Rotational Motion and
Angular Momentum
Unit 6
Lesson 1 : Angular Position, Velocity,
and Acceleration
When a rigid object rotates about its axis, at
any given time different parts of the object
have different linear velocities and linear
accelerations.
A particle at point
P at a fixed
distance r from
origin is rotating
about axis O.
P is at polar coordinate (r, q)
q is measured counterclockwise
from some reference line (q = 0)
s = rq
s
q=
r
q is measured in radians (rad)
One radian is the angle subtended by an
arc length equal to the radius of the arc.
q=
360o =
2pr
s
r
rad = 2p rad
r
1 rad =
360o
2p
= 57.3o
Converting from Degrees to Radians
p
q (rad) =
q (deg)
o
180
90o = p/2 rad
60o = p/3 rad
45o = p/4 rad
270o = 3p/2 rad
Angular Displacement (Dq)
As particle moves
from A to B in time
interval Dt, the
reference line of length
r sweeps out an angle
Dq = qf - qi
Average Angular Speed (w)
Ratio of the angular displacement of
a rigid object to the time interval Dt.
w=
qf - qi
tf - ti
=
Dq
Dt
The rad/s is the unit for angular speed.
w is positive when q increases
(counterclockwise motion)
w is negative when q decreases
(clockwise motion)
Instantaneous Angular Speed (w)
dq
Dq
w = lim
=
Dt  0 Dt
dt
Average Angular Acceleration (a)
a=
wf - wi
tf - ti
=
Dw
Dt
Instantaneous Angular Acceleration (a)
dw
Dw
a = lim
=
Dt  0 Dt
dt
The rad/s2 is the unit for angular speed.
a is positive when object rotates
counterclockwise and speeds up
OR
when object rotates clockwise and
slows down
Direction of Angular Velocity and
Acceleration Vectors
w and a are vector quantities with
magnitude and direction
The directions of w and a are
along the axis of rotation.
Right-Hand Rule
Wrap four fingers
of the right hand in
the direction of
rotation.
Thumb will point in
the direction of
angular velocity
vector (w).
Example 1
A rigid object is rotating with an angular
speed w < 0. The angular velocity vector
w and the angular acceleration vector a
are antiparallel. The angular speed of the
rigid object is
a) clockwise and increasing
b) clockwise and decreasing
c) counterclockwise and increasing
d) counterclockwise and decreasing
Example 2
During a certain period of time, the angular
position of a swinging door is described
by q = 5.00 + 10.0 t + 2.00 t2, where q is in
radians and t is in seconds. Determine the
angular position, angular speed, and
angular acceleration of the door at
a) at t = 0
b) at t = 3.00 s
Lesson 2 : Rotational Kinematics with
Constant Angular Acceleration
dw
a=
dt
dw = a dt
wf = wi + at
dq
Since w =
dt
dq
wi + at =
dt
(by integrating)
qf = qi + wit + ½ at2 (by integrating)
Eliminating t from
previous two
equations,
wf2 = wi2 + 2a(qf – qi)
Eliminating a from
previous two
equations,
qf = qi + ½ (wi + wf) t
q
Position
x
w
Velocity
v
a
Acceleration a
Example 1
A wheel rotates with a constant angular
acceleration of 3.50 rad/s2.
a) If the angular speed of the wheel is
2.00 rad/s at ti = 0, through what
angular displacement does the
wheel rotate in 2.00 s ?
b) Through how many revolutions has
the wheel turned during this time
interval ?
c) What is the angular speed of the wheel
at t = 2.00 s ?
Example 2
A rotating wheel requires 3.00 s to rotate
through 37.0 revolutions. Its angular
speed at the end of the 3.00 s interval is
98.0 rad/s. What is the constant angular
acceleration of the wheel ?
