Download Rotational motion and torque notes

Torque
Why is a door knob located as far as possible from the door hinge? When you want to
push open a door, you apply a force. Where you apply the force and in what direction you
push are also important. If you apply the same force at two points, one twice as far from
the point of rotation as the other, the force that is twice as far away has twice the torque
and produces twice the amount of angular acceleration.
Torque
in circular motion, the force applied at a radial distance that changes the direction
of motion of a rotation; torque can stop, start, or change the direction of circular
motion; it is the "unbalanced force" of circular motion
=Fd
where  is the torque in Newton meters ( or N m), F is the perpendicular
component of the applied force, and d is the radial distance (note: you may also
think of d as being r, for radius)
Tangential and radial components of force A force F acts at at an angle at a
point P on a rigid body that is free to rotate about an axis through O a distance r
from the axis of rotation. Only the tangential component of F or FT can have any
effect on the rigid body. FR (the radial component) passes through the axis of
rotation and cannot cause the object to rotate.
The angle F makes with the angle of rotation is called 
Torque () Torque is the "turning agent," or what causes the rotation about an
axis
 = r F sin
Moment Arm (rperpendicular) Perpendicular distance of the line of action of the
force from the axis of rotation.
 = F rperpendicular
Or, you may find the component of the force causing the torque.
Resolve the force into its x and y components. Only the component that is
perpendicular to the lever arm, or moment arm, causes a torque. In the image
below, the vertical component F sin  is perpendicular to the moment arm and
thus causes a torque. The horizontal component F cos  is parallel to the moment
arm and does not cause a torque.
When a torque is applied, rotation occurs around a pivot point, or fulcrum. When
more than one torque acts on a body, the acceleration produced is proportional to
the net torque
Center of gravity
the point at which the entire weight of an object seems to act
See-saw Torque
http://www.explorelearning.com/index.cfm?method=cResource.dspDetail&Resou
rceID=41
Uniform
if an object is considered uniform, its center of gravity is at its geometric center
Rotational equilibrium
an object is said to be in rotational equilibrium when all the torques acting on it
are balanced (or,   = 0). A torque can cause a counterclockwise (cc) or a
clockwise rotation (cw).
cw = cc
where  stands for "sum of"
Center of Mass (CM) The point where all the mass of an object seems to act. For
example, if you look at the motion of a high jumper, there is one special spot that
moves in a parabolic path. It would be the same point where you could balance
that person. An object is in equilibrium as long as its CM is over its base level. An
object is considered uniform when the CM is its geometric center. The CM
position is given by:
where M is the total mass, mi is the mass of the particle and xi is the distance from
the origin
Static Equilibrium
An analysis of static equilibrium is very important in engineering. The design
engineer must identify and isolate all external forces and torques acting on the
structure. With a good design and a correct choice of materials, structures can
support loads. Landing gear of aircraft survive the shock of rough landings and
bridges don't collapse under traffic loads and the wind.
Translational Equilibrium An object is in translational equilibrium (its
momentum is constant) if the sum of the forces acting on it is zero.
 Fx = 0
 Fy = 0
 Fz = 0
Rotational Equilibrium An object is in rotational equilibirum (its angular
momentum is constant) if the sum of the torques acting on it is zero.
An object will be in equilibrium if it is suspended from its center of gravity or its
center of gravity is below the suspension point.
Elasticity The branch of physics which deals with how objects deform when
forces are applied to them.
Elastic Limit The point where the material being deformed suffers permanent
deformation and will not return to its original shape.
There are three ways an object can change its dimensions when forces act on it:
1. An object can be deformed by shearing forces. It will behave like the
pages of a book when under shear. An example: the movement of layers of
rock in an earthquake.
2. An object can be deformed by stretching or compressing forces. An
example: tensile forces that stretch the length of a string until it breaks. An
example: Piling weights on a cylinder until it breaks.
3. An fluid can be deformed by bulk forces. An example: A fluid under high
pressure can be compressed on all sides resuliting in a volume change.
A stress due to forces produces a strain or a deformation. Stress is proportional
to strain and that proportionality cosntant is called its modulus. Stress is the
product of modulus and strain. Stress is the ratio of the force exerted on the object
to the cross-sectional area over which the force is exerted. Strain is the resulting
deformation, whether it is the ratio of change in length to original length, change
in height to original height, or change in volume to original volume.
where F is the force, A is the cross-sectional area in m2, E is Young's modulus, L
is the original length, and  L is the change in length.
where F is the force, A is the cross-sectional area in m2, G is shear modulus, h is
the original height, and  h is the change in height.
where F is the force, A is the cross-sectional area in m2, B is bulk modulus, V is
the original volume, and  V is the change in volume.
