Download Work-Kinetic Energy Theorem for Rotational Motion

PHYSICS SEMESTER ONE
UNIT 7
UNIT 7: ROTATION AND EQUILIBRIUM
Relationship to Labs: This unit is supported by Lab 6: Torque and Rotational Equilibrium (kit-based lab).
So far, we have covered motion of objects moving in a straight line, changing direction or going around a
curve. Now, imagine trying to apply the linear motion analysis to a spinning object. All of the parts of
the spinning object are moving in different directions, but the object itself may be stationary. We need
a new analysis system to cover items that are spinning about an axis. Rotation is the motion of an
object about a fixed axis. Rotation covers everything from spinning wheels and planets, to the use of a
wrench on a bolt. When we get to the equilibrium portion of this section, we will see that we can use
the linear and rotation properties to solve complicated problems where nothing is rotating.
Rotation has several terms that have similar properties to terms in linear motion. Distance gets
replaced with angle, speed with angular speed, acceleration with angular acceleration, mass with
moment of inertia, force with torque, kinetic energy with rotational kinetic energy, and momentum with
angular momentum. The relationships between the rotational terms are identical to the relationships
between the linear motion terms. Furthermore, we can often convert linear motion expressions to
rotational motion expressions; though expressions in one system are often simpler than the other.
The axis of rotation is the point about which all other points rotate. A
point, distance r from an axis of rotation, will move through an arc of
length s. The arc length for the motion of the point rotating through
s
the angle θ (in radians) is
s  r
Imagine an object that moves in a complete circle. The object rotates
through the angle of
  2
θ
r
The arc length around the circle is the circumference
sC
 2 r
 r
This works for all angles. We can rearrange this to get
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PHYSICS SEMESTER ONE
UNIT 7

s
r
This shows that the angle is the distance (arc length) divided by the radius. The angle (in radians) is
considered dimensionless (unit-free) because it is a ratio of two lengths.
We will now define the angular speed ω as
ave 

, average angular speed
t
or
d
, the instantaneous angular speed
dt

Using our definition for the arc length (distance), the instantaneous speed is
v

ds
dt
dr
dt
r
d
dt
 r

v
r
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= 4.0 s-1
object is 2.0 m/s. Find the angular speed of the point moving around the axis.
= (2.0 m/s)/(0.50 m)
The distance from a point to its axis of rotation is 0.50 m and the speed of the
 = v/r
Quick Question
equation:
the units s-1, or radians per second.
r = 0.50 m, v = 2.0 m/s
quotient formed by the speed divided by the radius. The angular speed has
terms:
As in the relation between angle and distance, the angular speed is the
The point moves through 4.0 radians per second.
or
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PHYSICS SEMESTER ONE
UNIT 7
Units
Rotation rates are often expressed in cycles per second (hertz, Hz) or revolutions per minute (rpm). A
cycle is equivalent to a revolution. In one complete revolution, an object rotates through 2π radians.
Similarly, in rotating through 1 radian an object moves through 1/2π.
To convert from radians to revolutions or cycles:
multiply by 1/2π
To convert from revolutions or cycles to radians:
multiply by 2π
To convert from 7.00 rpm to radians per seconds
7.00 rpm  7.00
revolutions  2 radians   1 min 



min
 revolution   60 s 
 0.733 s 1
 0.733 radians per second
To convert from 2.0106 Hz to radians per seconds
2.0  106 Hz  2.0  106
 2.0  106
cycles
s
cycles  2 radians 


s  1 cycle 
 1.3  107 s 1
 1.3  107 radians per second
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PHYSICS SEMESTER ONE
UNIT 7
MOMENT OF INERTIA
Moment of inertia in rotating systems is the equivalent to mass in the linear motion system. We can
think of mass as a resistance to motion (the heavier the mass the lower the acceleration for the same
force). The moment of inertia is a resistance to rotation.
K  12 mv 2 for the kinetic energy of a mass m moving with speed v. The kinetic
We have the formula
energy of a point, with mass m, rotating with angular speed  at a radius of r about an axis is
K  12 mv 2
 12 m  r 
2
 12 m 2 r 2
If we have several points, the rotational kinetic energy is the sum of the kinetic energies of the individual
points. We will assume that the object is rigid, so all of the points have the same angular speed, .
K R   Ki
i
  12 mi ri2 2
i
1
2
and  are constant so remove from sum


 12   mi ri2   2


 i

We can define the term in brackets as the objects moment of inertia, I.
I   mi ri 2
i
The units of angular momentum are kg·m2. The fact that we sum over all of the masses illustrates that
the moment of inertia of an object is the sum of the moments of inertia of its parts.
I   Ii
i
The rotational kinetic energy then becomes
K R  12 I 2
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This has the same form as the linear kinetic energy where the moment of inertia replaces the mass and
the angular speed replaces the linear speed. The units of rotational kinetic energy are joules, J.
Problem 1
Two balls, both with a mass of 3.5 kg are connected by a mass-less rigid rod. The centre of mass of the
system is as the axis of rotation. The balls are 1.0 m from the centre of mass and rotate about it with an
angular speed of 12 radians per second.
a)
Find the moment of inertia of the system.
b)
Find the rotational kinetic energy of the system.
Problem 2
Three 3.6 kg balls are equally space about a circle of radius of 1.0 m forming an equilateral triangle. Find
the moment of inertial of the system if
a)
all of the balls rotate at a radius of 1.0 m about the centre of the
system.
b)
1.0 m
the balls rotate about the bisector of one of the triangle angles.
0.866 m
In the second problem, the masses and radii for each are the same for parts a) and b) but the moments
of inertia are different. The second problem was chosen to illustrate the fact that the moment of
inertia of a system depends on the axis of rotation.
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PHYSICS SEMESTER ONE
UNIT 7
We can calculate the moment of inertia of any shape by finding the sum of all of the products between
mass and the square of the radius for all of the little pieces.
I   mi ri 2
i
or, using calculus,
I   r 2 dm
(FYI only, you are not expected to do integration in this course)
For objects with constant density, we can write the derivative of the mass in terms of the density ρ and
the volume V.
dm

