Chapter I

Rotational Dynamics and Equilibrium

Blinn College - Physics 2325 - Terry Honan

I.1 - The Vector Nature of Rotational Quantities

Torque about an Axis

F F⊥ = F sin θ

θ
Axis r

r⊥ = r sin θ

We define torque as the rotational analog of force. Suppose you are trying to loosen a bolt. The axis of rotation is the center of the bolt. If
you are unable to give sufficient torque with your hand you grab a wrench. Take r as the vector from the axis to where the force F is applied.
Clearly the important part of the force is the component of the force perpendicular to the radial vector r. Moreover the larger r is the larger the
torque. This motivates the definition of torque
τ = r F⊥

If θ is the angle between r and F then we can write F⊥ = F sin θ. Similarly we can r⊥ = r sin θ as the component of r perpendicular to F. This
gives us other ways of writing the torque.
τ = r F⊥ = r F sin θ = r⊥ F
The sign of torque depends on the sign convention for kinematics. If a force tends to make something rotate in the positive direction then
the torque is positive and similarly negative torques tend to make things rotate in the negative direction.

Example I.1 - Torques on a Disk

2 | Chapter I - Rotational Dynamics and Equilibrium

40 N
35 °

0.4 m m 50 °
0.4
0.2 m 0.3 m
axis 0.2 m
50 N

70 N
30 N

Four forces act on a disk with a 0.4-m radius.

(a) What is the torque of each force, taking counterclockwise as the positive sense of rotation?

Solution
The three equivalent formulas for torque are

τ = r F⊥ = r F sin θ = r⊥ F.
For clarity, we will label the torques by τ30 , τ50 , τ40 , and τ70 . For τ30 the r is the hypotenuse of the two sides given, but we do
not need its value. The part of r perpendicular to the force is r⊥ = 0.30 m. Since counterclockwise is positive this torque is
negative since, acting by itself, it will make the disk rotate clockwise.
τ30 = - r⊥ F = -(0.30 m) 30 N = - 9.00 Nm
The 50-N force is perpendicular to the radial vector. The 50 ° angle is irrelevant here. It is also a clockwise and thus a negative
torque.
τ50 = - r F⊥ = -(0.40 m) 50 N = - 20.00 Nm
The component of the 40-N force perpendicular to the radial vector is F⊥ = (80 N) cos 35 °. Alternatively, we can identify the
angle between the radial vector and the force is 90 ° - 35 ° = 55 °. This force make it rotate in the counterclockwise sense.
τ40 = + r F⊥ = +(0.40 m) (40 N cos 35 °) = 13.11 Nm
(or τ40 = + r F sin θ = +(0.40 m) (40 N) sin55 ° = 13.11 Nm)
The 70-N force is perpendicular to the radial vector and is counter-clockwise and thus a positive torque.

τ70 = r F⊥ = (0.20 m) 70 N = 14.00 Nm

(b) What is the net torque on the disk, where net torque is the sum of the torques?

Solution
τnet = τ30 + τ50 + τ70 + τ80 = - 1.89 Nm
The negative sign means that the net torque will cause a clockwise rotation.

(c) How would the answers to parts (a) and (b) be different if clockwise were chosen as the positive sense of rotation?

Solution
If clockwise were our positive sense of rotation then all torques would change signs.

Angular Velocity and Torque as Vectors

A rotation has a magnitude, the angle of rotation, and a direction, along the direction of the axis. Rotations are not vectors, though. Vector
addition is commutative but rotations are not. It turns out that infinitesimal rotations do commute and are vectors. We can write an infinitesimal

rotation as ⅆ θ. Since angular velocity about an axis requires only an infinitesimal rotation, ω = ⅆ θ / ⅆ t, we can define the angular velocity vector
 Chapter I - Rotational Dynamics and Equilibrium | 3


ⅆθ
ω= .
ⅆt
There are two possible directions along an axis. We decide which direction by using the right hand rule. Wrap the fingers of your right hand in
the direction of rotation. The thumb points in the direction of the vector. If angular velocity is a vector then we can also make angular accelera-
tion a vector.
ⅆ
α= ω
ⅆt
Wit these considerations we can now make a vector out of the torque. We can assign its direction to the sense of rotation due to that torque.
r and F are vectors; we will define the cross product so that the cross product of r and F is the torque τ.
τ = r ⨯ F

Interactive Figure

The Cross or Vector Product

The dot product, or scalar product, is a way of multiplying of two vectors that gives a scalar. The cross product, also known as the vector
product, is a multiplication that gives a vector.
A ⨯ B is a vector.

