
 Angular Motion
 Angle in radians
 An angle in radians is defined as the arc it subtends
divided by the radius of the circle. In Fig. 1 below,
the angle is DQ, the
arc it subtends is Ds,
and the radius is R. DQ
= Ds/R Notice that
the unit of Ds is
meters and the unit of R is meters so the angle has no
unit. We say radians or degrees to imply in which
we are measuring the angle.
 Conversion of radians to degrees or degrees to radians
 The circumference of a circle equals 2pR.
In Fig. 2 below, s equals onequarter of a circle
or s = 2pR/4 =
pR/2. In radians
the angle = s/R = (pR/2)/R
= p/2. In Fig.
2, Q = 90^{o}.
Thus p/2 radians
= 90^{o} or
p radians = 180^{o}.
 In general, (angle in radians)/(angle in degrees)
= p/180.
 Descriptive Terms
 Speed v at any point = the magnitude of thevelocity
tangent to the path at that point.
 For circular motion the Period T = time for one complete
rotation = 2pr/v
 The frequency f = reciprocal of period = 1/T = v/2pr
 The angular velocity w
(omega not w) = the angle turned through per second or
for a varying angular velocity = dQ/dt.
In Fig. 1 above,
DQ = (1/R)(Ds)
(Equation
#1)
Dividing both sides of Equation #1 by Dt,
DQ/Dt
= (1/R)(Ds/Dt)
(Equation
#2)
Taking the limit of both sides of Equation # 2 as t approaches
zero gives dQ/dt =
(1/R) ds/dt.
Recognizing dQ/dt
as the angular velocity w
and ds/dt as the linear velocity v gives us the relation
between w and v. w
= v/R.
Since the frequency f = v/2pR
or v = 2pRf, w
= (2pRf)/R = 2pf
 Alternative expressions for centripetal or radial acceleration,
 a_{r} = v^{2}/R
 a_{r} = (2pR/T)^{2}/R
= 4p^{2}R/T^{2}
= (2pRf)^{2}/R
= (2pf)^{2}R
= w^{2}R
 Angular acceleration a
is the rate of change of angular velocity.
a = dw/dt.
Its unit is s^{2}.
For constant angular acceleration

Linear analogy

w(t)
= w_{0}
+ at

v(t) = v_{o }+ at

Q(t)
= Q_{0
} + wt
+ 1/2 at^{2}

x(t) = x_{0 } + v_{0}t
+ 1/2 at^{2}

w^{2}(Q)
= w_{o}^{2}
+ 2a(Q
 Q_{o})

v^{2}(x) = v_{o}^{2}
+ 2a(x – x_{o})

 Sample Problems in 104
Problem Set for Angular Motion: 47.
 Rigid Body
 All points on the object rotate with the same angular
velocity w.
 For every mass m_{1}, m_{2} m_{3},
etc. that makes up the body, v_{1} = wr_{1},
v_{2} = wr_{2},
v_{3} = wr_{3},
etc. (See Fig. 3 below)
 K = 1/2 m_{1}v_{1}^{2} + 1/2 m_{2}v_{2}^{2}
+ 1/2 m_{3}v_{3}^{2} + + + +
= _{1}^{N}S
(1/2) m_{i}v_{i}^{2}K = 1/2 m_{1}r_{1}^{2}w^{2
}+ 1/2 m_{2}r_{2}^{2}w^{2}
+ 1/2 m_{3}r_{3}^{2}w^{2}
+ + +
= _{1}^{N}S
(1/2) m_{i}r_{i}^{2}w^{2
}
 Define Moment of inertia I = _{1}^{N}S
m_{i}r_{i}^{2}. Then
K = (1/2)Iw^{2}.
 Compare with K = 1/2 mv^{2} for translation