dV
or
dm  dV
so
I   r 2 dV
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PHYSICS SEMESTER ONE
UNIT 7
The following table shows moments of inertia for different objects. If you go into mechanical
engineering, you will have a book with many more of these. M is the total mass of the object. Problems
on assignments and exams will involve shapes or combinations of shapes from this table.
1. hoop or thin
2. solid sphere
R
cylinder shell
I  MR
I  52 MR 2
R
2
3. solid cylinder or disk
4. thin spherical shell
R
I  12 MR2
I  23 MR 2
R
5. long thin rod rotating
6. long thin rod
about centre
rotating about end
I  121 ML2
I  13 ML2
L
7. rectangular plate rotating
8. rectangular plate
about centre

I  121 M L2  W 2
L
rotation about edge

L

I  13 M L2  W 2
W
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
L
W
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PHYSICS SEMESTER ONE
UNIT 7
Problem 3
For a long thin rod with total mass of 5.0 kg and length 1.2 m, find the moment of inertia
a)
about an axis of rotation in the centre with the rod perpendicular to the axis,
b)
about an axis of rotation in the end with the rod perpendicular to the axis.
c)
Find the moment of inertia of the rod in part b) if there is a ball attached to the rod 1.0 m from
the axis. The mass of the ball is 2.0 kg.
Part c) of problem 3 illustrates the point that the moment of inertia of an object is the sum of the
moments of inertia of the parts.
Parallel Axis Theorem
With a bit of work, we can show that moment of inertia of any object about an axis through its center of
mass is the minimum moment of inertia for any axis in that direction (check text). The moments for
shapes 1 through 5 have the axis through the centre of mass of the shape.
The parallel axis theorem enables us to expand on the moments of inertia for the shapes noted in the
table in the last page. The parallel axis theorem states that the moment of inertia about any axis
parallel to that axis through the center of mass is given by
I parallel axis  ICM  Md 2
where d is the distance from the axis through the centre of mass to the parallel axis.
axis through
centre of mass
parallel axis
total mass, M
centre of mass
d
There are several physics analyses where we can treat the distributed mass of some object as if all of the
mass were concentrated on the centre of mass. This leads to jokes about how physicists can
approximate a cow as a single point. The parallel axis theorem indicates that we can account for a shift
of an axis parallel to one through the centre of mass by adding the moment of inertia of the entire mass
located at the centre of mass to the moment of inertial about the centre of mass axis. The moment of
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PHYSICS SEMESTER ONE
UNIT 7
inertia about the centre of mass axis accounts for the shape of the object, while the moment of inertia
with the concentrated mass at the centre of mass accounts for the location of the actual axis relative to
the one through the centre of mass.
Problem 4
a)
Use the parallel axis theorem to verify that the moment of inertia of a long thin rod rotating
about its end is I  1 ML2 given that the moment of inertia for the rod rotating about its centre is
3
1 ML2 .
I  12
b)
Show that the moment of inertia of a tire (thin walled cylinder) rotating about its edge is
I  2ML2 where L is the radius of the tire.
TORQUE
We have drawn on the similarities between mass and moment of inertia, position and angle, speed and
angular speed, and rotational kinetic energy and linear kinetic energy. In rotational systems, we will also
talk about the angular acceleration and the torque. Angular acceleration and torque are similar to linear
acceleration and force, respectively.
Angular acceleration  is the rate of change of the angular speed.


d
dt
d 2
dt 2
with the average angular acceleration defined as
 ave 

t
Both of these have the same form as the linear acceleration. The units of angular acceleration are s-2 or
radians per s–2 (radians are considered unitless because they are the ratio of two lengths, the
circumference and the diameter). Like the angle and angular velocity the angular acceleration is related
to its linear counterpart using