The magnitude of this vector is A B sin θ. We will specify the direction with a unit vector u. The two vectors A and B define a plane; their cross
product is perpendicular to that plane. There are two unit vectors perpendicular to any plane; we use the right hand rule to find the correct one.
Put your right thumb in the direction of the first entry A and your fingers in the direction of the second entry B. The palm of your hand is in the

direction u, giving the direction of the cross product.
4 | Chapter I - Rotational Dynamics and Equilibrium

B B⊥ = Bsin θ
out
of

A⊥ = Asin θ
A⨯B θ

 
To find the direction of the cross product A ⨯B , put the thumb of your right hand

in the direction of the first entry A and your fingers in the direction of the second

entry B . The palm of your right hand points in the direction of the cross product.
 
A ⨯ B = A B sin θ u (u by right hand rule)

Into

Out of

Figure: The convention we use to represent the third dimension relative to some
two-dimensional figure is to use a dot to represent "out of" and an × to represent "into".
A useful way to remember this is with an arrow; if it points at you it is a dot and away an ×.

Properties of the Cross Product

A ⨯ B = - B ⨯ A (antisymmetry - not commutative)
A ⨯ B ⨯ C ≠ A ⨯ B ⨯ C (not associative)

c A ⨯ B = c A ⨯ B = A ⨯ c B (associative w.r.t. scalar mult.)

A + B ⨯ C = A ⨯ C + B ⨯ C and
A ⨯ B + C = A ⨯ B + A ⨯ C (distributive)

The Cross Product and Components

With the dot product we were able to write it in terms of components. We can do the same for the cross product. As before we have to find
the products of the basis unit vectors. Because of the antisymmetry property we get A ⨯ A = 0. It follows then that
     
0 = x ⨯ x = y ⨯ y = z ⨯ z.
Using our definition of the cross product we can see that the cross product of two perpendicular unit vectors is a third unit vector perpendicular
to the two.
   
A ⨯ B = A B sin θ u ⟹ v ⨯ w = 1 · 1 · 1 u
   
The cross product of the unit vectors x and y is thus a unit vector perpendicular to the xy plane. This is either z or - z. We insist that our
coordinate system is right-handed; this means that
  
x ⨯ y = z.
For the other combinations of unit vectors there is a simple rule to keep track of their cross products. Arrange x, y and z around a circle.
 Chapter I - Rotational Dynamics and Equilibrium | 5

x
z y
If the order of the three coordinates has the same sense of rotation as x, y, z it gains a positive sign. If opposite it gets a minus sign.
     
y ⨯ z = x, z ⨯ x = y ,
        
y ⨯ x = - z, x ⨯ z = - y and z ⨯ y = - x
We can put all this together and get the cross product in terms of components.
     
A ⨯ B = ( A x x + A y y + A z z) ⨯ ( B x x + B y y + B z z)
  
= x ( A y Bz - Az B y ) + y ( Az B x - A x Bz ) + z ( A x B y - A y B x )
The determinant method is a common way to write this. A determinant is a mathematical operation that completely antisymmetrizes a
square matrix. On a 2 × 2 matrix we have:
a b
 =ad-bc.
c d
On a higher order matrix we write it as an alternating sum of determinants of lower order. We will consider only the case of the cross product.
  
x y z
 A y Az  A x Az  Ax Ay
A ⨯ B = A x A y Az = x -y +z
B y Bz B x Bz Bx By
B x B y Bz
  
= x ( A y Bz - Az B y ) - y ( A x Bz - Az B x ) + z ( A x B y - A y B x )

Note that there is not a discrepancy between the differing signs of the middle term for the y-component, since the factor multiplying y has terms
in reverse order.

Example I.2 - Dynamics of a Particle - Net Torque

A particle of mass m moves along the path given by

r(t) = a t2 , b t, - c t3  ,
where a, b and c are constants.

(a) What is the net torque about the origin that acts on the particle as a function of time? The net torque is the torque due to the net force.

Solution
The net force is found from the acceleration and the acceleration by differentiation.

v(t) = ⅆ r(t) = 2 a t, b, - 3 c t2 
ⅆt
ⅆ v (t)
a(t) = = 〈2 a, 0, - 6 c t〉
ⅆt
F net (t) = m a(t) = m 〈2 a, 0, - 6 c t〉
The torque about the origin is found by the cross product definition.