Remember, the ratio of F to A is the pressure (P) of the fluid.
The unit for modulus is N/m2 or Pascals (Pa).
Rotational Motion
Rotational motion is the motion of an object around an axis. Up to this point, we
have only studied motion in a straight line (translational motion). Now we will
study motion about an axis, or rotational motion. Objects can move translationally
or rotationally or both. They can be in translational equilibrium (the sum of all the
forces acting on the object is zero), but not in rotational equilibrium (the sum of
all the torques acting on the object is zero), and vice versa. Or, they may be in
both rotational and translational equilibrium.
Rotational Motion The translational motion of a rigid body is analyzed by
describing the motion of its center of mass, as well as rotational motion about its
center of mass. Each particle of a rotating rigid body has, at any moment, a linear
velocity v and a linear acceleration a. The angular velocity is the same for every
point in the rotating body at any instant, but the linear velocity is greater for
points further from the axis of rotation.
Children on a merry-go-round all have different linear speeds (measured in m/s)
depending on how far they are from the axis of rotation. They all have the same
rotational speed (in rev/sec or rad/sec) no matter where they are located.
In the image below, a body is rotating about a fixed axis through its center. An
object placed on the rotating object at point A that rotates to point B rotates
through the same angle as an object placed at point a that rotates to point b. Both
traveled the same angular distance . They did not travel the same tangential
distance. One traveled the arc length AB in time t while the other traveled the arc
length ab in time t.
Angular displacement,
is the angle about the axis through which the object turns. It is measured in
degrees, revolutions, or the SI unit of radians.
1 revolution = 360 = 2  radians
 = d/r
where d is the tangential distance and r is the radius.
Note: in rotational motion, it is easy to use the radius to convert back and forth
between rotational and translational quantities. It is also easy to remember what to
do. Think of the units! If you have a distance in meters, what would you do with
the radius (also in meters) to convert it into radians? You would divide distance in
meters by the radius in meters. Meters cancel leaving radians. A radian is a unit
that serves as a place holder.
Angular Positon The object has rotated through some angle  when it travels the
distance l measured along the circumference of its circular path.
Radian One radian (rad) is defined as the angle subtended by an arc whose length
is equal to the radius. In other words, if l = r, then  is exactly equal to one radian.
Angular speed (or velocity), 
the rate at which an object rotates. The SI unit is rad/sec. In the image above, the
object rotates through angle  in time t. The angular speed or velocity is given by
(remember, in translational motion v = d/t)
 = /t
Angular speed (velocity) can be converted to the analogous translational speed
(velocity) using the radius.
 = v/r
where v is the tangential velocity
Angular velocity (speed) and its relationship to frequency When an object
rotates from some initial position i to some final position f, then its angular
velocity (or speed)  is equal to the change in the angular position  = f - i
divided by the change in time, or  =  / t The angular velocity can be related to
the frequency of rotation, f, where frequency is the number of complete
revolutions per second. Since one revolution per second corresponds to an angle
of 2 radians per second, f =  / 2
Angular acceleration, 
the rate at which a rotating object changes angular speed. The SI unit is rad/s2.
The angular acceleration is the change in angular speed divided by time
(remember, in translational motion, a = (vf - vi)/t)
 = (f - i)/t
Angular acceleration can be converted to the analogous translational acceleration
using the radius.
 = a/r
where a is the tangential acceleration and r is the radius
Angular Acceleration Angular acceleration is the change in angular velocity
divided by the time to make this change. The average angular acceleration,  = 
/t
Radial Component of the linear acceleration The total linear acceleration a is
the vector sum of the radial component of the acceleration and the tangential
component of the acceleration. The radial component of the linear acceleration (or
centripetal acceleration) can be written as aR = 2 r. Thus, the centripetal
acceleration increases the farther you are from the axis of rotation. The children
farthest out on a merry-go-round experience the greatest acceleration.
In the image below, the radial component of the acceleration, aR, represents the
centripetal acceleration. The tangential component of the acceleration, atan,
represents the acceleration that is measured tangent to the circle. The total linear
acceleration of a rotating object is the vector sum of the two components
The equations for linear (tangential or translational) motion can be transformed
into analogous rotational forms:
d=vt
=t
d = do+ vi t + ½ at2
 =  o + i t + ½  t2
vf = vi + at
f = i +  t
vf2 = vi2 + 2 ad
f2 = i2 + 2 
Just as an unbalanced force is required to change the motion of an object in linear
(translational) motion, a torque is required to change the motion of an object in
rotation motional.