Lesson 3 : Angular and Linear Quantities
Every particle of a rigid rotating object
moves in a circle whose center is the
axis of rotation.
ds
Tangential speed v =
dt
Since s = rq,
v= r
dq
dt
v = rw
Tangential Speed
v = rw
angular speed
perpendicular
distance of a point
from the axis of
rotation
Tangential
speed depends
on distance
from axis of
rotation
Angular speed
is the same for
all points
at =
Tangential Acceleration
dv
dt
Since v = rw,
at = r
dw
dt
at = ra
Centripetal Acceleration in terms of
Angular Speed
ac =
v2
r
Since v = rw,
ac = rw2
Total Linear Acceleration
a=
a=
a=r
at2 + ar2
r2a2 + r2w4
a2 + w 4
Example 1
In order to keep the tangential speed of the disc
surface at the location of the lens constant, the
disc’s angular speed must vary as the lens moves
radially along the disc. In a typical compact disc
player, the constant speed of the surface at the
point of the laser-lens system is 1.3 m/s.
a) Find the angular speed of the disc in rev/min
when information is being read from the
innermost first track (r = 23 mm) and the
outermost final track (r = 58 mm).
b) The maximum playing time of a standard music
CD is 74 min 33 s. How many revolutions
does the disc make during that time ?
c) What total length of track moves past the lens
during this time ?
d) What is the angular acceleration of the CD over
the 4,473 s time interval ? [Assume that a is
constant.]
Example 2
The drive train of a bicycle
is shown to the right. The
wheels are 67.3 cm in
diameter and pedal cranks
17.5 cm long. The cyclist
pedals at a steady angular
rate of 76.0 rev/min. The
chain engages with a front
sprocket 15.2 cm in
diameter and a rear
sprocket 7.00 cm in
diameter.
a) Calculate the speed of a link of the
chain relative to the bicycle frame.
b) Calculate the angular speed of the
bicycle wheels.
c) Calculate the speed of the bicycle
relative to the road.
d) What pieces of data, if any, are not
necessary for the calculations ?
Lesson 4 : Rotational Kinetic Energy
The total kinetic energy of a rotating rigid
object is the sum of the kinetic energies of
its individual particles.
KErot = SKEi = S ½mivi2
Since v = rw,
KErot = ½ S miri2wi2
Factoring out w2,
KErot = ½ (S miri2) w2
KErot = ½ (S miri2) w2
Moment of Inertia (I)
I = S miri2
The kg . m2 is the SI unit for moment of inertia.
Moment of inertia is a measure of the
resistance of an object to changes in
its rotational motion.
Substituting I,
KErot = ½ Iw2
Rotational Motion
Linear Motion
I
m
w
v
KErot = ½ Iw2
KE = ½ mv2
Example 1
Consider an oxygen molecule (O2) rotating in the
x-y plane about the z-axis. The rotation axis
passes through the center of the molecule,
perpendicular to its length. The mass of each
oxygen atom is 2.66 x 10-26 kg, and at room
temperature the average separation between the
two atoms is d = 1.21 x 10-10 m. (The atoms are
modeled as particles.)
a) Calculate the moment of inertia of the molecule
about the z-axis.
b) If the angular speed of the molecule about the
z-axis is 4.60 x 1012 rad/s, what is its
rotational kinetic energy ?
Lesson 5 : Calculation of Moments of Inertia
Moment of inertia of a rigid object is
evaluated by dividing the object into
many small volume elements, each
with mass = Dmi.
I = lim Sri2Dmi =
Dmi 0

r2 dm
Since r = m/V,
dm = r dV
I=

rr2 dV
Example 1
Find the moment of inertia of a uniform
thin hoop of mass M and radius R about
an axis perpendicular to the plane of the
hoop and passing through its center.
Example 2
Calculate the moment of inertia of a
uniform rigid rod of length L and mass
M about an axis perpendicular to the
rod (the y-axis) and passing through its
center of mass.
Example 3
A uniform solid cylinder has a radius R,
mass M, and length L. Calculate its
moment of inertia about its central axis
(the z-axis).
