Transcript L3 ROTATIONAL MOTION

ROTATIONAL MOTION
NCEA Level 3
Physics
ROTATIONAL MOTION
This unit is made up of the following:
 Rotational Kinematics (page 62 - 65)
 Force and Torque (page 66).
 Rotational Inertia (page 67 - 68).
 Rolling Motion (page 69 - 72).
 Rotational Kinetic Energy (page 78 - 81).
 Angular momentum (page 81 - 86).
ROTATIONAL KINEMATICS
This is important to understand with respect to
aspects of circular motion.
In orbital motion, the object that is rotating stays a
fixed distance from the centre of rotation.
In spinning motion the concept of radius has no
meaning because each position on a spinning object
is rotating with a different radius to an adjacent
position:
1
2
3
4
PURE TRANSLATION:
Movement of the CoM without any rotation about CoM. Force is
applied through the CoM.
PURE ROTATION:
Circular motion around CoM, while CoM remains stationary.
Push
CoM
Push
Both translation and rotation can occur on an object when a
resultant force does not act through the CoM. The object will
move through a straight line through the CoM while rotating
about the CoM.
CoM
Translational
motion
Push
Rotational motion
ANGULAR MEASUREMENTS
A1
To move from A to A1 requires an
object to move through certain
angle of the circle, .
B1
To move from B to B1 requires
the object to move same angle, .

B
A
However, the distance covered
from A to A1 is far greater than
the distance from B to B1.
 = angle of rotation
In order for A & B to leave their start position and finish at their
end positions they move through the same angle of rotation
over the same time but there is a difference in their motion.
What is it?
‘A’ moves faster, over a larger distance
ANGULAR DISPLACEMENT
IN RADIANS

r
1 Radian = 57o
1 revolution = 2 rads
d
length of the arc [ d ]
divided by the radius
[ r ] produces the
angular
displacement (θ)
d

r
IN A FULL CYCLE

d
r
2r

 2
r
Hence 3600 = 2 Radians
RADIANS
1800 =  radians
Example 1:
A bicycle wheel turns through 4½ revolutions:
a. What is the angular displacement of the wheel?
b. What distance does a point on the rim travel, if
the wheel has a radius of 40cm?
SOLUTION:
a.  = 4.5 revolutions
= 4.5 x 2
= 28 rad
b. Distance d travelled around
the rim is calculated by  =
d/r
d=xr
= 0.40 x 28.2743
= 11 m
ANGULAR VELOCITY

d
r
v
Suppose the distance, d,
from A to A1 takes a time t.
d  r
A1
d

r
d

r
t
t
v
[DIVIDING EACH
SIDE BY
t]
v  r
A

v = tangential
velocity

t

= the
ANGULAR
VELOCITY
v  r
v = tangential
velocity
v


t
 = the ANGULAR
VELOCITY in
v

RADIANS PER SECOND
Considering a
full circle:

r
2

T
  2f
T = Periodic time,
f = frequency
Example 2:
An aircraft propeller rotates at a speed of 4000 revolutions per
minute & has a radius of 0.80m. Find:
a. The angular velocity of the propeller.
b. The linear speed of the tip of the propeller blade
SOLUTION:
a. Angular speed  = 4000rpm
= 4000 x 2 rad per minute
[1rev = 2  rad]
= 4000 x 2  60 rad per second
= 420 rads-1
b. Linear speed
v=r
= 0.80 x 418.879
= 340ms-1
Approximately the speed of sound, that’s why propeller blades are
noisey
ANGULAR ACCERLERATION
If an object is moving around a circle then even if its
angular velocity is constant its direction is still
changing. Thus it must be accelerating and hence has
angular acceleration, .
 =  / t
Units = rads-2
Just like angular velocity we can combine angular
acceleration with translational acceleration using the
formula:
a = r
‘r’ is the distance from the centre of rotation.
Combining translational and rotational quantities
gives us the table below:
TRANSLATIONAL
ROTATIONAL
EQUATION
d
 = d/t
d = r
v = d/t
 = /t
v = r
a = v/t
 = /t
a=r
Example 3:

A pulley of radius 0.20m has a length of string
wrapped around it, attached to a mass. The
mass is released from rest & falls with an
acceleration of 3.0ms-2. Find:
0.20m
a. The angular acceleration of the pulley.
b. The angular velocity of the pulley after 5.0s.
SOLUTION:
a = 3.0ms-2
a. Since the rim of the pulley has the same accleration as the string,
the angular acceleration is:
 = a/r
= 3.0 / 0.20
 = 15rads-2
b. The change in angular velocity in 5.0 seconds is:
 = t
= 15 x 5.0
 = 75rads-1
[NB pulley started from rest f = ]
COMPLETE EXERCISES
PAGE 62 - 63
RUTTER
ROTATIONAL KINEMATIC EQUATIONS:
As we saw with translational motion graphs can be
drawn which resulted in our equations of motion. The
same graphs can be drawn for rotational motion.
Gradient = 

Gradient = 


Area = 
t
Constant 
t
t
Constant 
Constant 
Equations involve 5 variables:
Angular acceleration, 
Final angular velocity, f
Initial angular velocity, i
Angular displacement, 
Time taken, t.
EQUATIONS ARE:
f = i + t
 = it + ½t2
f2 = i2 + 2
 = ½ (f + i)t
Example 4:
A centrifuge can accelerate from rest at a constant angular
acceleration of 7.0rads-2, taking 3 minutes to reach top speed.
a. What is the final angular velocity.
b. What angle does it turn through during this time?
SOLUTION:
Info given i = 0;  = 7.0rads-2 & t = 3 minutes = 180s
a. Final angular velocity f :
f = i + t
= 0 + 7.0 x 180
f = 1300 rads-1
b. Angular dispalcement :
 = it + ½t2
 = 110000 rad
= 0 + ½ x 7.0 1802
COMPLETE EXERCISES
PAGE 64 - 65
RUTTER
FORCE & TORQUE
Torque, ‘’ is a turning force applied on an object pivoting
about the CoM. Force is applied perpendicular to pivot
point. Thus:
 = Fd
Unit = Nm
The effect of applying a torque causes rotational motion.
In pure rotation the torque is caused by a force couple,
e.g. on an axle where one force acts as a reaction force
through centre of rotation (CoR) and the other acts as a
torque force a certain distance away.
F = reaction force
r
F = torque force
Example 5:
On the rear wheel of a mountain bike the chain varies position from
one gear to another. At first the chain is on the largest gear wheel
(radius 7.0cm) and pulls with a force of 200N. The rider then changes
gear and the chain is shifted to the smallest gear wheel (radius
3.0cm).
a. Calculate the torque the chain applies to the rear wheel when it is
on its largest wheel.
b. Calculate the force needed to apply the same torque when the
chain is shifted to the smallest wheel.
SOLUTION:
a.  = Fr
b. F = /r
= 200 x 0.070m
= 14 / 0.030
= 14Nm
= 470N
Turning effects can be
caused by large forces
on small wheels or
small forces on large
wheels.
READ
PAGE 66
RUTTER
ROTATIONAL INERTIA
Unbalanced forces acting on an object causes the object
to accelerate. Newton's second law. What determines
how likely it is for an object to move is the mass.
Increase the mass the force has less effect on
accelerating it. The mass resists the force. We call this
linear inertia.
Likewise this is true for rotational motion. In this case
the force applied is a torque which generates an angular
acceleration. Hence:
 = 
Where  = rotationl inertia, or the resistance an object
has to change its angular velocity. Unit = kgm2.
Rotational inertia depends upon:
a. Its mass -  mass,  .
b. How far from the CoR the mass is – further away  .
Use the above information to explain why tightrope