a
r
and
a  r
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PHYSICS SEMESTER ONE
The relationship between the net torque  net , moment of inertia and angular acceleration is
UNIT 7
 net  I
which has the same form as the scalar form of the force in Newton’s second law.
Fnet  ma
Consider the following situation. A force is applied to a wrench, which transfers the force to a nut.
F
r
Fsinθ
θ
θ
Fcosθ
d
The component of the force that is along the wrench, Fcosθ, does absolutely nothing to turn the nut. It
pushes or pulls the nut away from the bolt.
The component of the force perpendicular to the wrench, Fsinθ, is the only portion if the force that
turns the nut. With this in mind, the torque provided to the nut is the product of the applied force, the
distance to the axis of rotation and the sine of the angle between the force and the line joining the place
where the force is applied and the axis of rotation.
  rF sin 
This torque is the same as that created by the force F applied a distance d from the nut with the force
perpendicular to d
  Fd
where d  r sin  is the moment arm of F. The force is perpendicular to the moment arm.
The units of torque are newton-meters, Nm. Dimensionally, 1 Nm is the same as 1 J but we make the
distinction because they convey different ideas. In work, W  Fd , the force and distance are in the
same direction, while in torque,   Fd , the force and distances are perpendicular to each other.
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PHYSICS SEMESTER ONE
UNIT 7
Problem 5
1
A force of 5.0 N is applied to a nut with a moment arm of 0.50 m. A second force of 2.0 N is
applied to the nut with moment arm of 1.2 m. Find the total torque on the following system
a)
assuming the torques produced are in the same direction.
b)
assuming the torques produced are in the opposite direction.
2
Find the angular acceleration of the nut in the last question if the momentum of inertia is 4.0
kg·m2.
a)
torques in same direction
b)
torques in opposite direction
Example
The following example is a modified version of problem 8 in the dynamics section. In the dynamics
problem, the string passed over a frictionless surface so the tension was the same on both sides. In this
example the frictionless surface is replaced with a less-than-perfect pulley that has a moment of inertia.
The pulley will have some angular acceleration as it moves with the string (no slippage). The angular
acceleration is caused by a difference in the torques due to the different tensions on the two sides of
the pulley, T1 and T2 .
A loonie (mL = 6.99 g) and a dime (mD = 2.09 g) are attached to the ends of a mass-free chain. The chain
goes over a pulley (r = 0.10 m, I = 0.00010 kg·m2). Initially motionless, the coins and gear move when
released.
a) Find the acceleration of the coins.
Draw FBDs
T1
+y
+
T2
+y
r
T1
T2
Pulley
FgD
FgL
Dime
Loonie
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PHYSICS SEMESTER ONE
UNIT 7
The FBDs give us the following.
The tensions are tangent to the pulley, perpendicular to the radius to the pulley axis.
 net  I
T2 r  T1r  I
a
r
1
For the dime,
Fnet , D  mD a
 2
T1  mD g  mD a
For the loonie,
Fnet , L  mL a
 3
mL g  T2  mL a
Equations 2 and 3 can be rearranged to give
T1  mD a  mD g
 2'
T2  mL g  mL a
3'
and
Substituting these into equation 1 gives
T2 r  T1r  I
a
r
 mL g  mL a  r   mD g  mD a  r  I
a
r
Rearranging gives
mL gr  mL ar  mD gr  mD ar  I
mD ar  mL ar  I
a
a
r
a
 mL gr  mD gr
r
mL gr  mD gr
I
mD r  mL r 
r
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PHYSICS SEMESTER ONE
UNIT 7
Substituting values (make sure all values are expressed in meters, kilograms and seconds (MKS) or
combinations of MKS, like newtons or joules),
a
 0.00699  9.8 0.10    0.00209 9.8  0.10 
 0.00209  0.10    0.00699  0.10  
0.00010
0.10
 2.52 m/s 2
The acceleration of the system is 2.5 m/s2.
In the dynamics section problem 8, the acceleration was 5.3 m/s2. The addition of the pulley slows the
system down.
b)
Find the tensions T1 and T2.
Substituting into equation 2’ gives
T1  mD a  mD g



  0.00209 kg  2.52 m/s 2   0.00209 kg  9.8 m/s 2

 0.0257 N
Substituting into equation 3’ gives
T2  mL g  mL a



  0.00699 kg  9.8 m/s 2   0.00699 kg  2.52 m/s 2

 0.0509 N
The tensions are 0.026 N for the dime side and 0.051 N for the loonie side.
In the dynamics section problem 8, the tensions were both 0.032 N. The addition of the not-so-perfect
needed non-zero net torque to have acceleration.
Problem 6
A 240 kg crate sits on a horizontal surface. The coefficient of kinetic friction between the crate and floor
is 0.50. A rope runs horizontally from the crate, over a pulley, to a 175 kg hanging mass. The pulley has
a radius of 0.50 m and a moment of inertia of 32 kg·m2. Find the acceleration of the crate.
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PHYSICS SEMESTER ONE
UNIT 7
WORK AND ROTATIONAL KINETIC ENERGY
Consider the small amount work done in rotating an object such that a point P on the object moves by
the small arc length displacement ds .
F
ds
dθ
r
ϕ
P
O
The small amount of work dW is
dW  F  d s
 F sin  rd
Here F sin  is the tangential component of the force vector F.
Recalling the definition of torque, we get
dW   Fr sin   rd
  d
With a bit of work we can get
dW  Id
where the I is constant. Integrating, or adding up all of the little bits of work gives us
W 
f
i
I d
 12 I  2f  12 I i2
 K Rf  K Ri
(once again, you are not expected to use integration
on any exams for this course, but it is a good example
when you do see integration)
The work is equal to the change in rotational kinetic energy (yet another similarity with the linear
relations).
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PHYSICS SEMESTER ONE
UNIT 7
Note that the rotational kinetic energy has the same form as the translational kinetic energy.
K R  12 I 2 
1
2
 mass term  velocity term 2
Work-Kinetic Energy Theorem for Rotational Motion
The net work done by external forces in rotating a symmetric rigid object about a fixed axis equals the
change in the object’s rotation energy.
W  12 I 2f  12 Ii2  K Rf  K Ri
Conservation of energy is also valid for rotational systems
We can also show that W   which is the angular equivalent of W  Fd . You are not responsible
for W   , but it is yet another relationship that is very similar to a linear equation.
Example
A fly-wheel is a mechanical device used to store energy. If 2.0106 J of work is done to increase the
angular velocity of a fly-wheel from 56 s-1 to 134 s-1, calculate moment of inertia of the fly-wheel.
terms:
W  2.0 106 J, i  56 s1,  f  134 s 1
The work is equal to the change in kinetic energy
W  K f  Ki
 12 I  2f  12 I i2
W
I
1
2