Parallel-Axis Theorem
The moment of inertia about any axis
parallel to and a distance D away from
this axis is
I = ICM + MD2
Example 4
Consider once again the uniform rigid rod
of mass M and length L. Find the moment
of inertia of the rod about an axis
perpendicular to the rod through one end
(the y’ axis).
Moment of Inertia of a Thin Cylindrical Shell (Hoop)
Moment of Inertia of a Hollow Cylinder
Moment of Inertia of a Solid Cylinder (Disk)
Moment of Inertia of a Rectangular Plate
Moment of Inertia of a Long Thin Rod
(Axis Through Center)
Moment of Inertia of a Long Thin Rod
(Axis Through End)
Moment of Inertia of a Solid Sphere
Moment of Inertia of a Thin Spherical Shell
Lesson 6 : Torque
The tendency of a force to rotate an
object about some axis is measured by a
vector quantity called torque.
The component
Fsinq tends to
rotate the
wrench about
point O.
t = r Fsinf = Fd
distance between pivot
point and point of
application of F
perpendicular distance
from pivot point to the
line of action of F
Torque should not be
confused with force.
Torque has units of force x length or N.m.
(Same as work but not called Joules.)
F1 tends to rotate
counterclockwise (+t)
F2 tends to rotate
clockwise (-t)
The net torque about axis 0 is
St = t1 + t2 = F1d1 – F2d2
Example 1
A one-piece cylinder is shaped as
shown with a core section
protruding from the larger drum.
The cylinder is free to rotate about
the central axis shown in the
drawing. A rope wrapped around
the drum, which has radius R1,
exerts a force T1 to the right on
the cylinder. A rope wrapped
around the core, which has radius
R2, exerts a force T2 downward on
the cylinder.
a) What is the net torque acting on the cylinder
about the rotation axis (z-axis) ?
b) Suppose T1 = 5.0 N, R1 = 1.0 m, T2 = 15.0 N,
and R2 = 0.50 m. What is the net torque
about the rotation axis, and which way
does the cylinder rotate starting from rest ?
Lesson 7 : Relationship between Torque and
Angular Acceleration
Ft = mat
t = Ftr = (mat)r
t = (mra)r = (mr2)a
Since I = mr2,
St = Ia
(rotational analog to Newton’s Second Law)
Example 1
A uniform rod of length L and mass M is
attached at one end to a frictionless pivot and
is free to rotate about the pivot in the vertical
plane. The rod is released from rest in the
horizontal position. What is the initial angular
acceleration of the rod and the initial linear
acceleration of its right end ?
Example 2
A wheel of radius R, mass
M, and moment of inertia I
is mounted on a
frictionless horizontal
axis. A light cord wrapped
around the wheel supports
an object of mass m.
a) Calculate the angular acceleration of the
wheel.
b) Calculate the linear acceleration of the
object.
c) Calculate the tension in the cord.
Lesson 8 : Work, Power, and Energy in
Rotational Motion
Work done by F on an
object as it rotates
through an infinitesimal
distance ds = r dq
dW = F.ds = (Fsinf)r dq
(The radial component of F does no
work because it is perpendicular to the
displacement.)
Since t = rFsinf,
dW = t dq
analogous to
dW = Fx dx in
linear motion
Rate at which work is being done as
an object rotates through angle dq in
time interval dt is
dW
dq
t
=
dt
dt
analogous
to P = Fv in
linear
motion
dW
P=
= tw
dt
(Power delivered to a rotating rigid object.)
dw
St = Ia = I
dt
dw
St = I
dq
dq
dt
dw
St = I
dq
(chain rule)
w
St dq = Iw dw
Since dW = St dq,
dW = Iw dw
Integrating to find total work done,
wf
SW =

Iw dw = ½ Iwf2 – ½ Iwi2
wi
Work–KE 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 rotational energy.
