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General Physics I(aka PHYS 2013)

PROF. VANCHURIN(AKA VITALY)

University of Minnesota, Duluth(aka UMD)

OUTLINE

CHAPTER 9: ROTATION OF RIGID BODIES

I Section 9.1: Angular velocity and accelerationI Section 9.2: Rotation with constant angular accelerationI Section 9.3: Relating linear and angular kinematicsI Section 9.4: Energy in rotational motionI Section 9.5: Parallel-Axis TheoremI Section 9.6: Moment-of-Inertia calculations

SECTION 9.1: ANGULAR VELOCITY / ACCELERATION

I Angular position is describe in terms of radians defined by

2π rad ≡ 360◦ = 1 rev (1)

or1 rad ≡ 360◦

2π=

12π

rev. (2)

I One can convert from one angular unit to another using,

1 =

(2π rad360◦

)1 =

(1rev

2π rad

)1 =

(360◦

1rev

)(3)


I Average linear (velocity and acceleration)

vavg ≡∆x∆t

aavg ≡∆v∆t

(4)

I Instantaneous linear (velocity and acceleration)

v ≡ lim∆t→0

∆x∆t

=dxdt

a ≡ lim∆t→0

∆v∆t

=dvdt

=d2xdt2 (5)

I Average angular (velocity and acceleration)

ωavg ≡∆θ

∆tαavg ≡

∆ω

∆t(6)

I Instantaneous angular (velocity and acceleration)

ω ≡ lim∆t→0

∆θ

∆t=

dθdt

α ≡ lim∆t→0

∆ω

∆t=

dωdt

=d2θ

dt2 . (7)


Example 9.1-9.2. The angular position θ of a 0.36 m-diameter flywheel isgiven by

θ(t) =(

2.0 rad/s3)

t3. (8)

a) Find θ, in radians and in degrees, at t1 = 2.0 s and t2 = 5.0 s.b) Find the total distance (not displacement) that a particle on the flywheelrim moves over the time interval from t1 = 2.0 s to t2 = 5.0 s.c) Find the average angular velocity, in rad/s and in rev/min over thatinterval.d) Find the instantaneous angular velocity at t1 = 2.0 s and t2 = 5.0 s.e) Find the average angular acceleration between t1 = 2.0 s and t2 = 5.0 s.f) Find the instantaneous angular acceleration at t1 = 2.0 s and t2 = 5.0 s.

SECTION 9.2: CONSTANT ANGULAR ACCELERATION

I Angular velocity and acceleration are vectors pointing along theaxis of rotation and direction determined by right-hand rule.

I There is an equivalence of linear and angular quantities

x → θ

vx → ωz

ax → αz.

and so for motion with constant acceleration we have

a(t) = a → α(t) = αz

v(t) = v0 + axt → ω(t) = ω0z + αzt

x = x0 + v0xt +12

axt2 → θ(t) = θ0 + ω0zt +12αzt2

as well useful relations

vx(t)2 = v20x + 2ax(x− x0) → ωz(t)2 = ω2

0z + 2αz (θ(t)− θ0)

x(t)− x0 =12

(v0x + vx(t)) t → θ(t)− θ0 =12

(ω0z + ωz(t)) t.


Example 9.3. You have finished watching a movie Blu-ray and the disk isslowing to a stop. The disc’s angular velocity at t = 0 is 27.5 rad/s, and itsangular acceleration is a constant −10.0 rad/s2. A line PQ on the disc’ssurface lies along the +x-axis at t = 0 s.

a) What is the disc’s angular velocity at t = 0.300 s?b) What angle does the line PQ make with the +x-axis at this time?


