 Chapter 8
 Rotational Equilibrium
 Rotational Dynamics
Rotational Equilibrium
 Torque
 Lever (moment) arm
 Center of gravity (mass)
 Rotational equilibrium
Rotational Dynamics
 Rotational inertia (Moment of inertia)
 Rotational kinetic energy
 Angular momentum
 Focus – what is the cause of the rotational motion
 To make an object to rotate – need force
 Important:
1) Point where force is applied
2) Direction of force
 Torque, τ
 Rotational equivalent of force
 Torque is the tendency of a force to rotate an object about some axis
τ = F r┴
τ - the torque ( ‘tau’ )
F - the force
r┴ - the moment (lever) arm
– perpendicular distance from
the axis of rotation to the line of force
 SI unit: Nm
 Moment Arm, r┴
 The moment arm, r┴, is the perpendicular distance from the axis of rotation to a line drawn along the
direction of the force
r┴ = r sin ϕ
 ϕ – angle between line of force and radius
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 Direction of Torque
 If f = 0° - (force parallel to the radius of the circle of rotation) – zero torque
 If f = 90° - (force perpendicular to the radius of the circle of rotation) – maximum torque
 An Alternative Look at Torque
 The force could be resolved into its
x- and y-components:
 The x-component, F cos Φ, produces
zero torque
 The y-component, F sin Φ, produces a non-zero torque
 From the components of the force or from the lever arm,
τ = F r sinϕ
 F is the force
 r is the distance from force to axis of rotation
 ϕ is the angle between F and r
Torque depends on the tangential force
Torque is completely unrelated to the centripetal force:
- Centripetal force is what keeps the
object moving in a circle
- Torque is related to whether the
angular speed is increasing or
decreasing
 ex 1
125 N is applied to a nut by a wrench (perpendicular to radius). The length of the wrench is 0.300 m.
What is the torque?
 ex 1a
To open a door a force F is applied at a distance r from the axis of rotation. If the torque required to open the
door is
3.1 Nm, what force must be applied if
(a) r = 0.94 m,
(b) r = 0.35 m?
 ex 1b
You push on the door as shown in the drawing. What is the torque?
 ex 1c
A torque of 857 Nm is applied to flywheel that has a radius of 45.5 cm. What is the applied force?
 ex 1d
How far out
on a diving
board would
to exert a torque of 1000 Nm on the board, relative to the left support post?
a 60-kg diver have
 Net Torque
 The net torque is the sum of all the torques produced by all the forces
 Need to account for the direction for rotation:
 Counterclockwise torques are positive
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 Clockwise torques are negative
 ex 2
Find the net torque exerted on the crate if its width is 1.0 m. Assume an axis of rotation through the center of
the crate.
 ex 3
What is the net torque on the door?
 ex 3a
Find the net torque about the axle of the wheel shown. Assume that a
of 0.40 Nm
opposes
the motion.
friction torque
 Torque by force of gravity:
 In finding the torque produced by the force of gravity, all of the weight of the object need to be
considered to be concentrated at one point - Center of Gravity (mass)
 The center of gravity is the point in an object at which all the torques due to gravity vanish:
 you can balance an object at the center of gravity

Στ = 0
 Calculating the Center of Gravity
 The object is divided up into a large number of very small particles of weight (mg)
 Each particle will have a set of coordinates indicating its location (x, y)
 The torque produced by each particle about the axis of rotation is equal to its weight times its lever arm
 The point of application of a single force, whose magnitude is equal to the weight of the object, and
whose effect on the rotation is the same as all the individual particles is called the center of gravity of
the object
 Center of Gravity of a Uniform Object
 The center of gravity of a homogenous symmetric body must lie on the axis of symmetry
 Often - the geometric center of the object
 Experimentally Determining the Center of Gravity
 The wrench is hung freely from two different pivots
 The intersection of the lines indicates the center of gravity
 A rigid object can be balanced by a single force equal in magnitude to its weight as long as the force is
acting upward through the object’s center of gravity
 ex 4
Find x- and y-coordinates of the center of gravity of a 20 cm (height) by 30 cm uniform (same density) sheet of
plywood with the lower right corner (10cm x 20cm) removed.