Example 1
A uniform rod of
length L and mass M
is free to rotate on a
frictionless pin
passing through one
end. The rod is
released from rest in
the horizontal
position.
a) What is its angular speed when it reaches
its lowest position ?
b) Determine the tangential speed of the
center of mass and the tangential
speed of the lowest point on the rod
when it is in the vertical position.
Example 2
Consider two cylinders having
different masses m1 and m2,
connected by a string passing
over a pulley. The pulley has a
radius R and moment of
inertia I about its axis of
rotation. The string does not
slip on the pulley, and the
system is released from rest.
Find the linear speeds of the
cylinders after cylinder 2
descends through a distance
h, and the angular speed of
the pulley at this time.
Example 3 : AP 2001 #3
A light string that is attached to a large block of mass 4m
passes over a pulley with negligible rotational inertia and
is wrapped around a vertical pole of radius r, as shown in
Experiment A above. The system is released from rest,
and as the block descends the string unwinds and the
vertical pole with its attached apparatus rotates. The
apparatus consists of a horizontal rod of length 2L, with
a small block of mass m attached at each end. The
rotational inertia of the pole and the rod are negligible.
a) Determine the rotational inertia of the rod-andblock apparatus attached to the top of the
pole.
b) Determine the downward acceleration of the
large block.
c) When the large block has descended a distance
D, how does the instantaneous total kinetic
energy of the three blocks compare with the
value 4mgD ? Check the appropriate space
below.
____ Greater than 4mgD
____ Equal to 4mgD
____ Less than 4mgD
Justify your answer.
The system is now reset. The string is rewound
around the pole to bring the large block back to its
original location. The small blocks are detached from
the rod and then suspended from each end of the rod,
using strings of length l. The system is again
released from rest so that as the large block
descends and the apparatus rotates, the small blocks
swing outward, as shown in Experiment B above.
This time the downward acceleration of the block
decreases with time after the system is released.
d) When the large block has descended a distance
D, how does the instantaneous total kinetic
energy of the three blocks compare to that in
part c) ? Check the appropriate space below.
____ Greater
____ Equal
____ Less
Justify your answer.
Lesson 9 : Rolling Motion of a Rigid Object
Cylinder rolling on a straight path.
Center moves in a straight line (green line).
A point on the rim moves in a path called a
cycloid (red curve).
Speed of CM of Cylinder Rolling without Slipping
ds
vCM =
dt
Since s = Rq,
dq
vCM = R
dt
vCM = Rw
for pure rolling motion only
Acceleration of CM of Cylinder Rolling without
Slipping
dvCM
aCM =
dt
aCM = R
dw
dt
aCM = Ra
Total Kinetic Energy of a Rolling Cylinder
KErot = ½ IPw2
Since IP = ICM + MR2, (parallel-axis theorem)
KErot = ½ (ICM + MR2)w2
OR
KErot = ½ ICMw2 + ½ MR2w2
Since vCM = Rw,
KErot = ½ ICMw2 + ½ MvCM2
KEtotal = ½ ICMw2 + ½ MvCM2
rotational KE
translational KE
The total kinetic energy of a rolling
object is the sum of the rotational
kinetic energy about the center of
mass and the translational kinetic
energy of the center of mass.
Example 1
Three objects of uniform density – a solid
sphere, a solid cylinder, and a hollow cylinder
– are placed at the top of an incline. They are
all released from rest at the same elevation and
roll without slipping. Which object reaches the
bottom first ? Which reaches it last ?
Example 2
For the solid sphere shown above,
calculate the linear speed of the center
of mass at the bottom of the incline and
the magnitude of the linear acceleration
of the center of mass.
Example 3 : AP 1986 #2
An inclined plane makes an angle of q with the
horizontal, as shown above. A solid sphere of radius
R and mass M is initially at rest in the position shown,
such that the lowest point of the sphere is a vertical
height h above the base of the plane. The sphere is
released and rolls down the plane without slipping.