Example. Supposed the disc in Example 9.3 was initially spinning at twicethe rate (i.e. 55.0 rad/s) and slowed down at twice the rate (i.e.−20.0 rad/s2).a) Compared to the situation in Example 9.3 how long would it take the discto come to a stop?(i) the same amount of time;(ii) twice as much time;(iii) four times as much time;(iv) one half as much time;(v) one fourth as much timeb) Compared to the situation in Example 9.3 through how many revolutionswould the disc rotate before coming to a stop?(i) the same number of revolutions;(ii) twice as many revolutions;(iii) four times as many revolutions;(iv) one half as many revolutions;(v) one fourth as many revolutions.

SECTION 9.3: RELATING LINEAR AND ANGULAR

KINEMATICS

The linear displacement of a given point P on rotating body is

s = rθ, (9)

where θ must be measured in radians.


KINEMATICS

This suggests that the (magnitude of) linear velocity of the point P is

v =

∣∣∣∣dsdt

∣∣∣∣ =

∣∣∣∣d(rθ)dt

∣∣∣∣ = r∣∣∣∣dθdt

∣∣∣∣ (10)

orv = rω. (11)

and the tangential (or parallel) component of linear acceleration

a|| =dvdt

= rdωdt

= rα. (12)

and the centripetal (or perpendicular) component

a⊥ =v2

r= ω2r. (13)


KINEMATICS

Example 9.4. An athlete whirls a discus in a circle of radius 80.0 cm. At acertain instant, the athlete is rotating at 10.0 rad/s and the angular speed isincreasing at 50.0 rad/s2. At this instant, find the tangential and centripetalcomponents of the acceleration of the discus and the magnitude of theacceleration.


KINEMATICS

Example 9.5. You are designing an airplane propeller that is to turn at2400 rpm (or revolutions per minute). The forward airspeed of the plane isto be 75.0 m/s, and the speed of the tips of the propeller blades through the airmust not exceed 270 m/s. (This is about 80% of the speed of the sound in air.If the speed of the propeller tips were greater than this, they would produce alot of noise.) (a) What is the maximum possible propeller radius? (b) Withthis radius, what is the acceleration of the propeller tip?


KINEMATICS

Example. Information is stored on a disc in a coded pattern of tiny pits. Thepits are arranged in a track that spirals outwards towards the rim of the disc.As the disc spins inside a player, the track is scanned at a constant linearspeed. How must the rotation speed of the disc change as the player’sscanning head moves over the track?(i) The rotation speed must increase.(ii) The rotation speed must decrease.(iii) The rotation speed must stay the same.

SECTION 9.4: ENERGY IN ROTATIONAL MOTION

I When a rigid body rotates around a fixed axis different parts ofit move with different linear velocities and so have differentkinetic energies:

Ki =12

miv2i =

12

mir2i ω

2. (14)

The total (rotational) kinetic energy is then

K =∑

i

Ki =∑ 1

2mir2

i ω2 =

12

(∑mir2

i

)ω2 =

12

Iω2 (15)

whereI ≡

∑mir2

i (16)

is known as the moment of inertia.

I Note that the rotational kinetic energy looks very similar to thetranslational kinetic energy

12

Iω2 ↔ 12

mv2. (17)


Example 9.6. A machine part consists of three discs linked by light-weightstruts.

(a) What is this body’s moment of inertia about an axis through the center ofthe disc A, perpendicular to the plane of the diagram?(b) What is its moment of inertia about an axis through the centers of disksB and C?(c) What is the body’s kinetic energy if it rotates about the axis through Awith angular speed ω = 4.0 rad/s?


For more complicated objects the moment of inertia must be obtained(as we shall see) by volume integration. Here we summarize someuseful results:


Example 9.7. We wrap a light, nonstretching cable around a solid cylinderof mass 50 kg and diameter 0.120 m, which rotates in frictionless bearingsabout a stationary horizontal axis. We pull the free end of the cable with aconstant 9.0−N force for a distance of 2.0 m; it turns the cylinder as itunwinds without slipping. The cylinder is initially at rest. Find its finalangular speed and the final speed of the cable.