=Fr
where r is the radius
Newton’s 2nd law can be transformed into its analogous rotational form:
F = ma
=I
Rolling motion of a wheel or sphere Rolling without slipping involves both
rotation and translation. Remember the relationship between angular velocity of
the rotating object and the linear speed of the axel, or  = v/r. At any moment
when the rotating wheel is in contact with the ground, at that point of contact the
wheel is momentarily at rest. The velcity of the axel equals v; the velocity at the
top of the wheel is 2v.
moment of inertia, I
the rotational inertia of a rotating body. It is analogous to mass in translational
motion. The rotational inertia not only depends upon the rotating body's mass, but
also the distribution of that mass.
Advanced look at moment of inertia: You can think of a rotating rigid body as
conisting of many particles located at different radial distances from the axis of
rotation. The moment of inertia of the rotating body is simply the sum of the
masses of each particle multiplied by the square of the distance of that particle
from the axis of rotation.
I = mr2 = m1r21 + m2r22 + m3r23 + mnr2n
Until this point, our study of physics has been concerned with translational
motion, or motion in the xy plane. The standard English alphabet supplies the
variables for this motion. We use the Greek alphabet for variables to distinguish
rotational motion from translational motion. The following table lists variables for
translational motion and the analogous rotational variable with their SI variables.
distance/displacement
d in m
 in
rad
speed/velocity
v in m/s
 in
rad/s
acceleration
a in m/s2
 in
rad/s2
force
F in Newtons
 in
Nm
mass
m in kg
I in kg m2
Rotational Dynamics
A force is required to make an object start rotating about an axis. The direction of
the force and where it is applied is important. A torque produces angular
acceleration. A torque is required to start the rotation of a body. A rotating rigid
body can be considered as consisting of many particles located at various
distances from the axis of rotation. The sum of the torques due to each of these
particles is just the total torque. The moment of inertia, I, tells how the mass of
the body is distributed about the axis of rotation. In rotational motion,  = I 
Some rotational moments of inertia for solid bodies:
Thin hoop (bicycle wheel or ring) of radius r
I = mr2
Solid disk (solid cylinder, record, or pulley wheel) of radius r
I = 1/2 mr2
Uniform sphere (star) of radius r
I = 2/5 mr2
Long, uniform rod of length L with its axis of rotation through its
center
I = 1/12
mr2
Long, uniform rod of length L with its axis of rotation through one
end
I = 1/3 mr2
Rotational Kinetic Energy
A body rotating about an axis is said to have rotational kinetic energy. It is
analogous to translational kinetic energy. Its SI unit is Joules. An object that
rotates about its center of mass undergoes translational motion that has both
translational and rotational KE, if the axis is fixed. For an object that rolls without
slipping down an incline, the original potential energy is equal to the sum of the
translational kinetic energy and the rotational kinetic energy.
1/2 m v2
1/2 I 2
An object that rotates while its center of mass (CM) translates will have both
rotational and translational kinetic energies. The total kinetic energy of such an
object is given by:
KE = 1/2 mv2CM + 1/2 ICM2
An object sliding down an incline (without rolling) transforms all of its potential
energy into translational kinetic energy. An object that rolls down the inlcine
transforms part of its potential energy to translational kinetic energy and part to
rotational kinetic energy. Consider several objects at the top of an incline: a box
which slides down the incline and a hoop, a solid cylinder, and a sphere which
roll down the incline. Which will reach the bottom first? The sliding box which
converts all its potential energy into translational kinetic energy. The hoop will be
last because it converts the greatest amount of its potential energy into rotational
kinetic energy. The more translational kinetic energy an object has, the faster it
reaches the bottom of the incline.
Angular Momentum and its Conservation
Linear momentum has its analogous quantity, angular momentum, L.
p=mv
L=I
Angular momentum is an important concept because it remains constant if no
external torques are acting. The law of conservation of angular momentum is one
of the conservation laws of physics. If zero net torque is acting on a body and the
body is rotating about a fixed axis, the law of conservation of angular momentum
states
I  = constant.
Many things can be understood when one analyzes their motion using
conservation of angular momentum. An ice skater spinning on the ice with her
arms extended has a large I and a small ; an ice skater spinning on the ice with
her arms pulled close to her body has a small I and a large . Since angular
momentum must be conserved, when her I decreased by bringing her arms close
to her body, her angular speed must increase.