1
2

2
f

 i2 I
W
 2f
 i2

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PHYSICS SEMESTER ONE
UNIT 7
Inserting values
I
2.0  106 J
134 s   56 s 

1
2
1
2
1
2


 269.9 kg·m 2
The moment of inertia of the fly-wheel is 270 kg·m2.
Problem 7
A uniform rod of length L = 1.0 m and mass M = 0.30 kg is free to rotate about a frictionless pin through
one end. Released from rest in the horizontal position, the rod accelerates due to gravity.
a)
What is the angular speed when the rod reaches its vertical position?
b)
Does the angular speed in part a) depend on the mass of the rod?
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PHYSICS SEMESTER ONE
UNIT 7
ROLLING BODIES
An object that is rolling has both linear and rotational kinetic energy.
The motion of a tire rolling along the ground can be represented by a combination of both rotational
and linear motion. CM is the centre of mass, and the radius of the tire is R.
translational motion
CM
vCM
vCM
vCM
rotational motion
v = Rω
CM
v = Rω
combination or translation and rotation
motion
v = vCM + Rω = 2vCM
CM
Here we have used the fact that
vCM
v = vCM – Rω = 0
rolling motion
v  R  vCM
Rolling motion is a combination of linear motion and rotation. The kinetic energy of a rolling object is a
combination of the linear kinetic energy and the rotational kinetic energy.
K  KR  KL
 12 I  2  12 mv 2
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PHYSICS SEMESTER ONE
UNIT 7
Example
We can use this to find the speed of a can rolling to the bottom of the ramp after starting from rest. The
height of the ramp is h. The mass, moment of inertia and radius of the can (thin shelled cylinder) are M,
and R.
h
We will assume we can use the thin shelled cylinder approximation to estimate the moment of inertia of
the shell.
I  MR2
Initial energy is
U gi  Mgh
The final kinetic energy is
K f  KR  KL
 12 I  2  12 Mv 2

1
2

1
2
 MR 
2
M R2 v2
R
2
2
v
2
1
   2 Mv
R
 12 Mv 2
 12 Mv 2  12 Mv 2
The rotational kinetic energy equals
the linear kinetic energy in this case!
 Mv 2
Employing the law of conservation of energy (no external force or friction), we can set the initial
gravitational energy to the final kinetic energy energies equal to each other.
K f  U gi
M v 2  M gh
v  gh
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PHYSICS SEMESTER ONE
UNIT 7
ROTATION DIRECTION
If we defined the rotation direction by the direction of the radius vector,
we find that the defined direction is constantly changing because the
object is rotating. Instead, we can define the direction along
the axis of rotation as the rotation direction. The nice thing about this
vector is that it doesn’t change as the other vectors rotate about it.
For rotation motion in the horizontal plane, we can define the rotation direction as being out of the
plane. Counter-clockwise in the horizontal plane is equivalent to up, and clockwise is equivalent to
down.
With this definition of rotation direction, we can create right-hand rule where the fingers (of the right
hand) curl in the motion of the particles in the rotation and the thumb (of the right hand) points in the
direction of rotation.
The direction notation is useful when looking at the vector products. We will look
at some here with torque and angular momentum. In these cases, we can often
sketch the situation to determine the direction of the final answer. In the second
semester physics courses, the magnetic field interactions are more difficult to
imagine so direction definitions are more useful.
Example
z
An object rotates with an angular speed of 2.5 s-1. If the object is rotating
counter-clockwise in the x-y plane, what is the direction of rotation
(besides counter-clockwise in the x-y plane).
y
In 2D, we usually draw the 3D x-y-z axes with the z-axis vertical and x-y
plane slightly tilted from horizontal (the regular x-y plane can be seen
when viewed from above). Applying our right hand rule we see that the
x
rotation is in the z-direction, 2.5 s-1. The angular velocity is ω  2.5 s1 kˆ .
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PHYSICS SEMESTER ONE
UNIT 7
VECTOR PRODUCT AND TORQUE
Now that we have a convention for the direction of rotation, we can define our torque, angle, angular
velocity, angular acceleration, and angular momentum (later) as vectors. We will look primarily at
torque.
When we looked at energy, we had the “scalar product” or “dot product” of the force and displacement
in the definition of work
W  F d
 Fx d x  Fy d y  Fz d z
 Fd cos 
where  is the angle between the force and displacement. The product, work, is a scalar.
There is another product of vectors. This one produces a vector result and is known as the vector
product. The vector product (also called the “cross product”) is defined as follows







A  B  Ay Bz  Az By ˆi   Az Bx  Ax Bz  ˆj  Ax By  Ay Bx kˆ
This has some interesting properties