SECTION 9.4: ENERGY IN ROTATIONAL MOTIONExample 9.8. We wrap a light, non-stretching cable around a solidcylinder with mass M and radius R. The cylinder rotates withnegligible friction about a stationary horizontal axis. We tie the freeand of the cable to a block of mass m and release the block from rest atdistance h above the floor. As the block falls, the cable unwindswithout stretching or slipping. Find expression for the speed of thecylinder as the block strikes the floor.

SECTION 9.5: PARALLEL-AXIS THEOREM

I There are (infinitely) many axis passing through a given object,and thus there are infinitely many moments of inertia.

I However, some of them are related to each other.

I Consider two axis of rotation:I axis O: passing through the center of mass andI axis D: parallel to the first one, but displaced by distance d.

I Let’s choose z-axis to point along the direction of the first axisand with origin at the center of mass.

I Then the O axis of rotation has coordinates

(x, y, z) = (0, 0, z) (18)

and the D axis of rotation has coordinates

(x, y, z) = (a, b, z) (19)

whered =

√a2 + b2. (20)


I There are (infinitely) many axis passing through a given object,and thus there are infinitely many moments of inertia.

I However, some of them are related to each other.

I Consider two axis of rotation:I axis O: passing through the center of mass andI axis D: parallel to the first one, but displaced by distance d.

I Let’s choose z-axis to point along the direction of the first axisand with origin at the center of mass.

I Then the O axis of rotation has coordinates

(x, y, z) = (0, 0, z) (21)

and the D axis of rotation has coordinates

(x, y, z) = (a, b, z) (22)

whered =

√a2 + b2. (23)


I Then the two moments of inertia are given by

IO =∑

i

mi(x2

i + y2i)

(24)

andID =

∑i

mi

((xi − a)

2+ (yi − b)

2). (25)

I The second equation can be expanded as

ID =∑

i

mi(x2

i + y2i)− 2a

∑i

mixi − 2a∑

i

mixi +(a2 + b2)∑

i

mi

where there third and forth terms vanish.

I If we denote by total mass M =∑

i mi then the final expression is

ID = IO + Md2. (26)

also known as the parallel-axis theorem.


Example 9.9. A part of mechanical linkage has a mass of m = 3.6 kg. Itsmoment of inertia IP about an axis l = 0.15 m from its center of mass isIP = 0.132 kg ·m2. What is the moment of inertia Icm about a parallel axisthrough the center of mass?

SECTION 9.6: MOMENT-OF-INERTIA CALCULATIONS

I In general moment-of-inertia is defined by replacing summationwith integration, i.e.

I =∑

i

r2mi → I =

∫r2dm. (27)

I In the case when the density (or mass per unit volume)

ρ =dmdV

(28)

is independent of position the moment of inertia

I =

∫r2ρdV (29)

orI =

∫r2ρ dx dy dz (30)

where the limits of integration are set by the geometry of thebody.


Example 9.10. Consider a hollow cylinder of uniform mass density ρ withlength L, inner radius R1, and outer radius R2. Find its moment of inertiaabout its axis of symmetry.


Example 9.11. Find the moment of inertia of a solid sphere of uniform massdensity ρ (like a billiard ball) about an axis through its center.

CHAPTER 10: DYNAMICS OF ROTATIONAL MOTION

I Section 10.1: TorqueI Section 10.2: Torque and Angular Acceleration for Rigid

BodyI Section 10.3: Rigid-body Rotation About a Moving AxisI Section 10.4: Work and Power in Rotational MotionI Section 10.5: Angular MomentumI Section 10.6: Conservation of Angular Momentum

SECTION 10.1: TORQUE

I When force acts on an object it can change its translational as wellas rotational motion.

I The effect on the rotational motion depends not only on themagnitude of the applied force, but also to which point the forceis applied.

I For example:


I The relevant physical quantity which measures the twistingeffort of some force ~F applied to point A with respect to someother point B is called torque.

I If we place origin of coordinates to point B then the torque canbe defined as

~τ ≡~r× ~F. (31)

where

~F − force applied to object at point B~r − position of B measured with origin at A.