τnet (t) = r(t) ⨯ F net (t) = m a t2 , b t, - c t3  ⨯ 〈2 a, 0, - 6 c t〉

  
x y z
τnet = m a t2 b t - c t3 = m - 6 b c t2 - 0, -- 6 a c t3 + 2 a c t3 , 0 - 2 a b t
2 a 0 -6 c t
= m - 6 b c t2 , 4 a c t3 , - 2 a b t

Angular Momentum of a Particle and Torque

We previously defined the torque as

τ = r ⨯ F.
We can similarly define the angular momentum of a particle as
6 | Chapter I - Rotational Dynamics and Equilibrium

L = r ⨯ p.
Both of these expressions are relative to an origin; r is the position vector. It is from the origin to the position to where the force is applied in
the case of torque. It is from the origin to the position of the particle for the angular momentum.
The net torque on a particle (a point) is the torque due to the net force. If the particle is at position r then the net torque is τnet = r ⨯ F net . It is
straightforward to verify that the cross product satisfies the usually product rule for differentiation. Using this we can differentiate the angular
momentum for a particle. This gives:
ⅆ ⅆ  ⅆ
L= r ⨯ p + r ⨯ p = v ⨯ m v + r ⨯ F net = 0 + τnet .
ⅆt ⅆt ⅆt
This gives the analog of the momentum form of the second law F net = ⅆ p  ⅆ t. This is
ⅆ
τnet = L.
ⅆt

Example I.3 - Dynamics of a Particle (continued) - Angular Momentum

We extend the previous example. A particle of mass m moves along the path given by

r(t) = a t2 , b t, - c t3  ,
where a, b and c are constants.

(b) What is the angular momentum of the particle about the origin as a function of time?

Solution
ⅆ r(t)
v(t) = = 2 a t, b, - 3 c t2 
ⅆt
The angular momentum about the origin is found by the definition.

L(t) = r(t) ⨯ p(t) = r(t) ⨯ mv(t)

  
x y z
L(t) = m a t2 b t - c t3 = m - 3 b c t3 + b c t3 , -- 3 a c t4 + 2 a c t4 , a b t2 - 2 a b t2 
2 a t b - 3 c t2
= m - 2 b c t3 , a c t4 , - a b t2 

(c) For the results of parts (a) and (b) show that τnet = ⅆ L  ⅆ t.

Solution
ⅆ L(t)
= m - 6 b c t2 , 4 a c t3 , - 2 a b t
ⅆt

A System of Particles
 Chapter I - Rotational Dynamics and Equilibrium | 7

F 12
m1
ext
F 13 
F1 r 13
F 31
m3
r1   F3
ext
r 12 r3
F 32

r 23
r2 F 23
m2
ext
F 21 F2

Interactive Figure

In the preceding chapter we considered a three particle system with masses m1 , m2 and m3 at positions r1 , r2 and r3 . As before, we write the
ext
forces on m1 as a sum of internal forces F 21 and F 31 and external forces F 1 . The cross products of this with r1 gives the net torque. The
torques for m2 and m3 break up similarly.
ext ⅆ
τnet,1 = r1 ⨯ F 1 + r1 ⨯ F 21 + r1 ⨯ F 31 = L1
ⅆt
ext ⅆ
τnet,2 = r2 ⨯ F 2 + r2 ⨯ F 12 + r2 ⨯ F 32 = L2
ⅆt
ext ⅆ
τnet,3 = r3 ⨯ F 3 + r3 ⨯ F 13 + r3 ⨯ F 23 = L3
ⅆt
To concentrate on the bulk motion of our system we sum over these expressions. In the previous case with forces the internal forces canceled
due to Newton's third law. Here we need to make an additional assumption that the forces are central forces; this is that F 12 , the force of mass 2
on mass 1, is directed parallel (or antiparallel) to the line between the masses.
F 21 ∥ (r1 - r2 ) ⟺ (r1 - r2 ) ⨯ F 21 = 0
Now when we sum the net torques we get a cancellation of the internal torques. The internal torques cancel for all pairs of masses. The
cancellation between m1 and m2 follows from
r1 ⨯ F 21 + r2 ⨯ F 12 = r1 ⨯ F 21 + r2 ⨯ - F 21  = (r1 - r2 ) ⨯ F 21 = 0.
The other internal forces cancel similarly. We end up with
ⅆ
τext ext ext
1 + τ2 + τ3 =  L1 + L2 + L3  .
ⅆt
It should be clear how this could be generalized to four, or an arbitrary number, of particles. This gives the very fundamental result that for a
system of particles
ⅆ
τext
net = Ltot .
ⅆt