The moment of inertia of the sphere about an axis
through its center is 2/5 MR2. Express your answers
in terms of M, R, h, g, and q.
a) Determine the following for the sphere when it
is at the bottom of the plane.
i. Its translational kinetic energy
ii. Its rotational kinetic energy
b) Determine the following for the sphere when it
is on the plane.
i. Its linear acceleration
ii. The magnitude of the frictional force
acting on it
The solid sphere is replaced by a hollow sphere of
identical radius R and mass M. The hollow sphere,
which is released from the same location as the
solid sphere, rolls down the incline without
slipping.
c) What is the total kinetic energy of the hollow
sphere at the bottom of the plane
d) State whether the rotational kinetic energy of
the hollow sphere is greater than, less
than, or equal to that of the solid sphere at
the bottom of the plane. Justify your
answer.
Lesson 10 : Angular Momentum
From linear motion :
dp
SF =
dt
Take cross product of each side with r :
dp
r x SF = r x
dt
Net Torque
dp
St = r x
dt
dr
dt
x p is zero since
dr
dt
= v,
and v and p are parallel.
Add
Analogous to
dp in
translational
motion
St =
dr
dt
x p to right-hand side :
dp
dr
St = r x
xp
+
dt
dt
d(r x p)
dt
angular momentum (L)
L=rxp
Instantaneous Angular Momentum
L=rxp
The instantaneous angular
momentum L of a particle relative to
the origin O is defined by the cross
product of the particle’s
instantaneous position vector r and
its instantaneous linear momentum p.
The SI unit of angular momentum is kg . m2/s.
Since L = r x p,
St =
dL
dt
rotational
analog of
Newton’s
second law
The torque acting on a particle is equal
to the time rate of change of the
particle’s angular momentum.
The direction of L is always perpendicular
to the plane formed by r and p.
Since p = mv,
L = mvr sinf
Example 1
A particle moves in the xy plane in a circular path
of radius r, as shown above. Find the magnitude
and direction of its angular momentum relative to
O when its linear velocity is v.
Angular Momentum of a System of Particles
The total angular momentum of a
system of particles about some point is
defined as the vector sum of the angular
momenta of the individual particles.
Ltot = L1 + L2 + …. + Ln = SLi
Differentiating with respect to time :
dLtot
dLi
= S
dt
dt
St
St =
dLtot
dt
The net external torque acting on a system
about some axis passing through an origin
in an inertial frame equals the time rate of
change of the total angular momentum of
the system about that origin.*
* This theorem applies even if the center of
mass is accelerating, as long as t and L are
evaluated relative to the center of mass.
Example 2
A sphere of mass m1 and a block of mass m2 are
connected by a light cord that passes over a pulley, as
shown above. The radius of the pulley is R, and the
mass of the rim is M. The spokes of the pulley have
negligible mass. The block slides on a frictionless,
horizontal surface. Find an expression for the linear
acceleration of the two objects, using the concepts of
angular momentum and torque.
Lesson 11 : Angular Momentum of a Rotating
Rigid Object
Each particle of this rigid
object rotates in the xy
plane about the z axis
with angular speed w.
Since v = rw,
L = mr2w
Since I = mr2,
L = Iw
L = Iw
Differentiating with respect to time :
(I is constant for a rigid object)
dL
dt
= I
dw
= Ia
dt
Since St =
St = Ia
dL
dt
rotational form
of Newton’s
second law
Example 1
Estimate the magnitude of the angular
momentum of a bowling ball spinning at
10 rev/s.
Example 2
A father of mass mf and
daughter of mass md sit
on opposite ends of a
seesaw at equal
distances from the
pivot at the center. The
seesaw is modeled as a
rigid rod of mass M and
length l and is pivoted
without friction. At a
given moment, the
combination rotates in
a vertical plane with an
angular speed w.
a) Find an expression for the magnitude of the
system’s angular momentum.
b) Find an expression for the magnitude of the
angular acceleration of the system when the
seesaw makes an angle q with the horizontal.