walkers carry a long bar?
A tightrope walker will often carry a long bar ‘for
balance’. If the tightrope walker falls, it will be because
he has rotated too far about his foot (the CoR).
Increasing his reluctance to rotate would be a great
advantage! Carrying the bar means that the system of
man and bar not only has more mass, but also has some
of its mass a long way from the CoR, thus increasing his
rotational inertia
Often a good way to find the rotational inertia of an object is
to suspend a mass by a string around the object. This allows
the torque and angular acceleration to be measured. From
this using  = ,  can be calculated.
Example 6:
A mass of 0.18kg is used to accelerate
a wheel of radius 0.20m. The mass
accelerates downwards at 1.2ms-2.
a. Calculate the tension force in the
string.
b. Calculate the torque on the wheel.
c. Calculate the angular acceleration of
the wheel.
d.Calculate the rotational inertia of the
wheel.
0.20m
1.2ms-2
0.18kg
SOLUTION:
a. Consider the forces acting on the mass. The
tension force in the string acts upwards &
the gravity acts down wards. The mass is
accelerating downwards, and so the
resultant unbalanced force must be
downwards, & is given by:
FU = Fg – FT
FT = Fg – FU
The gravity force Fg = mg and unbalanced
force is found from the acceleration of the
mass FU = ma
Fg = 0.18 x 9.8 = 1.764
FU = 0.18 x 1.2 = 0.216
FT = 1.764 – 0.216
FT = 1.5N
FT
0.18kg
FU
Fg
b. Consider the forces acting on the wheel. The only force acting to
turn the wheel is the tension force in the string, and so the torque
on the wheel is given by:
 = FT x r
= 1.548 x 0.20
 = 0.31Nm
c. a = r therefore  = a/r
= 1.2 / 0.20
 = 6.0 rads-2
d.  =  therefore  =  / 
= 0.3096 / 6.0
 = 0.052 kgm2
COMPLETE EXERCISE
PAGE 67 - 68
RUTTER
ROLLING MOTION
As with any form of motion energy must be involved and
obviously rotational motion is no exception. Place a
boulder up a hill and let it go. If you are coming up the
hill in the path of the boulder you will get squished!!!
Meeting the boulder at the bottom compared to halfway
up or at the top leads to an increase in squisheyness!!!
Why? – more energy is involved.
Energy when spinning is rotational kinetic energy, Ekr.
Measured in joules and transformed from one form into
another. Made up of similar components to Ek.
Ekr = ½2
When a rolling object is moving it has two forms of
energy:
1. Translational Ek = ½mv2 &
2. Rotational Ekr = ½2
These two combine to give the total Ek of a rolling
object.
Ekr = ½2
+
=
CoM Ek = ½mv2
Total
gravitational
potential energy
When rolling shapes down a slope not all shapes will
move the same. What is important is how the mass is
distributed. The further away the mass is from the centre
the more inertia it has, thus is harder to move. A hollow
cylinder is harder to move than a ball bearing of the
same mass. This is due to the cylinder having its mass
further away from the centre thus increasing its
rotational inertia.
Check out the practical on rolling objects.
For a body that is both translating and rotating:
Calculate the Ek using the velocity of the CoM.
Calculate the Ekr using the  about the CoM.
Add these two to obtain the total kinetic energy.
Example 7
In an experiment, a heavy roller,
M, is accelerated as a mass, m,
falls to the ground. A timer is
used to record the speed of the
roller after it has moved distance
h.
Sample results are:

v
M
timer
m
Mass of roller, M = 0.70kg
Hanging mass, m =0.10kg
h
m
Height h = 0.60m
Final speed of roller v = 1.0ms-1
a. Calculate the rotational kinetic
energy of the roller when its
speed is 1.0ms-1.
The roller has a radius of 2.0cm.
b. Calculate the rotational inertia
of the roller.
SOLUTION:
The gravitational potential energy lost by the hanging mass is:
Ep = mgh
= 0.10 x 9.8 x 0.60
Ep = 0.588J
This gravitational potential energy is changed into linear kinetic
energy of the mass and of the roller plus rotational kinetic energy
of the roller; Ep = Ek + Ekr.
The linear kinetic energy gained by the roller and hanging mass is:
Ek = ½(M + m)v2
= ½ x (0.70 + 0.10) x 1.02
Ek = 0.40J
The rotational kinetic energy of the roller is:
Ekr = Ep – Ek
Ekr = 0.19J
= 0.588 – 0.40
b. When the roller is moving forward at a speed of 1.0ms-1, it is also
rotating with an angular speed, , given by:
=v/r
= 1.0 / 0.020m
 = 50rads-1
The rotational inertia of the roller can be calculated from:
Ekr = ½2
 = 2 x 0.188 / 502
 = 1.5 x 10-4kgm2
0.188 = ½ x  x 502
COMPLETE EXERCISES
PAGE 69 - 72
RUTTER
ROTATIONAL ENERGY & MOMENTUM PRACTICAL
AIM: To investigate rolling different shapes
down a slope.
h
If you drop different masses through the
same height, they all take the same time to
fall. But, when you roll them down a slope
they take different times.
Obtain the rolling cylinder set. They all
have the same radius and the same mass.
But will they take the same time to roll
down a slope?
Set up a ramp. Time each cylinder a few
times down the slope. Record the average
time, t, for each one.
Measure and record the mass, m and the
radius, r. Measure and record the two
distances, h and d.
d
h
Prove using equations of
motion that vf = 2d/t
Using equations of motion to derive final velocity:
vf2 = vi2 + 2ad
But vi = 0
vf2 = 2ad
Substituting an equation of motion in for ‘a’
d = vit + ½at2
But vi = 0
d = ½at2
a = 2d/t2
Substituting in ‘a’ into vf2 = 2ad
vf2 = 2 x 2d/t2 x d
vf2 = 4d2/t2
vf = 2d/t
[ both sides]
RESULTS:
Trial
Wood cylinder
Brass cylinder
1
2
3
4
5
Average Time
h = _______ m
d = _______ m
Use your slowest object for these calculations:
• Calculate the cylinders final speed, v, using v = 2d/t, which is
the final speed of the cylinder.
• Calculate its final linear kinetic energy, Ek(linear)
• Calculate the initial gravitational potential energy, Ep(grav)
• Hence find the final rotational kinetic energy, Ek(rot)
• Use v = r to calculate the cylinders final angular speed, 
• Use your value for the rotational kinetic energy to calculate
the rotational inertia, I, of the cylinder.
• Explain why the cylinders take different times. Why would a
small ball bearing be the fastest?
ROTATIONAL ENERGY
As work is done on an object that is rotating the energy put in is
converted into rotational kinetic energy. We know that
EK = ½mv2
but this is only translational energy it also has rotational energy
EROT = ½I2
Note: these two energies can be added together for an object that is
moving translationally and rotationally.
ETOT = EK + EROT = ½mv2 + ½I2
This basically obeys the law of energy of conservation
COMPLETE EXERCISES
PAGE 78 - 80
RUTTER
ANGULAR MOMENTUM
Just like all linear motion has speed, acceleration and momentum so
does all rotational motion. We call rotational momentum angular
momentum as it is mass moving in a circular motion. Thus angular
momentum (L) is made up of a mass (m) and angular velocity ().
However instead of mass we now talk about rotational inertia ().
Hence we get the formula:
L = 
Units = kgm2rads-1
Just like angular velocity and acceleration, angular momentum can
also be linked to linear momentum. This would occur when an
object having linear momentum moves into circular motion e.g. like
a child running onto a roundabout. Hence:
L = mvr
This obeys the rules that all momentum must be conserved.
The law of energy of conservation states that:
a) The total angular momentum before the collision equals
b) The total angular momentum after the collision.
Just like we learnt at level 2.
Try the activity on Page 82 RUTTER
Example 8:
A turntable (1 = 0.090kgm2) spins freely with an angular
velocity I = 4.0 rads-1. A disc (2 = 0.030kgm2) is dropped
onto the turntable. Calculate the new angular velocity f.
SOLUTION:
Since the turntable is spinning freely, the total angular momentum is
conserved
1I + 2 x 0 = (1 + 2)f
[1 + 2 is the rotational inertia of turntable + disc]
0.090 x 4.0 = (0.090 + 0.030) x f
f = 3.0 rads-1
Example 9:
A spacecraft is stationary in space. In oredr to turn the spacecraft,
two small rockets are fired for a fraction of a second. The spacecraft
is a cylinder with a radius of 2.0m and a rotational inertia, , of
1600kgm2.
The small rockets each fire 0.40Kg of gas at a speed of 100ms-1.
Calculate the angular speed of the spacecraft.
SOLUTION:
The linear momentum, , of the gas from the rocket is given by:
 = mv
= 0.40 x 100
 = 40kgms-1
The angular momentum, L, of the gas from each rocket is:
L = mvr
L = 80kgm2s-1
= 0.40 x 100 x 2.0
The two rockets produce a total angular momentum of 160Kgm2s-1
clockwise about the spacecraft.
Since angular momentum is conserved, the spacecraft will rotate
anticlockwise qith an equal angular momentum of 160kgm2s-1.
The angular speed of the spacecraft is:
=L/
= 160 / 1600
 = 0.10rads-1
To stop the spacecraft rotating would require an equal burst
of gas from the rockets in the opposite direction to the
original.
Try to explain the applications of conservation of
rotational angular momentum. This may include:
 Helicopter rotors
 Ice-skaters spinning
 Bicycle wheels.
COMPLETE EXERCISES
PAGE 81 - 86
RUTTER