Conservation of Angular Momentum

The conservation of angular momentum follows from the expression above. If there are no external torques on a system then the total
angular momentum of the system is conserved.
ⅆ
τext
net = 0 ⟹ Ltot = 0 ⟹ Δ Ltot = 0
ⅆt
This derivation mirrors the conservation of linear momentum.

This is a very fundamental result. It has deep implications on the very large scale; in astrophysics it is crucial in the dynamics of planets,
stars, solar systems and galaxies. It is also important on the very small scale; in particle accelerators where elementary particles are collided and
created, angular momentum is always conserved.
8 | Chapter I - Rotational Dynamics and Equilibrium

This is a very fundamental result. It has deep implications on the very large scale; in astrophysics it is crucial in the dynamics of planets,
stars, solar systems and galaxies. It is also important on the very small scale; in particle accelerators where elementary particles are collided and
created, angular momentum is always conserved.

I.2 - More on Rigid Bodies

Axes and Origins

We began with a discussion of rigid bodies rotating about a fixed axis. Then we considered quantities like angular velocity, angular
acceleration, torque and angular momentum as vectors. How are the two points of view related? Torque and angular momentum vectors are
relative to an origin, where the position vector r is based at the origin. If the origin is chosen as some point on the axis then the vector relative to
the axis is just the component in the direction of the axis. The torque about some origin is the vector
τ = r ⨯ F.
The torque about the z axis is just the z component of this τz = τ where

τ = r F⊥ = r F sin θ = r⊥ F.

Example I.4 - Axis and Origin

z

Axis/Hinge

r⊥ = y
F

θ τz r =〈0,y,z〉
τ
θ
y
Origin

Consider a door in the yz-plane with the hinge as the axis in the z-direction. The origin is at the base of the door. A force of of
magnitude F pushes into the door, in the negative-x direction.

F = -F x
The force acts at a point given by the position vector r = 〈0, y, z〉 from the origin.

(a) What is the torque relative to the axis?

Solution
The torque relative to an axis is τ = r⊥ F = y F.

(b) What is the torque relative to the origin?

Solution
The torque relative to an origin is τ = r ⨯ F.
      
τ = r ⨯ F = (y y + z z) ⨯ (- F x) = - y F y ⨯ x - z F z ⨯ x
 
= y F z - z F y = 〈0, - z, y〉 F
(c) Show the the z-component of the torque about the origin is the torque about the axis.
 Chapter I - Rotational Dynamics and Equilibrium | 9

Solution
It follows that τz = y F which is the torque about the z-axis.

Similarly, the angular momentum of a particle relative to an origin

L = r ⨯ p
can be written relative to an axis. If the axis is the z axis then L about the axis is just the z component of L about the origin. Lz = L where

L = r p⊥ = r p sin θ = r⊥ p.

Angular Momentum of a Rigid Body

As before, we view our rigid body as a collection of point masses where the perpendicular distance form the axis to mi is ri . Since all the ri
are fixed we get the momentum related to the tangential velocity, which is then related to the angular velocity.
pi⊥ = mi vit = mi ri ω
The angular momentum of the ith mass becomes

Li = ri pi⊥ = ri mi vit = mi ri2 ω

The total angular momentum is the sum over all these terms L = ∑i Li . Using I = ∑i mi ri2 we get the angular momentum of a rotating
rigid body
L = I ω.
This is the result we had in our table relating rotations about a fixed axis to one dimensional linear motion.

Example I.5 - Angular Momentum of the Earth

The mass of the earth, the radius of the earth and the earth-sun distance are:

ME = 5.97 × 1024 kg , RE = 6.38 × 106 m and RES = 1.50 × 1011 m.

Here assume a circular orbit.

(a) Estimate the rotational angular momentum of the earth, assuming it is a uniform sphere?

Solution
2
I= ME R2E = 9.72 × 1037 kg m2
5
The angular velocity can be found from its rotational period of 1 day.
2π 2π 2π
ωrot = = = = 7.27 × 10-5 s-1
T 1day 24 × 3600 s
The estimated angular momentum can then be found.