B  A  B y Az  Bz Ay ˆi   Bz Ax  Bx Az  ˆj  Bx Ay  B y Ax kˆ




  Ay Bz  Az B y ˆi   Az Bx  Ax Bz  ˆj  Ax B y  Ay Bx kˆ
 A  B
From here, we can get
A  A  A  A
( A  A , the colour is used to show the position swap)
The only value that can equal it negative is 0 so
A A  0
We can also use what is known as the determinant of a matrix to find the cross product. A matrix is a
rectangular array of numbers or variables. Examples of a 2 by 2 matrix K, a 3 by 3 matrix L, and a 3 by 2
matrix M are shown below.
1 2 
K
,
 4 3
a c
L   d e
 g h
b
f  ,
i 
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 j k
M   5 6 
8 7 
20
PHYSICS SEMESTER ONE
UNIT 7
For the cross product using the determinant (symbolized by replacing [ ] with | |), place the direction
vectors in the top row, the first vector in the middle row, and the second vector in the bottom row.
ˆi
ˆj
kˆ
AB  Ax
Ay
Az
Bx
By
Bz
To find the î components (x-direction):
1.
Block out the row and column containing î .
ˆi
2.
ˆj
kˆ
AB  Ax
Ay
Az
Bx
By
Bz
Match the unblocked terms along the diagonals.
ˆi
3.
ˆj
kˆ
AB  Ax
Ay
Az
Bx
By
Bz
Ay Bz
and
Az B y
The positive product is the one with the second row term to the right of the blocked off
column.
If there is no column to the right, use the far left column.







A  B  Ay Bz  Az By ˆi   Az Bx  Ax Bz  ˆj  Ax By  Ay Bx kˆ
4.
Repeat for the other directions.
For the ĵ direction (y-direction)
ˆi
ˆj
kˆ
AB  Ax
Ay
Az
Bx
By
Bz

A  B  Ay Bz  Az By ˆi   Az Bx  Ax Bz  ˆj  Ax By  Ay Bx kˆ
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PHYSICS SEMESTER ONE
UNIT 7
For the k̂ direction (z-direction)
ˆi
ˆj
kˆ
AB  Ax
Ay
Az
Bx
By
Bz




A  B  Ay Bz  Az By ˆi   Az Bx  Ax Bz  ˆj  Ax By  Ay Bx kˆ
Right Hand Rule II
Taking the right hand, if we extend the fingers in the direction of the first vector (vector
A). Then curl them towards the second vector (vector B) (alternately, the line coming out
of the palm is in the direction of B). The direction of the extended thumb is in the
direction of the cross product A × B.
Example
The cross product of A  1 ˆi with B  1 ˆj ,




A  B  Ay Bz  Az By ˆi   Az Bx  Ax Bz  ˆj  Ax By  Ay Bx kˆ
  0  0  0  1 ˆi   0  0  1  0  ˆj  1  1  0  0  kˆ
 1 kˆ
If we have two vectors, we can always find a plane that has
both vectors A  1 ˆi with B  1 ˆj are both in the x-y plane.
A×B
B
The cross product is always perpendicular to the plane of the
two vectors in the cross product.
A
If we measure the angle between the two vectors in
the plane, the magnitude of the cross product is the
product of the magnitude of the two vectors times the
|A × B| = AB sin 
sine of the angle between them.
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B

Aφ
22
PHYSICS SEMESTER ONE
UNIT 7
For the torque vector, we have
τ  rF




 ry Fz  rz Fy ˆi   rz Fx  rx Fz  ˆj  rx Fy  ry Fx kˆ
For force and radius in two dimensions (in x-y plane), this simplifies greatly to
τ  rF


  rx Fy  ry Fx  kˆ


 ry  0  0  Fy ˆi   0  Fx  rx  0  ˆj  rx Fy  ry Fx kˆ
 rF sin  kˆ
Where  is the angle between the radius and force vectors, measured from the radius vector.
Examples
A force of F  8.66 N ˆi  5.00 N ˆj is applied to an object r  1.00 m ˆi  1.73 m ˆj from the axis of
rotation.
a)
Find the torque using the cross product.
τ  rF
 1.73  0  0  5.00  ˆi   0  8.66  1.00  0  ˆj  1.00  5.00  1.73  8.66  Nm kˆ
  5.00  15.00  Nm kˆ
 10.0 Nm kˆ
b)
Find the magnitude of the torque using the magnitudes of the force and radius, and the sine
of the angle from the radius vector to the force vector, -30°.
F  5.002  8.662  10.0 N ,
r  1.732  1.002  2.00 m
  rF sin 
  2.00 m 10.0 N  sin  30 
 10.0 Nm
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PHYSICS SEMESTER ONE
UNIT 7
Your choice of method for calculating the torque depends on the form of the original values. Example b)
is easier when you have the magnitude-direction form of the vectors, while example a) is easier with the
component form vectors.
Alternate Direction Rules (please, ignore if confusing)
Please note, if you insist on using your left hand, you can paint (draw, tattoo …) an arrow on your thumb
pointing towards your palm, to point in the cross product direction. Alternatively, you can use the
opposite of the direction indicated by the left thumb (not the arrow), or multiply the direction indicated
by the left thumb by -1. It is usually easier to just use your right hand.
If you imagine a clock showing 3:00, you can think of the fist vector as the minute hand pointing at the
12 (XII) and the second vector along the hour hand pointing at the 3 (III). The direction of the cross
product is into the face of the clock.
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PHYSICS SEMESTER ONE
UNIT 7
Problem 8
Find the torque for the following
a)
F  2.6 N ˆi  5.0 N ˆj, r  1.8 m ˆi  1.1 m ˆj
b)
F = 12.5 N, r = 3.4 m,   85 , magnitude only
c)
F = 12.5 N, r = 3.4 m,   5 , magnitude only
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PHYSICS SEMESTER ONE
UNIT 7
ANGULAR MOMENTUM
Going back to the affect of torque, we had the relationship
 net  I
Recalling the definition of the angular acceleration
 net  I
I
I


