Ch. 10 Dynamics of Rotational
Motion
AP Physics C
December 7, 2009
Vector Multiplication – Cross Product
A  B  A B sin 
ˆi  ˆi  ˆj  ˆj  kˆ  kˆ  0
ˆi  ˆj  kˆ
A  A x ˆi  A y ˆj  A z kˆ
B  Bx ˆi  By ˆj  Bz kˆ
ˆj  kˆ  ˆi
kˆ  ˆi  ˆj
ˆi
A  B  Ax
ˆj
Ay
kˆ
Az
Bx
By
Bz
Right Hand Rule
ˆi
C  A  B  Ax
Bx
ˆj
Ay
By
kˆ
Az
Bz
Torque
• Definition: a measure of
how much a force acting
on an object causes that
object to rotate
• τ= r x F = r F sin(θ)
Direction of torque:
• Using the right hand rule,
we can find the direction of the
torque vector. If we put our
fingers in the direction of r,
and curl them to the direction
of F, then the thumb points in
the direction of the torque
vector.
Applications of torque:
• Tire Testing - Torque
sensors utilized in tire
development by monitoring
torque in the determination of
friction of tread design.
• Vehicle Safety Testing Wheel torque sensors are
designed for use in automotive
safety development.
• Mining Equipment Equipment design requires a
very high torque capacity.
• Aircraft Engine Design
• Golf Club Design
• Use of a screwdriver
• Use of a wrench
Sample Problem #1
• The pivot point is at the hinges of the door,
opposite to where you were pushing the door.
The force you used was 50N, at a distance 1.0m
from the pivot point. You hit the door
perpendicular to its plane, so the angle between
the door and the direction of force was 90
degrees.
• What is the torque
(magnitude and direction)?
Sample Problem #2
• Several children are playing in the park. One
child pushes the merry-go-round with a force of
50 N. The diameter of the merry-go-round is 3.0
m. What torque does the child apply?
• The child applied a torque of
A. 70 N m into the screen
B. 70 N m out of the screen
C. 141 N m into the screen
D. 141 N m out of the screen
Right-Hand Rule Example #1
• In the following diagram, use the right hand rule
to find the direction of torque.
• The direction of torque is:
A. into the screen
B. out of the screen
C. to the left
D. to the right
E. upwards
F. downwards
Right-Hand Rule Example #2
• In the following diagram, in which direction is
the torque?
• The direction of torque is:
A. into the screen
B. out of the screen
C. to the left
D. to the right
E. upwards
F. downwards
Net Torque
• There may be more than one force acting on an
object, and each of these forces may act on
different point on the object.
• Each force will cause a torque.
• The net torque is the sum of the
individual torques.
Rotational Equilibrium:
• Rotational Equilibrium is analogous to
translational equilibrium, where the sum of the
forces are equal to zero.
• In rotational equilibrium, the sum of the
torques is equal to zero.
• In other words, there is no net torque on
the object.
Sample Problem #3
• The figure shows an object with several forces
acting on it. The pivot point is at O. F1 = 10 N,
and is at a distance of 0.25 m from O, where θ =
80o , F2 = 7.0 N, acting perpendicular to the
object, at a distance of 1.25 m from O, F3 = 12 N,
is 0.60 m from O, and acts at θ = 40o from the
horizontal. Find the total
(net) torque on the object.
Torque and Angular Acceleration
• Imagine a force F acting on some object at a
distance r from its axis of rotation. We can break
up the force into tangential (Ftan), radial (Frad)
• From Newton's Second Law,
Ftan = m atan
• Then, Ftan = m r α
• Multiply both sides by r
(the moment arm), the
equation becomes Ftan r = m r2α
Newton’s 2nd Law of Motion for RM:
• Multiply both sides by r
(the moment arm), the
equation becomes Ftan r = m r2α
• Newton’s 2nd Law of Motion:
Sample Problem #4
• A cable is wrapped around a uniform, solid
cylinder of radius 'R' and mass 'M'. The cylinder
rotates about its axis, and the cable unwinds
without stretching or pulling. If the cable is
pulled with a force of 'F' Newtons, what is its
acceleration?
Sample Problem #5
• You spin a bicycle wheel (diameter of 0.85 m,
mass of 4.5 kg), applying a force of 24 N
tangentially. Assume that the wheel is a thinwalled cylinder.