L = I ωrot = 7.07 × 1033 kg m2  s

(b) Is the estimated result in part (a) too large or too small?

Solution
Because the earth is denser at its core the estimated moment is too large and thus the estimated angular momentum is too large.

(c) What is the orbital angular momentum of the earth as it orbits the sun?

Solution
Now we consider the angular momentum of a particle. First we find the speed from the orbital angular velocity
10 | Chapter I - Rotational Dynamics and Equilibrium

2π 2π 2π
ωorbit = = = = 1.992 × 10-7 s-1 .
T 1yr 365.24 × 24 × 3600 s
The tangential velocity gives the momentum, which is perpendicular to the radial vector.

v = vt = r ω ⟹ L = r p⊥ = r p = r m v = m r2 ω
L = ME R2ES ωorbit = 2.67 × 1040 kg m2  s
Note that an alternative solution can be found using L = I ω and I = ∑i mi ri2 = m r2 .

Example I.6 - The Rotating Figure Skater

A figure skater spins about a vertical axis. With her arms out she has a moment of inertia of Iout and rotates at ωout . When she brings
his arms in, her moment is smaller, Iin . The moments are about the vertical axis of rotation.
(a) What is ωin , her angular velocity with her arms in?

Solution
If the stool is frictionless then there is no net external torque acting, so angular momentum is conserved. The angular
momentum is L = I ω. It follows that
Iout
Lout = Lin ⟹ Iout ωout = Iin ωin ⟹ ωin = ωout
Iin
Since Iin < Iout it follows that he rotates faster ωin > ωout .

(b) Compare the kinetic energies Kin and Kout .

Solution
The kinetic energy is K = (1 / 2) I ω2 . Using L = I ω we can write K in terms of L and I; since L is conserved this is useful.

1 L2
K= I ω2 and L = I ω ⟹ K =
2 2I
Since Lin = Lout = L, it follows that Kin > Kout .

L2 L2
Iin < Iout ⟹ Kin = > = Kout
2 Iin 2 Iout
Where did the extra energy come from when she brings her arms in? View this from the perspective of the non-inertial rotating
frame where there is the false centrifugal force acting outward. To bring her arms in she must do work against the centrifugal
force; that is the source of the extra energy.

Example I.7 - Bullet Shot in Door

 Chapter I - Rotational Dynamics and Equilibrium | 11

A bullet of mass m is shot at speed v toward a door. The bullet’s velocity is perpendicular to the door and it hits the door at a distance d
from the door’s hinge. The door has mass M, height h and width w; assume that it swings without friction about the hinge. If the door is
initially at rest then what is its angular velocity after the bullet embeds in it.

Solution
Given there is no friction in the hinge, there is no external torque about the hinge and angular momentum is conserved.
Initially, there is no angular momentum in the door but the bullet does have angular momentum. We use the angular
momentum of a particle:
Li = r⊥ p = d m v.
After the bullet embeds in the door we have a rotating rigid body. The moment of inertia consists of the door’s moment added
to the bullet’s contribution. The door is the same as a rod of length w; its height is irrelevant.
1 1
Idoor = M L2 = M w2
3 3
The moment of inertia of the bullet after embedding comes from the moment for a discrete distribution.

Ibullet =  mi ri2 = m d2
i
The final angular momentum is L f = I f ω f where I f = Idoor + Ibullet . Conservation of angular momentum gives ω f .
1 dmv
Li = L f ⟹ d m v = M w2 + m d2 ω f ⟹ ω f =
3 1
3
M w2 + m d2

The Second Law

We can now, finally, derive the rotational equivalent of the second law τnet = I α. Start with the momentum form of the second law.
ⅆ
τext
net = Ltot
ⅆt
Now take the component along the axis of rotation. When the system is the rigid body then the net external torque on the rigid body is just the
torque on it. Similarly, the total angular momentum is just I ω the angular momentum of the body. We get
ⅆ
τnet = L.
ⅆt
Using L = I ω and α = ⅆ ω / ⅆ t we get our result.

τnet = I α
12 | Chapter I - Rotational Dynamics and Equilibrium

The Torque Due to Gravity

We saw earlier that to calculate the potential energy due to gravity we treat the object as if all the mass is at the center of mass. The same is
true for finding the torque due to gravity.
τgrav = rcm ⨯ M g
It is straightforward to verify this. Write the torque as the sum over the torques on all the masses in the body. Then use the definition of center
of mass to get the result.