Example 3 : AP 1983 #2
A uniform solid cylinder of mass m1 and radius R is
mounted on frictionless bearings about a fixed axis
through O. The moment of inertia of the cylinder
about the axis is I = ½ m1R2. A block of mass m2,
suspended by a cord wrapped around the cylinder
as shown above, is released at time t = 0.
a) On the diagram below draw and identify all of
the forces acting on the cylinder and on the
block.
b) In terms of m1, m2, R, and g, determine each of
the following.
i) The acceleration of the block.
ii) The tension in the cord.
iii) The angular momentum of the disk as a
function of t.
Example 4 : AP 1982 #3
A system consists of two small disks, of masses m and 2m,
attached to a rod of negligible mass of length 3l as shown
above. The rod is free to turn about a vertical axis through
point P. The two disks rest on a rough horizontal surface; the
coefficient of friction between the disks and the surface is m.
At time t = 0, the rod has an initial counterclockwise angular
velocity wo about P. The system is gradually brought to rest
by friction. Develop expressions for the following quantities
in terms of m, m, l, g, and wo.
a) The initial angular momentum of the system
about the axis through P.
b) The frictional torque acting on the system
about the axis through P.
c) The time T at which the system will come to
rest.
Example 5 : AP 1996 #3
l
M
Consider a thin uniform rod of mass M and
length l , as shown above.
a) Show that the rotational inertia of the rod
about an axis through its center and
perpendicular to its length is Ml 2/12.
The rod is now glued to a thin hoop of mass M
and radius R = l /2 to form a rigid assembly, as
shown above. The centers of the rod and hoop
coincide at point P. The assembly is mounted
on a horizontal axle through point P and
perpendicular to the page.
b) What is the rotational inertia of the rod-hoop
assembly about the axle ?
Several turns of string are wrapped tightly around
the circumference of the hoop. The system is at
rest when a cat, also of mass M, grabs the free
end of the string and hangs vertically from it
without swinging as it unwinds, causing the rodhoop assembly to rotate. Neglect friction and the
mass of the string.
c) Determine the tension T in the string.
d) Determine the angular acceleration a of the
rod-hoop assembly.
e) Determine the linear acceleration of the cat.
f) After descending a distance H = 5 l /3, the cat lets
go of the string. At that instant, what is the
angular momentum of the cat about point P ?
Lesson 12 : Conservation of Angular
Momentum
The total angular momentum of a system
is constant in both magnitude and
direction if the resultant external torque
acting on the system is zero, that is, if the
system is isolated.
Since St =
dLtot
dt
= 0, Ltot = constant
Lbefore = Lafter
Lbefore = Lafter
Since L = Iw,
Iiwi = Ifwf = constant
When arms are
moved inward, I
decreases.
Since Iw remains
constant, as I
decreases, w must
increase.
Example 1
A horizontal platform in the shape of a circular disk
rotates freely in a horizontal plane about a frictionless
vertical axle. The platform has a mass M = 100 kg and a
radius R = 2.0 m. A student whose mass is m = 60 kg
walks slowly from the rim of the disk toward its center.
If the angular speed of the system is 2.0 rad/s when the
student is at the rim, what is the angular speed when he
reaches a point r = 0.50 m from the center ?
Example 2
In a favorite classroom
demonstration, a student holds the
axle of a spinning bicycle wheel
while seated on a stool that is free
to rotate. The student and stool are
initially at rest while the wheel is
spinning in a horizontal plane with
an initial angular momentum Li that
points upward. When the wheel is
inverted about its center by 180o,
the student and stool start rotating.
In terms of Li, what are the
magnitude and the direction of L for
the student plus stool ?
Example 3
A 2.0 kg disk traveling at 3.0 m/s strikes a 1.0 kg stick of
length 4.0 m that is lying flat on nearly frictionless ice.
