
A, B, and C have the same angular velocity.
They all trace out the same number of
revolutions for a given time interval.
Do points A, B, and C have the same linear
velocity?
A, B, and C have different linear velocities.
A has the smallest, v = 0 m/s, B has the next
largest, v = R/2 × ω, and C has the largest, v =
R × ω.
The formula, v = ω × r, relates the linear
velocity (m/s) to the angular velocity (rad/s)
provided that ω is expressed in rad/s.
r is the distance from the axis of rotation.
What is the direction of the angular velocity?
Using the right hand rule, curl the fingers of
your right hand in the direction of the rotation,
and your thumbs points straight out
(perpendicular to the plane of the page) which
is the direction of ω.
Do points A, B, and C have the same linear
acceleration?
The linear acceleration equals the centripetal
acceleration and is given by,
a c = ω2 × r
where ω is in units of rad/s2 and r is the
distance from the axis of rotation.
Point A has an ac = 0, Point B has an
ac = R/2 × ω2, and Point C has an ac = R × ω2.