I It is assume that both vectors ~F and~r lie in the plane orthogonalto the axis of rotation, then torque would point along the axis ofrotation determined by right-hand rule.

I It is a convention to use right hand coordinate system and so thetorque is positive if the force causes counterclockwise rotationand negative if the force cause clockwise rotation.


I Moreover the magnitude of torque is

τ = rF sin θ. (32)

I First Interpretation: If we define lever-arm as distance betweenline of action of force, i.e.

l ≡ r sin θ (33)

where θ is the angle between vectors ~F and~r then the torque isproduct of force and lever-arm

τ = Fl. (34)

I Secondly Interpretation: If we define tangential component ofthe force as

F|| ≡ F sin θ (35)

then torque is product of tangential force to radius of rotation

τ = F||r. (36)


Example 10.1. To loosen a pipe fitting, a weekend plumber slips a piece ofscrap pipe (a “cheater”) over his wrench handle. He stands on the end of thecheater, applying his full 900 N weight at a point 0.80 m from the center ofthe fitting. the wrench and cheater make an angle of 19◦ with horizontal.Find the magnitude and direction of the torque he applies about the center ofthe fitting.

SECTION 10.2: TORQUE FOR RIGID BODY

I Consider a single particle in a rigid body whose displacementfrom axis of rotation is r1, mass is m1 and force acting on it F1.

I Since the only component of the force which contributes torotation is parallel, the second law for this force gives

F1,|| = m1a1,||. (37)

But since the parallel acceleration is

a1,|| = r1αz (38)

and torque isτ1,z = F1,||r1 (39)

we getτ1,z = m1r2

1αz. (40)

I The same applies to every point-like particle in a rigid body, i.e.∑i

τi,z =∑

i

mir2i αz = Iαz. (41)


Example 10.2. (setup as in Example 9.7) We wrap a light, nonstretchingcable around a solid cylinder of mass 50 kg and diameter 0.120 m, whichrotates in frictionless bearings about a stationary horizontal axis. We pullthe free end of the cable with a constant 9.0−N force for a distance of 2.0 m;it turns the cylinder as it unwinds without slipping. The cylinder is initiallyat rest. What is the cable’s acceleration?


Example 10.3. (setup as in Example 9.8) We wrap a light, non-stretchingcable around a solid cylinder with mass M and radius R. The cylinderrotates with negligible friction about a stationary horizontal axis. We tie thefree and of the cable to a block of mass m and release the block from rest atdistance h above the floor. As the block falls, the cable unwinds withoutstretching or slipping. What are the acceleration of falling block and thetension in the cable?

SECTION 10.3: ROTATION ABOUT A MOVING AXIS

I More generally a given rigid body can have both rotationalmotion (about some axis passing through center of mass) andtranslation motion (of the center of mass).

I In this case the total kinetic energy is a sum of rotational andtranslational kinetic energies, i.e.

K =12

Mv2cm +

12

Icmω2. (42)

I This can be shown by imagining the rigid body made up ofmany particles with masses mi moving with velocities ~vi so thatthe total kinetic energy is

K =12

∑i

mi (~vi · ~vi) . (43)

If we now define a relative velocity of each particle with respectto center of mass as

~v′i = ~vi − ~vcm (44)then (42) follows.


I An important example is rolling without slipping.

I Then the center of mass velocity is related to angular velocity

vcm = Rω (45)

I Then the motion is purely rotational about point 1, or arotational around point 0 and translational motion of 0.


I To see that two views give the same kinetic energy we can usethe parallel-axis theorem, i.e.

K =12

I1ω2

=12(Icm + MR2)ω2

=12

Icmω2 +

12

MR2ω2

=12

Icmω2 +

12

Mv2cm. (46)

I If in addition the potential energy changes (as in the case ofgravitational potential energy), then the total mechanical energy

K =12

Mv2cm +

12

Icmω2 + Mgycm (47)

and thus one might think of potential energy coming from apoint particle with mass M located at the body’s center of mass.