AP Multiple Choice Questions
1. Common questions involve simple torque calculations:
o A plank rests on the edge of a table. How far from the edge can a
weight be placed before the plank tips?
o Determine an unknown weight using torque calculations.
o Use torque calculations to determine where an object should be
suspended to achieve rotational equilibrium.
2. Also, questions that are based upon the concepts asked in AP Torque
Class Problems, problem number one, are asked.
3. Angular momentum is asked as a concept. Usually it is asked as one of the
options for what quantities are conserved.
AP Free Response Questions
1. I have never seen a torque free response question. I have only seen them
asked as multiple choice questions.
Physics AP B Rotational Motion Objectives AP Physics B - Rotational
Motion Objectives

Students should understand the concept of torque so they can:
1. Calculate the magnitude and sense of the torque associated with a given
force.
2. Calculate the torque on a rigid body due to gravity.


Students should be able to analyze problems in statics so they can:
1. State the conditions for translational and rotational equilibrium of a rigid
body.
2. Apply these conditions in analyzing the equilibrium of a rigid body under
the combined influence of a number of coplanar forces applied at different
locations.
Students should understand angular momentum conservation so they can
recognize the conditions under which the law of conservation is applicable and
relate this law to one-and two- particle systems such as satellite orbits or the Bohr
atom.
Torque and Rotational Motion Sample Problems Torque Sample
Problems
1. An 8 N weight hangs at the end of a 4 m uniform board weighing 2 N. The board
pivots at its center. How far must a 10 N weight be hung from the pivot to achieve
equilibrium?
2. A uniform 3 m long board with a mass of 50 kg pivots at its center. A 30 kg mass
is hung 1.6 m from one end. Where must a 40 kg mass be hung to achieve
rotational equilibrium?
3. A 20 N force is applied at the end of a 4 m long massless pole. It is applied at an
angle of 25 degrees. What torque is produced?
4. Now a 35 N force is applied at an angle of 35 degrees at the end of a 3 m long
massless pole. What torque does it produce?
5. A tapered log 3 m long weighs 50 N and has its center of gravity 1 m from its
thick end. It is supported at its center. Where must a 25 N weight be placed to
establish equilibrium?
6. A nonuniform pole 2.5 m long and has its center of gravity 0.4 m from one end.
When a 30 N weight is hung at the opposite end, the pole can pivot at its center.
What does it weigh?
7. A 100 m long uniform bridge weighing 200 N is supported at each end by piers A
and B. A 3500 N weight is 20 m from A and a 9000 N weight is 30 m from B.
How much weight is supported by each pier?
8. Adam and Bob support a uniform pole weighing 8 N that is 1.5 m long. A 18 N
weight is hung 0.4 m from Adam; a 12 N weight is hung 0.55 m from Bob. How
much weight does each support?
9. Find the magnitude of the tension in the wire: It is a uniform bar weighing 12 N.
10. Repeat the previous problem using a 53 degree angle and using a bar which has a
weight of 16 N.
Rotational Motion Sample Problems
1. A wheel spins at 400 rpm. What is its angular velocity in rad/s? If it has a radius
of 10 cm, what is its tangential velocity?
2. Convert 400 rad/sec into a speed in rev/min. Assume the rotating object has a
diameter of 10 cm, what is its tangential velocity?
3. A pulley with a diameter of 5 cm turns at 100 rad/s. Through what angular
distance does it turn in 10 sec? Through what tangential distance? If it is stopped
in 15 sec, what is its rate of angular acceleration? Tangential acceleration?
Through what angular distance did it turn as it stopped? Through what tangential
distance?
4. A wheel with a diameter of 8 cm turns at a constant rate of 80 rad/s for 20
seconds. Through what angular distance did it turn? Through what tangential
distance? Now assume that the wheel is stopped over 5 sec. What was its rate of
angular acceleration? What was its rate of tangential acceleration? Calculate the
angular distance through which it turned while it was being stopped. Calculate the
tangential distance through which it turned while it was being stopped.
5. A wheel turning at 80 rad/s is brought to a stop in 10 sec. What is its rate of
angular acceleration? If the radius is 5 cm, through what tangential distance did it
turn?
6. A 5 N force is applied to a wheel with a moment of inertia of 0.080 kg m2 that has
a radius of 5 cm. What angular acceleration did it cause?
7. A 10 N force is applied to a wheel with a radius of 3 cm causing an acceleration
of 12 rad/s2. What is the moment of inertia of the wheel? Through what tangential
distance did it turn in 5 seconds if it started from rest?