τgrav =  ri ⨯ mi g =  mi ri ⨯ g = (M rcm ) ⨯ g = rcm ⨯ Mg

i i

Example I.8 - Swinging Disk

Axis
R
θ

Center

α=?

A uniform disk of radius R swings without friction about a perpendicular axis through its rim. What is its angular acceleration as it
swings through a position where the center is at an angle of θ from vertical, as shown?

Solution
We first need to draw a free-body diagram. When we draw free-body diagrams for torques we must draw the forces into the
diagram carefully showing where they act. Here, the only contact force is at the axis; this force gives zero torque, since r is
zero. The only torque comes from the weight mg, which acts at the center. The angle between the radial vector and the force is
θ, so we can use the τ = rF sin θ formula. We choose our sense of rotation, clockwise, as positive, so the torque is positive.
τnet = τgrav = rF sin θ = R mg sin θ
The moment of inertia can be found using the parallel-axis theorem.
1 3
I = Icm + m d2 = m R2 + m R2 = m R2
2 2
The rotational second law gives us the angular acceleration.
3 2g
τnet = I α ⟹ R mg sin θ = m R2 α ⟹ α = sin θ
2 3R

Example I.9 - Atwood’s Machine with a Massive Pulley

a=?
m2

m1

In chapter D we solved Atwood’s machine with an ideal pulley. Recall that an ideal pulley was frictionless and light, where light means
that the pulley’s mass is small compared to the other masses in the system. Now we will consider a pulley with mass; it will still be
frictionless. With an ideal pulley the tension on both sides is the same. Here, with a massive pulley the tensions on either side are
different. The different tensions are responsible for the angular acceleration of the pulley.
 Chapter I - Rotational Dynamics and Equilibrium | 13

In chapter D we solved Atwood’s machine with an ideal pulley. Recall that an ideal pulley was frictionless and light, where light means
that the pulley’s mass is small compared to the other masses in the system. Now we will consider a pulley with mass; it will still be
frictionless. With an ideal pulley the tension on both sides is the same. Here, with a massive pulley the tensions on either side are
different. The different tensions are responsible for the angular acceleration of the pulley.
m1 and m2 are two masses connected by a light string over a frictionless pulley as shown. The pulley is a uniform disk of mass M. Take
m1 < m2 . What is the downward acceleration of m2 ?

Solution
With our constrained system the motion of each mass is related. Let Δ x1 be the upward displacement of mass 1 and Δ x2 be the
upward displacement of 2. The assumption of tension is that the rope or string does not stretch, so these must be equal. We also
assume that the string does not slide on the pulley. This relates the rotational motion of the pulley to the linear motion of the
hanging masses; the arc length R Δθ, where R is the pulley’s radius, must equal the hanging masses displacements.
Δ x1 = Δ x2 = Δ x = R Δθ
Taking derivatives we can relate the velocities v1 = v2 = v = R ω and accelerations.

a1 = a2 = a = R α
Note that R was not given. We introduce it in our solution, so it must cancel.

We need to draw a free-body diagram for each mass and for the pulley. For the pulley we draw the free-body diagram into the
diagram, showing where the forces act. For the hanging masses this is the same as what we saw in Chapter D, except that the
tensions are now different.

Faxis
m1 m2

α
R R

M T2
T1

T1
T2
a a
Mg

m1 g

m2 g
m2 a

a m1

Applying the second law to the hanging masses gives a pair of equations. Here we choose the directinos of the accelerations as
positive.
Fnet,1 = m1 a1 ⟹ T1 - m1 g = m1 a
Fnet,2 = m2 a2 ⟹ m2 g - T2 = m2 a
In Chapter D, where the tensions were equal this gave two linear equations with two unknowns. Now we have three unknowns,
a and the two tensions. There are four forces acting on the pulley, the two tensions, the pulley’s weight Mg and an upward
force Faxis acting at the axis; since these two forces act at the axis they produce no torque. We choose clockwise as positive,
since that is the direction of our angular acceleration. The tensions are perpendicular to the radial vector so τ = r F⊥ = RF. We
now apply the rotational second law applied to the pulley.
1
τnet = I α ⟹ R T2 - R T1 = I α = M R2 α
2
Here we have added another equation but also added another unknown α. We can use a = R α to eliminate α in favor of a. We
can also use the fact that the pulley is a uniform disk.
1 a M
R T2 - R T1 = I α = M R2 ⟹ T2 - T1 = a
2 R 2
Adding this to the two second law expressions for the hanging masses we eliminate the tensions and we get our answer.
14 | Chapter I - Rotational Dynamics and Equilibrium