 Coordinates of the Center of Gravity
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 The coordinates of the center of gravity can be found from the sum of the torques acting on the
individual particles being set equal to the torque produced by the weight of the object
 Conditions of Mechanical Equilibrium:
1 The net external force must be zero
 This is a statement of translational equilibrium
 Necessary, but not sufficient, condition to ensure that an object is in complete mechanical
equilibrium
2 The net external torque must be zero
 To ensure total mechanical equilibrium, you need to ensure rotational equilibrium as well as
translational
 Mechanical Equilibrium
 First Condition of Equilibrium is satisfied
 Second Condition is not satisfied
 Both forces would produce clockwise rotations
 Notes About Equilibrium
 A zero net torque does not mean the absence of rotational motion
 An object that rotates at uniform angular velocity can be under the influence of a zero net torque
 This is analogous to the translational situation where a zero net force does not mean the object
is not in motion
 Axis of Rotation
 If the object is in equilibrium, it does not matter where you put the axis of rotation for calculating the
net torque
 Often the nature of the problem will suggest a convenient location for the axis
 When solving a problem, you must specify an axis of rotation
 Once you have chosen an axis, you must maintain that choice consistently throughout the
problem
 Solving Equilibrium Problems
 Draw a diagram of the system
 Isolate the object being analyzed and draw a free body diagram showing all the external forces acting on
the object
 Problem Solving, cont.
 Establish convenient coordinate axes for each object. Apply the First Condition of Equilibrium
 Choose a convenient rotational axis for calculating the net torque on the object. Apply the Second
Condition of Equilibrium
 Solve the resulting simultaneous equations for all of the unknowns
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 ex5
A child of mass m is supported on a light plank of length L by his parents, who exert forces of F1 and F2.
Find the forces required to keep the plank in static equilibrium. Use the right end of the plank as the axis of
rotation.
 ex6
A 5.00 m long diving board of negligible mass is supported by two pillars. One pillar is at the left end of the
diving board, the other is d = 1.50 m away. Find the forces exerted by the pillars when a 90.0kg diver stands at
the far end of the board.
 ex7
A cat walks along a uniform plank that is 4.00m long and has mass of M=7.00kg. The plank is supported by two
sawhorses, one 0.440m from the left end of the board and the other 1.50m from its right end. When the cat
reaches the right end, the plank just begins to tip (F1 = 0). What is the mass m of the cat?
 ex8
 A uniform 10.m long, 50.N ladder rests against a smooth vertical wall. The ladder is just on the verge of
slipping when it makes a 50° angle with the ground, find the coefficient of static friction between the
ladder and the ground.
 ex9
 If a 600. N person stands 1.50 m from the wall, find the tension in the cable (T) and the force exerted by
the wall on the beam (R).
 ex10
Find the upward force exerted by the bicepts (F) on the forearm and the downward force exerted by the
humerus (R) on the forearm, acting at the joint. Neglect the weight of the arm.
 ex (20.)
A hungry 700. N bear walks out on a uniform beam in an attempt to retrieve some “goodies” hanging at the end.
The beam weighs 200. N, and is 6.00 m long; the goodies weigh 80.0 N.
(a) When the bear is at x = 1.00 m,
components of the
force at the hinge.
find the tension in the wire and
(b) If the wire can withstand a
maximum tension of 900. N, what
the bear
can walk before the wire breaks?
the
is the maximum distance
 ex (25.)
An 8.00 m, 200. N uniform ladder rests against a smooth wall. The coefficient of static friction between the
ladder and the ground is 0.600, and the ladder makes a 50.0° angle with the ground.
How far up the ladder can an 800. N person climb before the ladder begins to slip?
 ex (28.)
One end of a uniform 4.0-m-long rod of weight w is supported by a cable. The other end rests against the wall,
where it is held by friction. The coefficient of static friction
between the wall and
the rod is μs = 0.50.
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Determine the minimum
additional
hung without causing the
distance x from point A
at which an
weight w (same as the
weight of the rod) can be
rod to slip at point A.
 Torque and Angular Acceleration
 When a rigid object has a net torque (τ≠0), it undergoes an angular acceleration a
 The angular acceleration is directly proportional to the net torque
 The relationship is analogous to ∑F = ma (Newton’s Second Law)
 Rotational Inertia, I
 The analogy of the mass in a rotating system:
rotational inertia (moment of inertia) of a point mass
 SI units: kg m2
α = a/r & a = F/m
 Newton’s Second Law for a Rotating Object
 The angular acceleration is directly proportional to the net torque and inversely proportional to the
moment of inertia (rotational inertia) of the object
 Difference between moment of inertia and mass:
Moment of inertia depends on:
 the quantity of matter (mass)
 the distribution of matter in the rigid object
 the location of the axis of rotation
 Moment of Inertia of a Uniform Ring with mass M
 The hoop is divided into a number of small segments (m1, m2, …)
 These segments are equidistant from the axis
 ex1
 Find the moment of inertia of two equal masses at equal distances from the axis of rotation
 ex2
 If the dumbbell-shaped object is rotated about one end, is its moment of inertia
less than, c) the same as the moment of inertia about its center?
a) more than, b)
 ex3
Consider a light rotating rod with 4 equal masses attached to it at distances 2 and 4, as shown. The axis of
rotation is thru the center of the rod, perpendicular to its length. The 4 masses are moved to new positions so
that all masses are now 3 units from the axis.