Torque Homework Torque Homework
1. Torque is used to determine the weight of an object. A 10 m long pole, weighing
20 N, has its center of gravity 2 m from one end. An object of unknown weight is
hung at the end opposite the center of gravity. With the object hanging at this end,
the pole now balances 2 m from the unknown weight. What is the weight of the
unknown weight? Ans: 60 N
2. A tapered log 5 m long has its center of gravity 1 m from the thick end. It weighs
20 N. Where must a 30 N weight be hung in order to balance it at its center? Ans:
1 m from center
3. A 10 m long pole balances at its center. A 20 N force is applied upward at one
end. A 35 N weight is hung 3 m from this 20 N force. A 40 N force is applied
upward at the other end. What is the magnitude of the unknown weight that is
hanging 2 m from the 40 N force that causes equilibrium? Ans: 56.67 N
4. A uniform, hinged beam, weighing 20 N and 8 m long is attached to a wall. A 100
N weight is hung at the free end. A rope is attached to beam 2 m from the free
end. If the rope is attached perpendicular to the ceiling and to the beam, what is
the tension in the rope? Ans: 146.67 N
5. Now the rope in number 4 is attached at an angle of 20 to the ceiling. What is the
tension in the rope? Ans: 428.82 N
6. Two men, A and B, carry a 3 m massless pole with its ends horizontally on their
shoulders. A 480 N weight is hung on the pole 1 m from A. A 320 N weight is
hung on the pole 0.6 m from B. How much weight is carried by each man? Ans:
384 N; 416 N
7. A tapered log, 15 long and weighing 300 lb, has its center of gravity 5 ft from the
thick end. The log is supported at its ends. A 150 lb weight is hung from the log 3
ft from its thin end. How much force must be exerted by each of the end supports?
Ans: 230 lb; 220 lb
8. A 100 N force is applied at the end of a massless pole 3 m long. It is applied at an
angle of 30 to the pole. What is the torque produced? Ans: 150 N m
Rotational Motion Homework Rotational Motion Homework
1. What is the angular velocity in rad/s of a flywheel spinning at the rate of 4820
rev/min? Ans: 504 rad/s
2. If a wheel spins at a rate of 625 rad/s, what is its angular velocity in rev/min? Ans:
5970 rev/min
3. A fan blade spins at the rate of 94 rad/s. The blade is 20 cm long. What is the
tangential speed of the fan blade? Through what angular distance does it turn in 5
sec? Through what tangential distance does it turn in 5 sec? Ans: 18.8 m/s; 470
rad; 94 m
4. A flywheel accelerates uniformly from rest to an angular velocity of 94.25 rad/s in
6.0 sec. What is its angular acceleration? Ans: 15.71 rad/s2
5. A solid ball has a moment of inertia of 0.0314 kg m2. It is rotated by applying a
4.7 N force tangentially to it. The radius is 14 cm. What is the angular
acceleration of the ball? Ans: 20.96 rad/s2
AP Torque Problems AP Torque Sample Problems
1. Calculate the torque (magnitude and direction) about point O due to the force F in
each of the situations sketched below. In each case, the object to which the force
is applied has a length of 4 feet and the force is 50 lb. Problems are in order of a
through d (first row, left to right) and e through f (second row, left to right). Ans:
200 ftlb, cc; 173.20 ftlb, cc; 100 ftlb, cc; 86.6 ftlb, cw; 0; 0
2. Calculate the weight of the uniform beam subjected to the forces shown below.
The beam is in rotational equilibrium. Ans: 20.8 N
3. A bowler holds a bowling ball whose mass is 7.2 kg in the palm of his hand. The
upper arm is vertical and his lower arm is horizontal. What forces must the biceps
muscle and the bony structure of the upper arm exert on the lower arm? The
forearm and hand together have a mass of 1.8 kg. The ball is 33 cm from the
elbow; the center of mass of the lower arm is 15 cm from the elbow; and the point
of insertion (where the bicips muscle attaches to the lower arm) is 4 cm from the
elbow. The elbow serves as the pivot. Ans: 648.27 N
4. Three weights are hung from a pole that is 7 m long. A 8 N weight is hung from
one end and a 12 N weight is hung at the other end. The pole pivots 4 m from the
8 N weight and 3 m from the 12 N weight. An unknown weight is hung 2 m from
the pivot on the 8 N side. What is the magnitude of the unknown weight? Ans: 2
N
5. A uniform bar 3 m long and weighing 6 N is attached by a hinge to a wall as
shown below. A 5 N weight is attached to its end. An unknown force is applied at
its end at the angle shown below. What is the magnitude of the unknown force?
What is the value of R, the force with which the wall pushes on the bar? Ans:
12.45 N; 10.00 N