Example 10.4. You make a primitive yo-yo by wrapping a massless stringaround a solid cylinder with mass M and radius R. You hold the free end ofthe string stationary and release the cylinder from rest. The string unwindsbut does not slip or stretches the cylinder descends and rotates. Usingenergy considerations, find the speed of the center of mass of the cylinderafter it has descended a distance h.


In a special case when the axis of rotation

I is the axis of symmetry and

I does not change direction,

the combined rotational and translational dynamics can be describedin context of second law, i.e.∑

I

~FI = M~acm∑I

τzI = Icmαz (48)

where the sum in both equations goes only external quantities (i.e.forces and torques).


Example 10.6. You make a primitive yo-yo by wrapping a massless stringaround a solid cylinder with mass M and radius R. You hold the free end ofthe string stationary and release the cylinder from rest. The string unwindsbut does not slip or stretches the cylinder descends and rotates. Finddownward acceleration of the cylinder and the tension in the string.

SECTION 10.4: WORK AND POWER IN ROTATIONAL

MOTION

I Similarly to how translational work is given by an integral

W =

∫ xf

xi

Fx(x)dx (49)

the rotational work is given by

W =

∫F||ds =

∫F||Rdθ =

∫ θf

θi

τz(θ)dθ (50)

I In the case of force independent of angle we have

W = τz(θf − θi) = τz∆θ. (51)


MOTION

I Moreover, the power associated with rotational work is

P =dWdt

= τzdθdt

= τzωz. (52)

I And the integral of (50) can also be rewritten by applying the(rotational) second law,

W =12

Iω2f −

12

Iω2i (53)

i.e expressing work as a change of rotational kinetic energy.


MOTION

Example 10.8. An electric motor exerts a constant τ = 10−N ·m torqueon grindstone which has moment of inertia of I = 2.0 kg·m2 about its shaft.The system starts from rest. Find the work W done by the motor in t = 8.0 sand the grindstone kinetic energy K at this time. What average power Pav isdelivered by the motor?

SECTION 10.5: ANGULAR MOMENTUM

I A rotation analog of linear momentum is angular momentum

~L ≡~r× ~p =~r× (m~v) . (54)

I Then the rate of change change of angular momentum

d~Ldt

=d~rdt× (m~v) +~r×

(m

d~vdt

)= ~r× (m~a) =~r× ~F = ~τ (55)

is nothing but torque of the net force acting on a particle.


I For a rigid body the same is true for every particle and if wechoose the axis of rotation to point along z axis, then

Lz =∑

i

Lzi =∑

i

ri (mivi) =∑

i

ri (miriωz) = ωz. (56)

I If the axis of rotation is also a symmetry axis, then

~L = I~ω. (57)

I If we apply (55) to every particle in a rigid body, then∑I

~τI =d~Ldt

where the sum on the left-hand-side is over all external torques.

I Moreover if the rigid body rotates around a symmetry axis then∑τz = Iαz.


Example 10.9. A turbine fan in a jet engine has a moment of inertia ofI = 2.5 kg·m2 about its axis of rotation. As the turbine starts up, its angularvelocity is given by

ωz =(

40 rad/s3)

t2.

(a) Find the fan’s angular momentum as a function of time and find its valueat t = 3.0s.(b) Find the net torque on the fan as a function of time, and find its value att = 3.0 s.

SECTION 10.6: CONSERVATION OF ANGULAR

MOMENTUM

I When the net external torque acting on a system is zero, the totalangular momentum of the system is constant.

~Lf = ~Li

I This does not mean that the change of angular momentum ofpart of a given body must not changes.


