Transcript Rotational dynamics

Advanced Higher Physics
Unit 1
Rotational Dynamics
Using moments
The spanner exert a moment or turning effect on the nut.
Turning point
distance from force to
turning point
force
If the moment is big enough, it will unscrew the nut.
If not, there are two ways of increasing the moment.
Using moments-increasing the moment
1. Increase the distance from the Force to the pivot-apply a force
at the end of the spanner or use a longer spanner.
Turning point
distance from
force to
turning point
force
If the same force is applied over a greater distance, a larger moment
is produced.
Using moments-increasing the moment
2. Increasing the Force applied-push/pull harder or get someone
stronger to do it!
Turning point
distance from
force to
turning point
force
If a greater force is applied over the same distance,
a larger moment is produced.
Moment of a force
The moment of a force is given by:
moment  Fd
F is the force applied, measured in Newton (N)
d is the distance from turning point, measured in metres (m)
The moment of a force is therefore measured in
Newton metres (Nm).
Torque
F
r
A force is applied to the rim of
a disc which can rotate around
its centre axis. In this case the
moment of a force is called the
Torque.
T  Fr
In data booklet
F is the force applied, measured in Newton (N)
r is the radius of the circle, measured in metres (m)
T is the Torque associated with force F,
measured in Newton metres (Nm).
Inertia
Inertia can be defined as resistance to change in motion.
In linear motion, MASS is a measure of an object’s inertia
(since a large mass needs a large force to produce an acceleration).
In angular motion, we use MOMENT OF INERTIA.
Moment of inertia
The moment of inertia of an object is its resistance to change
in angular motion.
The moment of inertia depends on:
•The mass of an object
•How the mass is distributed about the axis of rotation.
r
Consider a mass m at a distance r from
the axis of rotation. The moment of
inertia can be calculated using:
m
I  mr
2
In data booklet
(additional relationships)
m is the mass of the object, measured in kg
r is the distance from the axis of rotation, measured in m
I is the moment of inertia measured in kgm²
Moment of inertia : simple situation
All the mass can be considered to be at the same distance from
the axis.
Masses on a very light rod
r
r
Wheel with heavy rim and very light spokes
In these cases I=mr² where m is the total mass.
Moment of inertia: rods
rod about centre
1
I  ml 2
12
With:
l total length of the rod
m total mass of the rod
rod about end
1 2
I  ml
3
The moment of inertia for a rod rotating about end is 4 times bigger
than the moment of inertia for a rod rotating about centre as it is
harder to do so.
This is because there are now more particles at a greater distance
from the axis of rotation.
Moment of inertia: solid disc about centre
r
Where
1 2
I  mr
2
In data booklet
(additional relationships)
m is the total mass of the disc
r is the radius of the circle
Moment of inertia: Sphere about centre
2 2
I  mr
5
Where
m is the total mass of the sphere
r is the radius of the sphere
(all the moment of inertia formulas can be found
in the data booklet in Additional Relationships)
Newton 2nd Law
An unbalanced Torque will produce an angular acceleration.
T  I
With
In data booklet
I, the moment of inertia in kgm²
α, the angular acceleration in radsˉ²
T, the Torque in Nm
Angular momentum
The angular momentum is defined as the moment of the linear momentum.
r
m
w
v
For this particle of mass m:
The linear momentum p = mv
The angular momentum = the moment of p = mvr = mr²w, since v=rw.
L  mvr  mr 2 w
With
In data booklet
L angular momentum measured in kgm²sˉ¹.
Angular momentum of a rigid body
A rigid body is an object in which all the individual parts have the
same angular velocity w.
The angular momentum of this body is the total of the angular
momenta of its particles:
L   mr 2 w
w is constant as all particles must be rotating at the same rate.
L  w mr 2
L  Iw
In data booklet
Conservation of angular momentum
In the absence of external Torque, the total angular momentum
before impact equal the total angular momentum after impact.
I A w0 A  I B w0 B  I A wA  I B wB
Not in data booklet
Example: lump of mud stuck to a bike wheel
Before:
After:
Total angular momentum before = total angular momentum after
Iwheelw0 = w (Iwheel + Imud)
The moment of inertia of the wheel and the mud after impact is larger
than the moment of inertia of the wheel before impact.
Therefore the angular velocity of the wheel is smaller after impact.
Example: pupil spinning on a chair
Before:
Pupil pushes arms out
After:
Pupil draws arms in
Total angular momentum before = total angular momentum after
Ioutw1 = Iinw2
Iin is smaller than Iout because the particles are closer to the axis
of rotation.
Therefore w2 is larger than w1.
Example: mass dropped on a turntable
Before:
After:
Axis of
rotation
Total angular momentum before = total angular momentum after
Idiscw1 = w2(Idisc+Imass)
The moment of inertia of the disc and the mass after impact is larger
than the moment of inertia of the disc before impact.
Therefore the angular velocity of the disc is smaller after impact.
Rotational Kinetic energy
Erot
With
1 2
 Iw
2
In data booklet
I moment of inertia, measured in kgm²
w angular velocity, measured in radsˉ¹
Erot rotational kinetic energy, measured in J
Work done
E w  T
With
Not in data booklet
T, Torque measured in Nm
θ, angular displacement in rad
Ew, work done in J
Conservation of energy
In the absence of frictional torque:
Ew  Ek
1 2 1 2
T  Iw  Iw0
2
2
Not in data booklet
Objects rolling down an inclined plane
Ep = mgh
h
Erot = ½Iw²
w
Ek = ½mv²
v
Potential energy at top = total of linear and angular kinetic energy
mgh
= ½mv²
+
½Iw²
Comparison of linear and angular motion
The equations of angular motion are similar to those of linear motion.
linear motion
angular motion
dv
F  m  ma
dt
dw
T I
 I
dt
mAu A  mBuB  mAvA  mBvB
I A w0 A  I B w0 B  I A wA  I B wB
1 2 1
Fs  mv  mu 2
2
2
1 2 1 2
T  Iw  Iw0
2
2
Example 1:
10 kg
6m
  4rads
2
Frictional torque = 1000 Nm
time = 5s
Find:
a) Total I
b) Unbalanced T
c) Driving T
d) Driving F
e) Final w
f) Deceleration when driving T
removed
g) Time taken to move to rest.
Example 2:
r
F
Find:
a) Driving Torque
Solid disc
Axle
mass = 20 kg
b) Moment of inertia of solid
disc
Radius = 3m
c) Angular acceleration
radius = 1cm
Force applied to axle = 4000 N
Frictional Torque = 30 Nm
Cord length = 50 cm
d) Angular displacement
e) Angular velocity
Example 3:
A turntable of mass 5kg and radius 25 cm is rotating at 10 radsˉ¹.
A metal ring of mass 2 kg and radius 10 cm is dropped over the centre
of the turntable.
a) Find the new angular velocity of the system.
b) Using rotational energy determine whether this is an elastic or an
inelastic situation.