• Find the angular
acceleration of the wheel.
Rigid Body Rotation about a Moving
Axis
• Complex motion since it is combined translation
and rotation.
• Key to Understanding:
▫ Can be represented as a combination of
translational motion of the center of mass and
rotation about an axis through the center of mass
Rigid Body Rotation about a Moving
Axis:
• The total kinetic energy of the
rolling rigid body is
K 
1
I P 2
2
• Applying the Parallel-Axis
theorem,
1
1 2
2
K  I CM   Mv CM
2
2
Motions of a rolling object:
• Pure translational
▫ All points move forward with the same velocity,
vCM
Motions of a rolling object
• Pure Rotation
▫ All points on the edge of the rolling object have the
same velocity
▫ The CM’s velocity is zero.
Motions of a rolling object
• Combination of translation and rotation
▫ The CM’s velocity is vCM.
▫ The velocity of the point at the bottom is zero.
▫ The velocity of the point at the top is 2vCM
Motions of a Rolling Object:
Angular Momentum
• The product of an object's moment of inertia (its
rotational mass) and its angular velocity.
• A vector quantity represented by the variable, L.
• L = Iω
• The units for angular momentum are: (kg
m2)(radians/sec) = kg m2/sec.
Direction of Angular Momentum
• Use of right-hand
rule.
• When your fingers
curl in the direction
of the object's angular
velocity, your thumb
points in the direction
of the object's angular
momentum.
Angular Momentum for a Point Mass
• L = Iω where I = mr2
and ω = v/r
• L = (mr2)(v/r)
• Lpoint mass = mvr
• Note: r in these equations
represents the radial distance
from the axis of rotation to the
center of mass of the point
mass
Sample Problem #6
• Suppose a turntable rotates at 78 rev/min. What
would be its angular velocity in radians/sec?
• What would be the linear velocity in m/sec of the
ceramic horse located 15 cm from the center of
the record?
• What would be the moment of inertia of this 10gram ceramic horse?
• What is its angular momentum?
Law of Conservation of Angular
Momentum:
• Angular momentum is conserved within a system
whenever there are no external forces exerting torques
on the objects in the system.
• An example of this occurs in skating. A skater spinning
with arms out has a greater moment of inertia, but a
smaller angular velocity compared to when she is
spinning with her arms folded in (I is small, ω is large).
Unless she drags her blades against the ice, her angular
momentum is a constant.
Sample Problem #7
• During her final spin, an ice skater can reduce
her moment of inertia to 33% of its original
value. If her initial rate of rotation is 1 rev/min
when her hands are outstretched (I = 4 kg m2),
then what would be her maximum angular
velocity during her final spin?
Kepler’s Second Law
• Conservation of angular momentum also
justifies the relationship shown in Kepler's
Second Law: a line from the planet to the sun
sweeps out equal areas of space in equal
intervals of time.
Conservation of Angular Momentum
• Remember that gravity is an internal force
within its gravitational system.
• Thus, the satellite’s speed is inversely
proportional to its average distance from the
sun.
Sample Problem #8
• During its trip around the sun, the Earth's
aphelion radius is 1.52 x 108 km and its
perihelion radius is 1.47 x 108 km. If its velocity
at the aphelion is 29.3 km/sec, what is its speed
as it passes through its perihelion position?
Work done by a Constant Torque:
• When the force applied to a particle in rotational
motion is perpendicular to the radius of the
particle.
• The work done is simply W = Fs, where s is the
arc length that the force acts through in a given
period of time.
• Recall that s = rθ; therefore, W = Frθ.
• Or W = τθ
Sample Problem #9
• A single particle of mass 1 kg, starting from rest,
experiences a torque that causes it to accelerate
in a circular path of radius 2 m, completing a full
revolution in 1 second. What is the work done by
the torque over this full revolution?
Work Done by a Varying Torque:
• Determine the area under the torque-angular
displacement graph:
W    d
Power
• W = τθ
• The rate of doing work is defined to be power.
• P = τω
Sample Problem #10
• Often revolving doors have a built in resistance
mechanism to keep the door from rotating
dangerously quickly. A man pushing on a door of
100 kg at a distance of 1 meter from its center
counteracts the resistance mechanism, keeping
the door moving at a constant angular velocity if
he pushes with a force of 40 N. If the door moves
at a constant angular velocity of 5 rad/s, what is
the power output of the man over this time?