t
 f  i
t
I  f  I i
t
It is convenient to introduce the angular momentum L as the product of the moment of inertia and the
angular speed.
L  I
This is the counterpart to the linear momentum. For a given speed, an object with a large mass will
have a higher momentum than object with a small mass. For a given angular speed, an object with a
large moment of inertia will have a larger angular momentum than an object with a small moment of
inertia. The units of angular momentum are kg·m2/s.
We can rewrite the average net torque on an object as
 net , ave 

L f  Li
t
L
t
Using calculus, we can use this to find the instantaneous net torque as
 net 
dL
dt
The net torque acting on the system is equal to the time rate of change of the object’s angular
momentum.
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PHYSICS SEMESTER ONE
UNIT 7
As with linear momentum and energy, angular momentum is conserved under specific conditions.
Angular momentum is conserved when there is no net torque on the system. If
 net  0
then
L f  Li .
The classic demonstration of this is the spinning stool demonstration (a similar, less artistic, version
occurs in figure skating and dancing). A person sits on a stool with weights extended out from his/her
body. Using whatever method they have, they start rotating. Keeping the feet off the ground, the
person will continue to rotate and a fairly constant speed (slowing a bit due to friction). By pulling their
arms in their angular speed can be increased greatly. Alternately, a person with a high rotation can slow
by spreading their arms and legs out.
Once started, the person is rotating without external torques. Conservation of angular momentum tells
us that the angular momentum is constant.
If  net  0 then L f  Li or I f  f  I ii .
When the person is spread out they have a large moment of inertia. When the person tucks in, their
moment of inertia decreases. Their angular momentum must increase to keep the angular momentum
constant.
f 
Ii
i   1i
If
In figure skating, a skater exploits this property when they spin. Dancers can leap while spinning, then
extend their arms and appear to freeze in mid air before tucking in and speeding up as they land.
Problem 9
A skater, with a moment of inertia of 9.0 kg·m2, is rotating at 3.0 radians per second with arms
extended.
a)
Find the skater’s angular momentum.
b)
Find the skater’s angular speed if she pulls her arms and legs in, reducing her moment if inertia
to 3.2 kg·m2.
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PHYSICS SEMESTER ONE
UNIT 7
The vector form of the angular momentum is expressed by the equation
L  rp
Like the translational momentum, angular momentum is actually a vector, and like torque, we use the
cross product to find the angular momentum vector. Here we used the fact that
dL
 τ net
dt
 r  Fnet
 r
dp
dt
To be truly consistent with the product rule in calculus, we really need
dL
dp dr
 r
 p ,
dt
dt dt
but we will considering cases where r as constant.
Note that we don’t need an actual rotating object with this definition. Consider the following example:
A 920 kg car is driving along a straight highway with a velocity of v  32 m/s ˆi . Bob is located 15 m at
an angle perpendicular to the highway. The car a 100 m to Bob’s left. Find the moment of inertia of the
car relative to Bob (axis).
v  32 m/s ˆi
r  120 m ˆi  15 m ˆj
We can use the sketch to show that r   120 m  ˆi  15 m ˆj .
Noting that p  mv , the angular momentum is
L  rp




 120 m ˆi  15 m ˆj   920 kg  32 m/s ˆi 


 4.4  105 kg  m 2 /s kˆ
The angular momentum is 4.4  105 kg  m 2 /s kˆ or 4.4105 kg·m2/s in the downwards direction.
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PHYSICS SEMESTER ONE
UNIT 7
We are usually just interested in the magnitude. If we know the magnitudes of and angle between the
position and momentum vectors,
L  rp sin 
 rmv sin 
Example
A 920 kg car is driving along a straight highway 32 m/s. Bob is located 121 m from car at an angle of 7.1°
from the highway. Find the moment of inertia of the car relative to Bob (axis).
terms:
m  920 kg, v  32 m/s, r  121 m,   7.1, L  ?
equation:
L  rmv sin 
L  121 m  920 kg  32 m/s  sin 7.1
calculate:
 440300 kg
m2
s
The car has an angular momentum of 4.4105 kg·m2/s relative to Bob.
Like the moment arm with torque, we really only need to consider the radius perpendicular to the
momentum because
r sin   r
so
L  r p
 r mv
In the last example, the radius perpendicular to the direction of the car is
r  r sin 
 121 m  sin 7.1
 15.0 m
keep extra decimal place for intermediate term
The angular momentum is then
L  r mv
 15.0 m  920 kg  32 m/s 
 4.4  105 kg·m 2 /s
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PHYSICS SEMESTER ONE
UNIT 7
CENTRE OF GRAVITY
A few sections ago we had the centre of mass of an object defined as
xCM 
m x  m x
M
m
i i
i i
yCM 
i
m y  m y
M
m
i
i
i
i
i
We will consider a two dimensional object. The arguments can be easily extended to 3D.
Gravity acts on the all of the individual masses in a body. We can define a centre of gravity xCG using the
following equation for the total torque acting on the object due to gravity.
 net , g  gMxCG
Gravitational forces are parallel to the y-axis so only the x-components of each position vector
contribute the net torque. The torque for an the ith individual mass is
xi
τ i  ri  Fi
 ri Fi sin i
yi
 xi mi g
ri
 Fgi = mig