Example 10.10. A physics professor stands at the center of a frictionlessturntable with arms outstretches and a m = 5.0− kg dumbbell in eachhand. He is set rotating about the vertical axis, making one revolution inT = 2.0 s. Find his final angular velocity is he pulls the dumbbells in to hisstomach. His moment of inertia (without the dumbbells) is Ii = 3.0 kg·m2

with arms outstretched and If = 2.2 kg ·m2 with his hands at his stomach.The dumbbells are ri = 1.0 m from the axis initially and rf = 0.20 m at theend.

CHAPTER 11: EQUILIBRIUM AND ELASTICITY

I Section 11.1: Conditions for EquilibriumI Section 11.2: Center of GravityI Section 11.3: Solving Rigid-Body Equilibrium ProblemsI Section 11.4: Stress, Strain and Elastic ModuliI Section 11.5: Elasticity and Plasticity

SECTION 11.1: CONDITIONS FOR EQUILIBRIUM

I An object is said to be in equilibrium if two conditions aresatisfied:

I First condition for equilibrium. The sum of all externalforces acting on the body is zero:∑

~F = 0. (58)

I Second condition for equilibrium. The sum of all torques(due to external forces about any point) on the body is zero:∑

~τ = 0. (59)

I Although we shall only apply these equilibrium conditions tobodies at rest, the same conditions applies to bodies in uniformtranslational motion (without rotation).


Example. Which situation satisfies both the first and second conditions forequilibrium?(i) a seagull gliding at a constant angle below the horizontal and at constantspeed;(ii) an automobile crankshaft turning at an increasing angular speed in theengine of a parked car;(iii) a thrown baseball that does not rotate as it sails through the air.

SECTION 11.2: CENTER OF GRAVITY

I Consider a collection of particles which make up a given object,then their center of mass is then given by

~rcm =m1~r1 + m2~r2 + m3~r3 + ...

m1 + m2 + m3 + ...=

∑i mi~ri

M, (60)

whereM =

∑i

mi. (61)

I Now let us consider gravitational torque caused by gravitationalforces with axis of rotation in the origin

~τ =∑

i

~τi

=∑

i

~ri × (mi~gi)

=∑

i

(~rimi

M

)× (M~gi) (62)


I In general the gravitational acceleration changes with attitude.For example, the center of gravity of 452-m-tall Petronas Towersin Malaysia is 2 cm below the center of mass.

I Most of the time the gravitational acceleration can be assumedto be constant and then the torque is given by

~τ =

(∑i mi~ri

M

)× (M~g)

= ~rcm × (M~g) . (63)

I However. it turns out that even in a more general case when thegravitational acceleration decreases with altitude

~τ =~rcg × (M~g) . (64)

where~rcg is the center of gravity.


I Finding the center of gravity (theoretically) is easy when theobject is symmetric, but what if the object has irregular shape?

I To find the center of gravity experimentally one can suspend anobject from a given point.

.I By repeating the same experiment for two distinct points a pair

of such lines can be constructed and then the point where thelines intersect should be the center of gravity.


Example 11.1. A uniform plank of length L = 6.0 m and mass M = 90 kgrests on sawhorses separated by D = 1.5 m and equidistant from the centerof the plank. Cousin Throckmorton wants to stand on the right-hand end ofthe plank. If the plank is to remain at rest, how massive can Throckmortonbe?


Example. A rock is attached to the left end of a uniform meter stick that hasthe same mass as the rock. In order for the combination of rock in meter stickto balance atop the triangular object (see figure below), how far from the leftend of the stick should the triangular object be placed?

SECTION 11.3: RIGID-BODY EQUILIBRIUM PROBLEMS

For problem in two dimensions we can choose a coordinate system sothat the vectorial equilibrium conditions (58) and (59) are replaced bythree scalar conditions ∑

i

Fi,x = 0 (65)∑i

Fi,y = 0 (66)∑i

τi,z = 0. (67)


Example 11.2. An auto magazine reports that a certain sports car has 53%of its weight on the front wheel. The distance between the axles isD = 2.46 m. How far in front of the rear axle is the car’s center of gravity?(Let’s denote is by Lcg)


Example 11.3. Sir Lancelot, who weights 800 N, is assaulting a castle byclimbing a uniform ladder that is 5.0 m long and weights 180 N. The bottomof the ladder rests on a ledge and leans across the moat in equilibriumagainst a frictionless, vertical castle wall. The ladder makes an angle of 53.1◦

with the horizontal. Lancelot pauses one-third of the way up the ladder.(a) Find normal and friction forces on the base of the ladder.(b) Find the minimum coefficient of static friction needed to prevent slippingat the base.(c) Find the magnitude and direction of the contact force on the base of theladder.