If distribution of mass is continuous then integrate
to find moment of inertia
I=∫dm
r^{2
}
 Moments of Inertia
 Hoop (Fig. 4a below) All points on the hoop equal the distance
R from the axis of rotation. The moment of inertia of a hoop
about its center of mass = MR^{2}.
 Circular Disk (Fig. 4b above) Now there is mass spread throughout
the object. The moment of inertia of a disk about its center
of mass is 1/2 MR^{2}, smaller than that for a hoop
because some of its mass is closer to the center of the disk
and r for those masses is less than the radius R of the disk
 A cylinder can be considered a series of disks. The moment
of inertia of a cylinder about its center of mass =1/2 MR^{2}.
 The moment of inertia of a sphere about its center of mass
I = 2/5 MR^{2}.
 Thin Rod
 Moment of Inertia about center of mass = ML^{2}
/12
 Moment of Inertia about one end = ML^{2}/3
 (Fig. 5 below) Parallel Axis theorem: I(parallel) = I(CenterMass)
+ Md^{2}, where d is the distance between the axes.
Example: Thin rod about one end. I = ML^{2}/12
+ M(L^{2}/4) = ML^{2}/3
 Sample Problems in 104
Problem Set for Angular Motion: 8, 9.
 Application of Conservation of Energy to Rotation
Object of mass M, radius R and moment of inertia I_{CM}is
initially at rest at the top of an incline. The distance between
the center of the object at the top of the incline and at the
bottom of the incline is h (Fig. 6 below).
It rolls without slipping down the incline. A frictional force
acts on the object, but the frictional force acts at a point,
not through a distance so the frictional force does no work and
we can use conservation of energy to solve this problem.
1) At the top, E_{top} = U + K = Mgh
+ 0
2) At the bottom since the sphere has both translational
and rotational kinetic energy, E_{bottom }= U + K = 0
+ (1/2 Mv^{2 }+ 1/2 I_{CM}w^{2})
3) Since w
= v/R, E_{bottom }= 1/2 Mv^{2 }+ 1/2 I_{CM}(v^{2}/R^{2})
4) From conservation of energy, Mgh = 1/2 Mv^{2
}+ 1/2 (I_{CM}/R^{2})v^{2}or v^{2}
= 2gh/(1 + I_{CM}/MR^{2})
Special Cases:
a. For hoop, I_{CM }= MR^{2},
v = (gh)^{1/2}
b. For disk or cylinder, I_{CM }= MR^{2}/2,
v = (4gh/3)^{1/2}
c. For sphere, I_{CM }= 2MR^{2}/5,
v = (10gh/7)^{1/2
}
 Work Done on Rotating Rigid Body
 Since w^{2 }
= w_{o}^{2}
+ 2a(Q
 Q_{o}), 1/2
I_{CM }w^{2
}  1/2 I_{CM} w_{o}^{2}
=
I_{CM}a(Q
 Q_{o}).
 Identify the lefthand side of the above equation as DK
and the right hand side as the work W = I_{CM}a(Q
 Q_{o}).
 Sample Problems in 104
Problem Set for Angular Motion: 1113.
 Torque t (Fig. 7 below)
 t = r
x F
 Magnitude of t = rF
sin r,F = (r
sin r,F)F=
 Direction of torque. For t
= r x F, point the fingers of your
right hand in the direction of r with the palm of your
hand ready to rotate in to F. Rotate your fingers into
F. The thumb of your right hand points out of the page.
The torque t is
out of the page (Figure 7a below).
 Sample Problems in 104
Problem Set for Angular Motion: 10, 14, 15.
 Angular Momentum L
 For a point particle of mass m moving with velocity v,
L = r x mv, where r is the distance from the
axis to the position of the mass m.
 For direction of L, point the fingers of your right
hand in the direction of r and rotate your fingers
into the direction of v, your thumb and the angular
momentum L point out of the page (Fig. 8 below).
 For a rotating rigid body L = Iw.
The direction of w is