The net torque on the object is the sum of all of the gravitational forces on all of the particles
 net , g   i
  mi xi g
If the gravitational field is constant within the object (good assumption for most things under 30 km in
diameter), the net gravitational torque simplifies to
 net , g  g  mi xi
 gM
 mi xi
M
 gMxCM
When the gravitational field is constant over the entire object, the centre of gravity is equal to the
centre of mass.
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PHYSICS SEMESTER ONE
UNIT 7
Problems 10
A torque of 13 Nm is applied to an object for 4.2 seconds, increasing its angular speed from 1.1 radians
per second to 3.4 radians per second. Calculate the moment of inertia of the object.
You can now revisit the PHeT torque and rotational motion applet to look at the relationships between
the angular acceleration, torque and moment of inertia for the lady bug merry-go-round,
http://phet.colorado.edu/new/simulations/sims.php?sim=Torque.
In the torque section, you can vary the applied force and radius of force (moment arm) to examine their
effects on the torque. Unfortunately, there are no examples where you can vary the angle between the
force and radius.
In the moment of inertia section, get the disk rotating by click-dragging the mouse on it. You can then
vary the moment of inertia by altering the inner and outer diameters of the disk. Notice how the
angular speed speeds up/slows down when the moment of inertia is reduced/increased. For some
reason, the angular acceleration does not register this change in angular speed.
In the angular momentum section, check out how the angular moment stays constant while the angular
speed changes when you alter the moment of inertia.
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PHYSICS SEMESTER ONE
UNIT 7
STATIC EQUILIBRIUM
static – lacking change
equilibrium – state in which opposing forces & torques are balanced
A system is in static equilibrium if the net external force and net external torque are zero.
Fnet  0
τ net  0
These sums combine with Newton’s Laws to give us the conditions where the linear acceleration and
angular acceleration must be zero.
a0
 0
The direction for the angular acceleration is not critical for this course. We can often define directions
with respect to the pivot points in a sketch. Torques that would cause rotations in one direction are
positive, the others are negative. In equilibrium questions, the sum of the torques on an object is zero
so there is really no rotation anyway.
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PHYSICS SEMESTER ONE
UNIT 7
Example
Dr. Evil has devised an “escape-proof” plan for eliminating Anne and Bob. Anne, mass 61 kg, is dropped
on one end of a 9.0 m see-saw balanced above a pool of killer goldfish. Bob, mass 79 kg, is dropped on
the other side 2.3 m from the axis of the see-saw.
a)
How far must Anne be from the axis to ensure that she and Bob are in static equilibrium, with
neither dipping into the pool.
mB = 79 kg
mA = 61 kg
?
2.3 m
9.0 m
Let’s draw the FBD for the see-saw.
mS-Sg
mAg
xA
FN
2.3 m
mBg
In this type of FBD, we account for the fact that the forces on the see-saw are at different locations. The
locations of the forces are included because they affect the torque on the see-saw.
Here, the unknown variables are mss , FN and x A . It turns out that we can ignore mss and FN if we
consider the see-saw pivot axis for our rotation (non-rotation). The net torque about the axis is
 net   Fi xi
 mB g  2.3 m   mA gxA  mss g  0   FN  0 
For static equilibrium
 net  0
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PHYSICS SEMESTER ONE
UNIT 7
Solving for xA,
0  mB g  2.3 m   mA gx A
mA g x A   mB g  2.3 m 
xA 