SECTION 11.4: STRESS, STAIN, AND ELASTIC MODULI

I So far we considered (mostly) rigid bodies.

I In reality all objects can deform when external forces are appliedto them, e.g. stretching, squeezing, twisting

I For each type of deformation there are two relevant quantities:I stress which characterizes the cause of the deformationI strain which characterizes the effect of the deformation

I Often the two quantities are directly proportional to each other.

I Coefficient of proportionality is known as elastic modulus,

StressStrain

= Elastic modulus (68)

I We have already seen an example of this (or Hooke’s) law

Fx

= k. (69)

SECTION 11.4: STRESS, STAIN, AND ELASTIC MODULITensile/compressive stress. For stretching/squeezing deformationsthe tensile/compressive stresses are defined as

σ =F⊥A

(70)

and the tensile/compressive strains are defined as

ε =l− l0

l=

∆ll. (71)

The units of the stress is

[Stress] =[Force]

[Distance]2

and in SI system a unit of stress is measures in Pascals,

1 Pa ≡ 1 N/m2.

but the strain is dimensionless.

SECTION 11.4: STRESS, STAIN, AND ELASTIC MODULIYoung’s modulus.

I In the case of tensile/compressive stress the elastic modulus isknown as Young’s modulus

Y =σ

ε=

F⊥/A∆l/l0

=F⊥l0A∆l

. (72)

I This definition insures that Young’s modulus is positive if theperpendicular component of force F⊥ is chosen to pointoutwards:

Tensile: F⊥ > 0 and ∆l > 0 ⇒ Y > 0Compressive: F⊥ < 0 and ∆l < 0 ⇒ Y > 0 (73)

(Note that since the strain was dimensionless, the Young’smodulus has the same units as stress.)

I For many materials the tensile and compressive Young’smodulus are the same and the Hooke’s law works very well.


Here is a table of the modulus for some of the materials

However for composite materials (such as concrete or stone) thecompressive stress can be quite large, while the tensile stress easilybreaks the object and thus the Hooke’s law does not hold.


Example 11.5. A steel 2.0 m long has a cross-sectional area of 0.30 cm2. It ishung by one end from a support, and a 550 kg milling machine is hung fromits other end. Determine the stress on the rod and the resulting strain andelongation.

SECTION 11.4: STRESS, STAIN, AND ELASTIC MODULIBulk stress.

I There situations when the same tensile/compressive stress isapplied to all spatial directions simultaneously.

I Such an isotropic (i.e. the same in all directions) stress is called(sometimes bulk stress) or pressure

p =F⊥A

(74)

and an isotropic strain is called bulk or volume strain

εbulk =∆VV0

. (75)

where ∆V = Vf − Vi.

I Units of pressure

1 Pa ≡ 1 N/m2 1 atm ≈ 105 Pa 1 psi ≈ 6900 Pa. (76)


I Confusing Point #1: The pressure is usually positive which is inconflict with our choice for perpendicular direction to pointoutwards of an object. This can be fixed by either stating that

p = −F⊥A. (77)

I Confusing Point #2: The bulk stress is defined not as pressure butas a negative change in pressure as an object is moved from onesubstance (e.g. air) to another (e.g. water), i.e.

