Download ANGULAR POSITION

ANGULAR POSITION
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To describe rotational motion, we define angular
quantities that are analogous to linear quantities
Consider a bicycle wheel that is free to rotate about its
axle
The axle is the axis of rotation for the wheel
If there is a small spot of red paint on the tire, we can
use this reference to describe its rotational motion
The angular position of the spot is the angle θ, that a
line from the axle to the spot makes with a reference
line
SI unit is the radian (rad)
θ > 0 – anticlockwise rotation: θ < 0 – clockwise rotation
A radian is the angle for which the arc length, s, on a
circle of radius r is equal to the radius of the circle
The arc length s for an arbitrary angle θ measured in
radians is s = r θ
1 revolution is 360°= 2π rad
1 rad = 360°/2π = 57.3°
4. Rotational Kinematics
and Dynamics
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ANGULAR VELOCITY
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As the bicycle wheel rotates, the angular position of the
spot changes
Angular displacement is ∆θ = θf – θi
Average angular velocity is ωav = ∆θ/∆t (rad/s)
Analogous average linear velocity vav = ∆x/∆t
Instantaneous angular velocity is the limit of ωav as
the time interval ∆t reaches zero
ω > 0 – anticlockwise rotation: ω < 0 clockwise rotation
The time to complete one revolution is known as the
period, T
T = 2π/ω seconds
4. Rotational Kinematics
and Dynamics
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ANGULAR ACCELERATION
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If the angular velocity of the rotating bicycle wheel
increases or decreases with time, the wheel
experiences an angular acceleration, α
The average angular acceleration is the change in
angular velocity in a given time interval
αav = ∆ω/∆t rad/s2
The instantaneous angular acceleration is the limit of
αav as the time interval ∆t approaches zero
The sign of angular acceleration is determined by
whether the change in angular velocity is positive or
negative
If ω is becoming more positive
(ωf > ωi), α is positive
If ω is becoming more negative
(ωf < ωi), α is negative
If ω and α have the same sign,
speed of rotation increasing
If ω and α have opposite signs,
speed of rotation decreasing
4. Rotational Kinematics
and Dynamics
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ROTATIONAL KINEMATICS
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Rotational kinematics describes rotational motion
Consider the pulley shown below, which has a string
wrapped around its circumference with a mass
attached to its free end
When the mass is released, the pulley begins to rotate
– slowly at first, then faster and faster
The pulley thus accelerates with constant angular
acceleration: α = ∆ω/∆t
If the pulley starts with initial angular velocity ω0 at time
t = 0, and at the later time t the angular velocity is ω
then α = ∆ω/∆t = (ω – ω0)/(t – t0) = (ω – ω0)/t
Thus the angular velocity ω varies with time as follows:
ω = ω0 + αt
Example: If the
angular velocity of the
pulley is -8.4rad/s at a
given time, and its
angular acceleration is
-2.8rad/s2, what is the
angular velocity of the
pulley 1.5s later?
4. Rotational Kinematics
and Dynamics
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LINEAR AND ANGULAR ANALOGIES
4. Rotational Kinematics
and Dynamics
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ROTATIONAL KINEMATICS:
EXAMPLE (1)
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To throw a curve ball, a baseball pitcher gives the ball
an initial speed of 36.0 rad/s. When the catcher gloves
the ball 0.595s later, its angular speed has decreased
(due to air resistance) to 34.2 rad/s. What is the ball’s
angular acceleration, assuming it to be constant? How
many revolutions does the ball make before being
caught?
4. Rotational Kinematics
and Dynamics
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ROTATIONAL KINEMATICS:
EXAMPLE (2)
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On a TV game show, contestants spin a wheel when it
is their turn. One contestant gives the wheel an initial
angular speed of 3.4 rad/s. It then rotates through1 ¼
revolutions and comes to rest on the BANKRUPT
space. Find the angular acceleration of the wheel,
assuming it to be constant. How long does it take for
the wheel to come to a rest?
4. Rotational Kinematics
and Dynamics
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TANGENTIAL SPEED OF A
ROTATING OBJECT
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Consider somebody riding a merry-go-round, which
completes one circuit every T = 7.5s
Thus ω = 2π/T = 0.838 rad/s
The path followed is circular, with the centre of the
circle at the axis of rotation
The rider is moving in a direction that is tangential to
the circular path
The tangential speed is the speed at a tangent to the
circular path, and is found by dividing the circumference
by T: vt = 2πr/T m/s
Because 2π/T = ω we have: vt = rω m/s
Example: Find the angular speed a CD must have to
give a linear speed of 1.25m/s when the laser beam
shines on the disk 2.50cm and 6.00cm from its centre
4. Rotational Kinematics
and Dynamics
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CENTRIPETAL ACCELERATION OF
A ROTATING OBJECT
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When an object moves in a circular path, it experiences
a centripetal acceleration, acp, which is always directed
toward the axis of rotation
acp = v2/r
However v = vt = rω, so acp = (rω)2/r = rω2 m/s2
Rotating devices known as centrifuges can produce
centripetal accelerations many times greater than
gravity, such as those used to train astronauts, or
microhematocrit centrifuges used to separate blood
cells from plasma
Example: In a microhematocrit centrifuge, small
samples of blood are placed in capillary tubes. These
tubes are rotated at 11,500rpm, with the bottom of the
tubes 9.07cm from the axis of rotation. Find the linear
speed of the bottom of the tubes. What is the
centripetal acceleration at the bottom of the tubes?