found by curling the fingers of your right hand in the direction
of the rotation. Your thumb points to the direction of w
and for a rotating rigid body to the direction of L.
 For point object, dL/dt = d(r x mv)/dt
= d(r)/dt x mv + r x d(mv)/dt
= [v x mv]+ r x d(mv)/dt = 0 +
r x F, because [v x v] = 0 and
F = d(mv)/dt. Thus dL/dt = r x
F = t . When the torque equals zero,
the angular momentum remains constant. When no external
torque acts on a system, angular momentum is conserved.
 For rotating rigid body,
 t = dL/dt = d(Iw)/dt
= Ia
 Work = W = (Ia)Q
= tQ
 Conservation of Angular Momentum1. If no net external torque
acts on a system, its total angular momentum L remains constant.
t = dL/dt.
If t = 0,
dL/dt = 0 and L remains constant.
Examples:
 Person on rotating table
 Planetary motion
 Gyroscope
 Wheel spins clockwise as viewed from support (Fig.
9a below).
L is to the right.
 The torque due to the weight mg is t
= r x mg points into the paper (Fig.
9b below).
 As seen from above L precesses slowly counterclockwise
(Fig. 9 c and d below). L + DL
= L + tDt,
where t
= dL/dt.
 dF = t
dt/L or W = dF/dt
= t /L = mgr/Iw.
Wheel spins at rate w
and precesses at rate W.
 d(L^{2})/dt = d(L ^{.} L)/dt
= L ^{.} dL/dt + L ^{.}
dL/dt = 2L ^{. } dL/dt
= 2L ^{. }t
= 2Lt cos
L
t =2Lt
cos 90^{o} = 0.
Since t is perpendicular
to L, there is no change in the magnitude of L, only
change is in its direction.
 Sample Problems in 104
Problem Set for Angular Motion: 16, 1820.
 Example. A piece of playground apparatus consists of two carts,
each of mass 50 kg, that are joined by very light (compared to
the mass of the carts) rigid rods, 2 m each in length, as shown
in Fig. 10 below.
To make this device rotate, a child applies a constant force of
10p N to the outer cart, directed
perpendicular to the rods. The apparatus is initially at rest.
Find:
 the moment of inertia of the apparatus
 the angular acceleration produced by the child
 how long it will take for the apparatus to make one complete
revolution
 at the time for (c), the angular velocity and the kinetic
energy of the apparatus, and
 the work done by the child.
 Find the direction and magnitude of DL/Dt
and the torque
 Solutions:
 Treating the two carts as point particles,
I = 50 kg(2 m)^{2} + (50 kg)(4 m)^{2} =
1000 kgm^{2}.
 t = r x
F. t =
(4 m)(10p N) sin 90^{o}
= 40p Nm = Ia
= 1000 kgm^{2}a.
a = 0.04p
s^{1}
 Angle turned through in one complete revolution =2p
= 1/2 at^{2}T
= [(4p)/a]^{1/2}
= [(4p)/4p
x 10^{2} s^{2}]^{1/2} = 10 s
 angular velocity:
w
= w_{o} + at
= 0 + (4p x 10^{2}
s^{2})(10 s) = 0.4p
s^{1}
kinetic energy:
K = 1/2 Iw^{2}
= 1/2 (1000 kgm^{2})(0.4p
s^{1})^{2} = 80p^{2}
J
 W = tQ = (40p
Nm)( 2p) = 80p^{2}
J = K_{f }– K_{I }= 80p^{2}
J – 0
 Since w_{o}
= 0, L_{o }= 0. Final L = Iw.
Since the apparatus rotates counterclockwise, w
is out of the page. After one revolution,
L = (1000 kgm^{2})(0.4p
s^{1}) = 4p
x 10^{2 }mkg/sDL/Dt
= (4p x 10^{2 }mkg/s)/(10
s) = 40p mkg/s^{2}
= t out of page
 Combination of Translational and Rotational Motion (Fig. 11
below)
For a sphere to roll without slipping down the incline plane there
must be a frictional force f. We need two equations of
motion, one for translation and one for rotation.
For translation:
F_{net }= ma
mg sin Q  f = ma
For rotation about the center of mass, neither N nor mg
produces a torque about the center of mass because their line