σbulk = ∆p. (78)

where ∆p = pf − pi

I Confusing Point #3: The corresponding elastic modulus (knownas bulk modulus) is defined with a minus sign

B =σbulk

εbulk=−∆p · V0

∆V. (79)

I Two out of three confusion (i.e. #1 and #3) could have beenavoided if the pressure was defined with a minus sign as in (77).


Example 11.6. A hydraulic press contains V0 = 0.25 m3 (250 L) of oil. Findthe decrease in the volume ∆V of the oil when it is subjected to a pressureincrease ∆p = 1.6× 107 Pa (about 160 atm). The bulk modulus of the oil isB = 5.0× 109 Pa (about 5.0× 104 atm) and its compressibility isk = 1/B = 20× 10−6 atm−1.

SECTION 11.4: STRESS, STAIN, AND ELASTIC MODULIShear stress. In addition to tensile/compressive deformations due toperpendicular component of the applied force, a parallel componentof the applied force can cause a deformation.For such deformations we defined shear stress as

σshear =F||A

(80)

shear strain asεshear =

xh

(81)

and shear modulus as

S =σshear

εshear=

F||/Ax/h

=F||hAx

. (82)

If we imagine a shear deformation of rectangular parallelepiped ofdimensions x, y, z then

S =Fxy

xz∆x. (83)


Example 11.7. Suppose a brass base plate of an outdoor sculptureexperiences shear force in an earthquake. The vertically oriented square plateis x = 0.5 cm thick. Whats is the force exerted on each of its edges is theresulting displacement is ∆x = 0.16 mm?


Example A copper rod of cross-sectional area A = 0.500 cm2 and length1.00 m is elongated by 2.00× 10−2 mm, and a steel rod of the samecross-sectional area but 0.100 m in length is elongated by 2.00× 10−3 mm.a) Which rod has greater tensile strain?

I the copper rod

I the steel rod

I the stain is the same

b) Which rod is under greater tensile stress?

I the copper rod

I the steel rod

I the stain is the same

SECTION 11.5: ELASTICITY AND PLASTICITYFor larger deformations the linear dependence of stress as functionsof stain (68) does not hold. A typical dependence for metals undertension is illustrated in

and for rubber under tension is illustrated by

CHAPTER 9: ROTATION OF RIGID BODIES

I Angular quantitiesI Angular position in terms of revs, degrees and radiansI Average angular velocity and accelerationI Instantaneous angular velocity and acceleration

I Constant angular accelerationI Angular velocity and position as functions of timeI Angular velocity as function of angular positionI Angular position as function of angular velocity

I Relating angular and linear quantitiesI Relating linear velocity from angular velocityI Relating tangential acceleration and angular accelerationI Relating centripetal acceleration and angular velocity

I Energy in rotational motionI Definition of rotational kinetic energyI Applications of parallel axis theoremI Calculation of moment of inertia

CHAPTER 10: DYNAMICS OF ROTATIONAL MOTION

I Torque

I Scalar and vectorial definitions of torqueI Lever-arm interpretation of torqueI Tangential interpretation of torque

I Torque for rigid body

I Rotational Newton’s second lawI Rotational and translational kinetic energyI Relation of velocities for motion without slipping

I Application of second law for rotations about a moving axis

I Definitions and calculations of rotational work and power

I Angular momentum

I Definition of angular momentumI Conservation of angular momentum of closed systemsI Change of angular momentum due to external torques

CHAPTER 11: EQUILIBRIUM AND ELASTICITY

I Equilibrium conditionsI Translational equilibrium conditionsI Rotational equilibrium conditionsI Identification of of equilibrium conditions

I Center of gravityI Definition of center of massI Calculation of torque due to gravitational forceI Finding center of gravity (theoretical/experimental)

I 2D equilibrium problemsI Drawing free-body diagram for an equilibrium objectI Applying two translational equilibrium conditionsI Applying a rotational equilibrium condition

I Stress/Stain/ModulusI Tensile,compressive, bulk and shear stressesI Tensile,compressive, bulk and shear strainsI Tensile,compressive, bulk and shear moduli

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