m2 - m1
m2 g - m1 g = (m1 + m2 + M / 2) a ⟹ a = g
m1 + m2 + M / 2

I.3 - Equilibrium

The Conditions for Equilibrium

If a body is in equilibrium then there is no acceleration and there is no angular acceleration. This implies that the net force and the net
torque must vanish.
F net = 0 and τnet = 0
For the examples we will consider all possible rotation will be in a plane and thus we only need to consider torques relative to an axis.

The Origin (or Axis) is Arbitrary

When considering an equilibrium problem sometimes the choice of axis is clear. Often in isn't clear, though, and there isn't a natural choice.
The key point is that the choice of origin or axis is arbitrary. When something is arbitrary then we have the luxury of making a choice that
simplifies the problem.
The basic result is this: If the torques balance about one origin and the forces balance then the torques balance about any origin. If the
vector from one origin to another is r0 . If r′i is the vector from the new origin at r0 to the mass mi and ri is from the first origin to the mass then
ri = r′i + r0 .

Take the net torques about these axes to be τnet and τ′net . If τnet = 0 and F net = 0 then τ′net = 0.

0 = τnet =  ri ⨯ F i =  r′i ⨯ F i + r0 ⨯  F i = τ′net + 0

i i i
This proves our result.

Example I.10 - Hanging Meter Stick

TL TR

A horizontal uniform meter stick of weight W hangs from vertical strings at the 20-cm and 60-cm lines. What are both tensions, TL and
TR ?

Solution
The net torque and net force are both zero. The two tensions are the unknowns. Setting the net force to zero gives one equation,
since all forces are vertical.
Fnet = 0 ⟹ TL + TR = W
 Chapter I - Rotational Dynamics and Equilibrium | 15

0.40m

0.30m TR
TL 0.10m

A C

W
+

We may choose the axis anywhere. Any force that acts at the origin produces no torque. If we choose the axis to be when an
unknown acts then the torque equation will not involve that unknown. We will choose the axis labeled A where TL acts.
Choosing clockwise as our positive sense of rotation we can write the torque equation and can solve for TR
3
0 = τnet,A = 0 + (0.30 m) W - (0.40 m) TR ⟹ TR = W
4
The force equation lets us find TL .
1
TL + T R = W ⟹ TL = W - T R = W
4
(Note that if we chose a different axis we would get the same answer. For instance, choosing the center C we get:
τnet,C = (0.30 m) TL - (0.10 m) TR . This leads to 3 TL = TR and with the force equation we get the same solution.)

Example I.11 - Leaning Ladder

A uniform ladder of length L leans against a frictionless wall, making an angle of θ with the floor. What is the normal force N of the
wall on the ladder and what are the horizontal and vertical components, H and V, of the force of the floor on the ladder?

N =?

V =?

H =?

Solution
The position of the axis is arbitrary but in this problem, given that two of the three unknowns act at the base of the ladder, that is
the natural axis to choose; those two unknowns will not appear in our torque equation.
16 | Chapter I - Rotational Dynamics and Equilibrium

L
L sin θ

V
W L/2

Axis
H
L
cos θ
2

This is a two-dimensional problem so the force condition gives two equations.

Fnet,hor = 0 ⟹ N = H
Fnet,ver = 0 ⟹ V = W
For torques about our axis at the base of the ladder, we have two forces to consider.

Since the weight W is vertical r⊥ is the horizontal part of r. Since we have r = L / 2. Chosing counterclockwise as positive the
torque due to the weight is positive.
L
τW = r ⊥ F = + cos θ W
2
The normal force of the wall N is horizontal, so r⊥ is the vertical part of r = L. It is clockwise and thus negative with our
convention.
τN = - r⊥ F = -(L sin θ) N
Our torque equation gives N and using H = N, gives H.
L
0 = τnet = τW + τN + τH + τV = + cos θ W - (L sin θ) N + 0 + 0
2
Our full answer follows.
W
H=N=
2 tan θ