Did the moment of inertia
a) increase,
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b) decrease,
c) remain the same?
 rolling without slipping
 hollow sphere (red),
cylinder (blue)
solid sphere (orange),
cylindrical ring (green), solid
 The time for each object to reach the finishing line depends on their moment of inertia
 ex4
The hanging blocks are released simultaneously from rest.
Is it observed that
a) the block at
the left lands
b) the block at
right lands first,
c) both at the
same time?
first,
 ex5
 A light rope wrapped around a disk-shaped pulley is pulled tangentially with a force of 0.53 N. Find the
angular acceleration of the pulley, given that its mass is 1.3 kg and its radius is 0.11 m.
 ex6
 A 0.31 kg cart (M) on a horizontal air track is attached to a string, which passes over a disk-shaped
pulley of mass 0.080 kg and radius 0.012 m, and is pulled vertically downward with a constant force of
1.1 N. Find a) tension in the string between the pulley
and the cart,
b) the acceleration
of the cart.
 Rotational Kinetic Energy, Krot
 An object rotating about some axis with an angular speed, ω, has rotational kinetic energy
Krot =½Iω2
Krot depends on the distribution of mass in the rotating object
The farther the mass from the axis of rotation, the greater its rotational kinetic energy
 Kinetic energy of rolling motion
K = ½mv2 + ½Iω2
 Total Energy of a System
 Conservation of Mechanical Energy
 This is for conservative forces, no dissipative (non-conservative) forces such as friction can be
present
 ex7
A solid sphere and a hollow sphere of the same mass and radius roll without slipping at the same speed. Is the
rotational kinetic energy of the solid sphere a) more than, b) less than, or c) the same as the kinetic energy of
the hollow sphere?
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 ex8
 A 1.20 kg disk with a radius of 10.0 cm rolls without slipping. If the linear speed of the disk is 1.41 m/s,
find the translational KE, rotational KE and the total KE.
 ex9
 Derive formula for v of a rolling solid sphere at the bottom of the ramp using conservation of
mechanical energy
 ex10
 A skater pulls in her arms, decreasing her rotational inertia by a factor of two, and doubling her angular
speed. Is her final kinetic energy a) equal to,
b) greater than, or c) less than her initial kinetic energy?
 Object of mass m moving with speed v has linear momentum:
p = mv
 If an object spins or rolls, it has an angular momentum
 The quantity angular momentum tells you how hard it is to stop a spinning object
 Angular Momentum, L
 Object moving with angular speed ω along a circle with radius r has angular momentum :
L=Iω
 SI unit: kg m2/s
 Angular Momentum, cont
 Linear:
 Angular:
 Recall:
 No external force acting on a system, linear momentum p = mv conserved in the system
 If a pluck of mass m slides on the ice, with a given direction and a given speed, it will conserve its
momentum - it will keep the same velocity.
 If 2 plucks collide, we can predict in which direction and at what speed they will go after the
collision because:
ptotal before = ptotal after
 Conservation of Angular Momentum
 The angular momentum of a system is conserved (remains constant) when the net external torque
acting on the systems is zero
Li = Lf
Iiωi = Ifωf
 Kepler: the planets have elliptical orbits around the Sun
II LAW: planets sweep equal area during the same amount of time: faster when they get closer to the
Sun and slower when they are farther away
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 Problem Solving Hints
 The same basic techniques that were used in linear motion can be applied to rotational motion.
 Analogies:
F becomes τ
m becomes I
a becomes α
v becomes ω
x becomes θ
 More Problem Solving Hints
 Techniques for conservation of energy are the same as for linear systems, as long as you include the
rotational kinetic energy
 Problems involving angular momentum are essentially the same technique as those with linear
momentum
 The moment of inertia may change, leading to a change in angular momentum
 The angular momentum is a vector:
is
not only the magnitude but also
applied
the
direction have to be
conserved if no external torque
 This can result in phenomena like the precession of a top, a bicycle wheel processing about a
string
 ex11
A student sits on a piano stool holding a sizable mass in each hand. Initially, he holds his arms outstretched and
spins about the axis of the stool with angular speed of 3.74 rad/s, his moment of inertia 5.33 kg m 2. He then
pulls his arms in to his chest, new moment of inertia 1.60 kg m2.
Find his new angular speed.
 ex12
 The Earth’s gravity exerts no torque on a satellite orbiting the Earth in an elliptical orbit. Compare the
motion at the point nearest the Earth (perigee) to the motion at the point farthest from the Earth
(apogee). At these two points:
a. the tangential velocities are the same.
b. the angular velocities are the same.
c. the angular momentum is the same.
d. the kinetic energies are the same.
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