mB  2.3 m 
mA
  79 kg  2.3 m 
 61 kg 
 2.98 m
Anne is at -3.0 m, 3.0 m on the side of the see-saw opposite Bob.
b)
If Anne can reach the end of her side of the see-saw, she can reach a switch that locks the seesaw in place. If Bob moves toward his end at a rate of 1.0 m/s, how fast must Anne go in order
to ensure that the see-saw is balanced while they move to their ends?
Bob’s position, in m, at time t is
xB  2.3  1.0t
Anne’s position, in m, at time t is
xA  3.0  vAt
At each location, the positions of Anne and Bob must satisfy the equilibrium net torque
condition.
mB gxB  mA gxA  0
Canceling out gravity and inserting other terms, we have
 79  2.3  1.0t    61 2.98  vAt   0
181.7  79t  181.7  61v At  0
79t  61v At  0
Solving for Anne’s speed
vA 
79t
61t
 1.295 m/s
Anne must move at about 1.3 m/s away from Bob to ensure static equilibrium is maintained.
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PHYSICS SEMESTER ONE
UNIT 7
Equilibrium Problem Solving Strategy
1.
Draw a sketch.
Include coordinates and a convenient rotation axis.
The choice of the rotation axis is arbitrary, though some choices are
better than others. The choice is arbitrary because the object in
equilibrium is not rotating about any axis. Whichever axis you select,
the object is not rotating about it.
A convenient rotation axis is usually one that eliminates the most
unknown terms. In the last example, choosing the rotation axis at the
pivot for the see-saw allowed us to ignore the normal force and weight
of the see-saw.
2.
Draw a FBD.
Show all external forces and distances. If there is more than one
object, draw separate diagrams.
3.
Apply  net  0 .
This generates an equation relating the forces and their positions.
It helps to note the direction or expected direction of each torque on
the FBD. This way you can account for the torque signs within the
equation.
4.
Apply Fnet  0 .
Apply this for all directions, ignoring the positions of the forces.
This yields an equation linking the forces for each direction.
5.
Solve.
We have two or three equations. Solve the system of equations for
unknown variables.
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PHYSICS SEMESTER ONE
UNIT 7
Example
A sign hangs from a beam outside Bob’s Butcher shop. The beam is 2.5 m long and weighs 32 N with a
centre of gravity 1.0 m from the wall. The beam’s centre of mass is the centre of the beam. The 85 N
sign is suspended from a point 1.5 m from the wall. A cable extends from the end of the beam to the
wall above the cable. The cable makes an angle of 37° with the beam. Find the tension in the cable and
the force applied by the wall in the x- and y-directions.
Terms:
1.
xT  2.5 m, T  ?,   37, xB  1.0 m, FgB  32 N, xS  1.5 m,
FgS  85 N, FWx  ?, FWy  ?
Sketch
y
x
θ = 37°
Bob’s
axis of
rotation
2.
Fine Meats
FBD for the beam
T
Fw
37°
FgB
FgS
1.0 m
1.5 m
2.5 m
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PHYSICS SEMESTER ONE
UNIT 7
Here, we have three unknowns are the x and y-components of the wall force and the magnitude of the
tension. The arrow indicating the force from the wall is only a guess. You can get a rough idea of the
direction and size knowing that it must balance the other forces.
3.
In the sketch, the axis of rotation is the place where the wall meets the beam. This choice
means that the radius for the wall force (and the wall torque) is zero. The system is in static
equilibrium so the total torque on the beam gives us
 net  FWy  0   FgB xB  FgS xS  TxT sin T
0  0   32 N 1.0 m    85 N 1.5 m   T  2.5 m  sin 37
Solving for T
1.505 m  T  32 Nm  127.5 Nm
T
159.5 Nm
1.505 m
 106 N
The tension on the wire is 110 N.
This stage doesn’t always simplify to a single value. Sometimes the result is an equation
with two or three variables.
4.
The big trick here is to ignore all of the distances. As far as the Fnet  0 analysis is
concerned, the FBD for the beam is
T
Fw
Ty
37°
Fwy
Tx
Fwx
FgB
FgS
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PHYSICS SEMESTER ONE
UNIT 7
Looking at the forces in the x-direction, we have
Fnet , x  0
FWx  Tx  0
FWx  Tx
 T cos T
 106 N  cos 37
 84.67 N
In the y-direction
Fnet , y  0
FWy  FgB  FgS  Ty  0
Solving for FWy
FWy   FgB  FgS  Ty
   32 N    85 N   T sin T
 119 N  106 N  sin 37
 55.2 N
The tension is 110 N and the force from the wall is 85 N in the x-direction and 55 N in the ydirection.
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PHYSICS SEMESTER ONE
UNIT 7
Problem 11
A two dimensional tractor sits on level ground. The distance between the
front and back wheel is 2.7 m. The 1.0 x 104 N weight of the tractor has a
centre of mass 1.0 m from the back of the tractor. Calculate the weight
on each wheel (two variables).
Problem 12
A 3.0 m ladder leans against a wall, making an angle of 51° with the ground. Fiction between the ladder
and ground keeps the ladder from slipping. There is no friction between the ladder and the wall so the
wall force is perpendicular to the wall. Calculate the normal force on the ladder, the friction force and
the force from the wall.
Check out equilibrium problems in your textbook.
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PHYSICS SEMESTER ONE
UNIT 7
SUMMARY OF ROTATIONAL MOTION AND COMPARISON WITH LINEAR MOTION
Rotational Motion
Angle
Link
Linear Motion

Position, Distance, Arc Length x, s

s
r
s on curve
θ
r
  d dt
Angular Speed
Speed v  dx
  vr
Angular Acceleration
dt
v perpendicular to r
θ
r
  d dt
Acceleration a  dv
  ar
dt
a perpendicular to r
θ
r
If  is constant
If a is constant
 f  i   t
v f  vi  at
 f  i  it  12 t 2
x f  xi  vit  12 at 2
 2f  i2  2  f  i 
v2f  vi2  2ax f  xi 
Moment of Inertia I
Mass m
Torque   Fr sin   Fd , d is the moment arm
Force F
Net Torque  net  I
Work W 
f

Net Force Fnet  ma
 d or W  
Work W 
i
xf
x
Fdx or
i
W  F x  Fd
Rotational Kinetic Energy K R  1 I  2
2
Kinetic Energy K  1 mv 2
2
Power P  
Power P  Fv
Angular Momentum
L  I
Net Torque II  net  dL
dt
Creative Commons Attribution 3.0 Unported License
Momentum p  mv
Net Force II
Fnet  dp
dt
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PHYSICS SEMESTER ONE
UNIT 7
NANSLO Physics Core Units and Laboratory Experiments
by the North American Network of Science Labs Online,
a collaboration between WICHE, CCCS, and BCcampus
is licensed under a Creative Commons Attribution 3.0 Unported License;
based on a work at rwsl.nic.bc.ca.
Funded by a grant from EDUCAUSE through the Next Generation Learning Challenges.
Creative Commons Attribution 3.0 Unported License
41