Assume that the collision is elastic and that the disk
does not deviate from its original line of motion. Find
the translational speed of the disk, the translational
speed of the stick, and the angular speed of the stick
after the collision. The moment of inertia of the stick
about its center of mass is 1.33 kg.m2.
Example 4 : AP 1987 #3
A 1.0 kg object is moving horizontally with a velocity of
10 m/s, as shown above, when it makes a glancing
collision with the lower end of a bar that was hanging
vertically at rest before the collision. For the system
consisting of the object and bar, linear momentum is not
conserved in this collision, but kinetic energy is
conserved. The bar, which has a length l = 1.2 m and a
mass m = 3.0 kg, is pivoted about the upper end.
Immediately after the collision the object moves with
speed v at an angle q relative to its original direction.
The bar swings freely, and after the collision reaches
a maximum angle of 90o with respect to the vertical.
The moment of inertia of the bar about the pivot is
Ibar = ml 2/3. Ignore all friction.
a) Determine the angular velocity of the bar
immediately after the collision.
b) Determine the speed v of the 1 kg object
immediately after the collision.
c) Determine the magnitude of the angular
momentum of the object about the pivot just
before the collision.
d) Determine the angle q.
Example 5 : AP 1992 #2
Two identical spheres, each of mass M and negligible
radius, are fastened to opposite ends of a rod of negligible
mass and length 2l. This system is initially at rest with the
rod horizontal, as shown above, and is free to rotate about
a frictionless, horizontal axis through the center of the rod
and perpendicular to the plane of the page. A bug, of mass
3M, lands gently on the sphere on the left. Assume that the
size of the bug is small compared to the length of the rod.
Express your answers to all parts of the question in terms
of M, l , and physical constants.
a) Determine the torque about the axis immediately
after the bug lands on the sphere.
b) Determine the angular acceleration of the rodspheres-bug system immediately after the bug
lands.
The rod-spheres-bug system swings about the
axis. At the instant that the rod is vertical, as
shown above, determine each of the following.
c) The angular speed of the bug.
d) The angular momentum of the system.
e) The magnitude and direction of the force that
must be exerted on the bug by the sphere to
keep the bug from being thrown off the
sphere.
Lesson 13 : Rotational Equilibrium
A system is in rotational equilibrium if the
net torque on it is zero about any axis.
St = 0
Since St = Ia,
a=0
St = 0 does not mean an absence of rotational
motion. Object can be rotating at a constant
angular speed.
Example 1
Consider the object
subject to the two
forces shown to the
right. Choose the
correct statement with
regard to this situation.
The object is in
____ force equilibrium but not torque equilibrium.
____ torque equilibrium but not force equilibrium.
____ both force and torque equilibrium.
____ neither force nor torque equilibrium.
Example 2
Consider the object
subject to the three
forces shown to the
right. Choose the
correct statement with
regard to this situation.
The object is in
____ force equilibrium but not torque equilibrium.
____ torque equilibrium but not force equilibrium.
____ both force and torque equilibrium.
____ neither force nor torque equilibrium.
Center of Gravity
To compute the torque due to the
gravitational force on an object of mass M,
we need only consider the force Mg acting
at the center of gravity of the object.
Center of gravity = center of mass if g
is constant over the object.
Example 3
A seesaw consisting of a uniform board of mass M and
length l supports a father and daughter with masses mf
and md, respectively. The support (called the fulcrum) is
under the center of gravity of the board, the father is a
distance d from the center, and the daughter is a distance l
/2 from the center.
a) Determine the magnitude of the upward force n
exerted by the support on the board.
b) Determine where the father should sit to
balance the system.
Example 4
A person holds a 50.0 N sphere in his hand. The forearm is
horizontal, as shown above. The biceps muscle is attached
3.00 cm from the joint, and the sphere is 35.0 cm from the
joint. Find the upward force exerted by the biceps on the
forearm and the downward force exerted by the upper arm
on the forearm and acting at the joint. Neglect the weight
of the forearm.