4. Rotational Kinematics
and Dynamics
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TANGENTIAL AND CENTRIPETAL
ACCELERATION
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When the angular speed of an object in a circular path
changes, so does its tangential speed
When tangential speed changes, a tangential
acceleration is experienced at
If ω changes by the amount ∆ω, with r remaining
constant, the corresponding change in tangential speed
is ∆vt = r∆ω
If ∆ω occurs in time interval ∆t, then the tangential
acceleration is at = ∆vt/∆t = r∆ω/∆t
Since ∆ω/∆t is the angular acceleration α, then the
tangential acceleration of a rotating object is given
by at = rα m/s2
Recall that at is due to a changing tangential speed,
and that acp is caused by a changing direction of motion
(even if at remains constant)
In cases where both tangential and centripetal
accelerations are present, the total sum is the vector
sum of the two
r
r
a t and a cp are at right angles, hence the magnitude of
the total acceleration is a = √(at2 + acp2)
The direction is given by φ = tan-1(acp/at)
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TANGENTIAL AND CENTRIPETAL
ACCELERATION: EXAMPLE
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Suppose the centrifuge above is starting up with a
constant angular acceleration of 95.0 rad/s2. What is
the magnitude of the centripetal, tangential and total
accelerations of the bottom of a tube when the angular
speed is 8.00 rad/s? What angle does the total
acceleration make with the direction of motion?
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and Dynamics
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TORQUE: WHEN FORCE APPLIED IS
TANGENTIAL
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Trying to loosen a nut by rotating a wrench
anticlockwise is easier when you apply the force as far
away from the nut as possible
Likewise to open a revolving door is easier when you
push further from the axis of rotation
The tendency for a force to cause a rotation increases
with the distance r from the axis of rotation to the force
Torque is a quantity that takes into account both the
magnitude of the force and the distance from the axis of
rotation, r
Torque: τ = rF Nm (Newton – metre)
This equation is only valid when the applied force is
tangential to a circle of radius r centred on the axis of
rotation
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and Dynamics
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TORQUE: WHEN FORCE APPLIED IS
NOT TANGENTIAL
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Consider pulling on a merry-go-round in a direction that
is radial (along a line that extends through the axis of
rotation)
Such a force has no tendency to cause a rotation, and
thus the axle simply exerts an equal and opposite force,
and thus the merry-go-round remains at rest
A radial force produces zero torque
If the force applied
is at an angle θ to the radial line, the
r
vector force F needs to be resolved into radial and
tangential components
Radial component magnitude: Fcosθ
Tangential component magnitude: Fsinθ
Only tangential component causes rotation, thus Fcosθ
=0
General definition of torque: τ = rFsinθ Nm
τ > 0 – anticlockwise angular acceleration
τ < 0 – clockwise angular acceleration
4. Rotational Kinematics
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TORQUE: EXAMPLE
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Two helmsmen, in disagreement about which way to
turn a ship, exert different forces on the ship’s wheel.
The wheel has a radius of 0.74m, and the two forces
have the magnitudes
r F1 = 72N and F2 = 58N. Find
r the
torque caused by F1 and the torque caused by F2 . In
which direction does the wheel turn as a result of these
two forces.
4. Rotational Kinematics
and Dynamics
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TORQUE AND ANGULAR
ACCELERATION
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A single torque, τ, acting on an object causes the object
to have an angular acceleration α
Consider a small object of mass m connected to an
axis of rotation by a light rod of length r
If a tangential force of magnitude F is applied to the
mass, it will move with an acceleration according to
Newton’s 2nd law, a = F/m
Linear and angular accelerations related by α = a/r
Combining: α = a/r = F/mr
Multiplying by r/r gives α = rF/mr2
Since torque τ = rF, we define a new quantity called the
moment of inertia: I = mr2
Thus α = τ/I or τ = Iα
In a system with more than one torque, we take the net
sum of all the torques acting: τnet = Στ = Iα
Above is Newton’s 2nd law
for rotational motion
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and Dynamics
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MOMENT OF INERTIA
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I = mr2 is general case for moment of inertia
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TORQUE AND ANGULAR
ACCELERATION: EXAMPLES
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A light rope wrapped around a disk shaped pulley is
pulled tangentially with a force of 0.53N. Find the
angular acceleration of the pulley, given that its mass is
1.3kg and its radius is 0.11m.
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A fisherman is dozing when a fish takes the line and
pulls it with a tension T. The spool of the fishing reel is
at rest initially and rotates without friction as the fish
pulls for time t. If the radius of the spool is R and its
moment of inertia is I, find the angular displacement of
the spool. Also find the length of line pulled from the
spool and the angular speed of the spool. Hint: make
use of θ = θ0 + ω0t + ½ αt2 and ω = ω0 + αt
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and Dynamics
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