of actions pass through the center of mass. Thus:
t_{CM } = I_{CM}a
fR sin 90^{0 }= (2/5 mR^{2})a
or since a = a/R, f = 2/5
ma
Substituting the second equation into the first:
mg sin Q  2/5 ma = ma or
a = 5g sinQ/7v^{2}
= v_{0}^{2 }+ 2as = 0 + (10/7)g(s sinQ)
= (10/7)gh, v = (10gh/h)^{1/2 }
as we found before using conservation of energy.
 Conditions for Equilibrium
 The sum of the external forces must be zero, or
S F
= 0
 The sum of the external torques must be zero, or
St
= 0
 Example (Fig. 12a below):
A uniform horizontal beam of length 8.00 m and weight 200
N is attached to a wall by a pin connection. Its far end is
supported by a cable that makes an angle of 53.0^{o}
with the horizontal. If a 600N person stands 2.00 m
from the wall, find the tension in the cable and the force
exerted by the wall on the beam.
 Solution to Example (Fig. 12b below):
 The forces acting on the beam are:
 The gravitational force of the earth on the beam,
its weight = 200 N acts at the center of the uniform
beam.
 The force of the man on the beam = 600 N
 The tension T in the rope
 The force R exerted by the wall at the pivot
 The torques acting on the beam about the axis at the
pivot point are:
 Torque due to R is zero because it acts at the
pivot point. Any force that passes through the axis
produces no torque (r = 0).
Note:
I chose the axis at the pivot point to eliminate one
of the unknowns in the torque equation, leaving only
T as an unknown.
 The torque due to the tension in the rope = 8.0
m(T) sin 53^{o
} out of the page.
 The torque due to the force of the man = 2.0 m(600
N)
into the page
 Since it is a uniform beam, the weight of the beam
acts at the center.
r = 4 m for the beam weight.
Torque = 4.0 m(200 N) into the page.
 S t
= 0
Taking out of the page as positive:
+8.0 m(T) sin 53^{o}  2.0 m(600
N)  4.0 m(200 N) = 0
8.0 m(T)(0.80) = 2000 Nm. T
= 313 N
 The vector sum of the forces must = 0.
Change this twodimensional problem into two onedimensional
problems, by taking the components of the tension T
and the force of the wall at the pivot R into X
and Y components.
As shown in Fig. 12b,
T_{x }= T cos 53^{o
} T_{y }= T sin 53^{o},
R_{x }= R cos Q,
and
R_{y }= R sin Q
S
F_{x} = 0 
S
F_{y} = 0 
R cos Q
 T cos 53^{o} = 0 
R sin Q
+ T sin 53^{o} – 600N – 200N
= 0 
R cos Q
= T cos 53^{o} 
R sin Q
=  T sin 53^{o} + 600 N + 200 N 
R cos Q
= 313(0.6) 
R sin Q=
313 N(0.8) + 800 N 
R cos Q
= 188 N (Equation
1) 
R sin Q=
550 N (Equation
2) 
Dividing Eq.2 by Eq. 1:
tan Q=
550 N/188 N = 2.92. Q
= 71.1^{0} R.
R sin Q= 550 N.
R = 550 N/sin Q.
R = 550 N/sin 71.1^{o} = 581 N.
 Sample Problems in 104
Problem Set for Angular Motion: 2224.
 Using Cross Product with Unit Vectors
 Definitions
 Magnitude of C = A
x B is C = ABcos A,
B.
 Point the fingers of your
right hand in the direction of A with your palm
toward B. Rotate your palm into B.
Your thumb points in the direction of C. In
Fig. 13 below, C is out of the page.
 Application to unit vectors
 i x i = iisin
i,
i = 11sin 0^{o}
= 0
j x j = jjsin j,
j = 11sin 0^{o}
= 0
k x k = kksin k,
k = 11sin 0^{o} = 0
 Magnitudes and Directions
 Magnitude: i
x j = ijsin
i,
j = 11sin 90^{o}
= 1 = j x i
Direction: i x j = k
j x i =  k
 Magnitude: i
x k = iksin
i,
k = 11sin 90^{o}
= 1 = k x i
Direction: i x k = j k
x i =j
 Magnitude: j
x k = jksin
j,
k = 11sin 90^{o}
= 1 = k • j Direction: j
x k = i k x
j =  I

