Center of GravityCenter of Gravity

The center of gravity is the point at which all of

an objects weight can be considered to beconcentrated.

The center of gravity can be considered to bethe average position of the weight distribution.

Any object will balance if allowed to pivot at itscenter of gravity.

The weight of an object, Fw, behaves as if it is concentrated at its cog.

Fw can be thought of as the resultant of an infinite number of points contributing to an objects weight.

For a uniform object, the cog is located at itsgeometric center.

The cog may be either inside or outside of an object.

The location of the center of gravity determines

the stability of an object.

If a line drawn from the center of gravity

straight down falls inside the base, the object will be stable.

Stable in the sense of not toppling over or

rotating about a point.

In football, lineman line up in a three point stance to become more difficult to tip

over.

Whenever the base of an object is made larger with its cog closer to the base, the object becomes more rotationally stable.

If a net force is applied at the cog, the object will

not rotate but accelerate in a straight line.

Torque is CheapTorque is Cheap

In order to make an object rotate, a force must

be applied but more specifically a torque must

be applied.

Torque is the twisting effect of a force and isgiven by:

Torque is a vector quantity.

T = F × d

where F is the force in N, is theperpendicular distance from the axis of

rotation,and T has units of N•m.

When it comes to objects rotating, think of thehands of a clock.

Tcw Tccw

The hands of a clock, a toy top, or a gyroscope are examples of rotary motion.

d

Torque ProblemTorque Problem

A bar 6.0 m long has its center of gravity 1.8 mfrom the heavy end. If it is placed on the edgeof a block 1.8 m from the light end and a weightof 650 N is added to the light end, the bar is inrotational equilibrium. What is the weight of thebar?

l = 6.0 m F1 = 650 NCOG 1.5 m from heavy end

.

Fw F1

1.8 m1.8 m

6.0 m

Δ

Fup

AORCOG

ΣF = 0

Fup = Fw + F1

TranslationalEquilibrium

ΣT = 0

Tcw = Tccw

RotationalEquilibrium

AOR at the fulcrum.

ΣT = 0

Tcw = Tccw

F1 × d1 = Fw × dw

650 N × 1.8 m = Fw × 2.4 m

Fw = 490 N

Talking About TorqueTalking About Torque

The AOR (axis of rotation) is an arbitrary pointthat can be placed on or off the object.

All perpendicular distances are measured from this point.

The COG (center of gravity) is the point on oroff the object that Fw is considered to be acting.

An example of the COG not found on the object is a hallow ring.

Remember, the distances measured from the

AOR must be perpendicular distances.

For example, if F1 had not been perpendicular to the bar, you would have to resolve F1 to find the component that is perpendicular.

You know the object is not symmetricalbecause the problem torques about a

heavyend.

To maintain translational equilibrium (motion in

a straight line) the ΣF must equal zero.

To maintain rotational equilibrium (rotary motion

about a point), the ΣT must equal zero.

If a force is applied at the COG and the object is

free to move, the object will move in a straight

line.

Another Torque ProblemAnother Torque Problem

A uniform board weighing 360 N and 8.0 m long

rests on a dock with 2.4 m of the boardextending out over the water. How far

from thedock could a 460 N boy walk out on the

boardwithout getting wet?

Fwb = 360 N l = 8.0 mFwboy = 460 N

.

Fwb Fwboy

2.4 m1.6 m

8.0 m

Δ

Fup

ΣF = 0

Fup = Fwb + Fwboy

ΣT = 0

Tcw = Tccw

=360 N + 460 N = 820 N

AOR at the fulcrum.

ΣT = 0

Tcw = Tccw

Fwboy × dwboy = Fwb × dwb

460 N × d = 360 N × 1.6 m

dwboy = 1.3 m

Beyond 1.3 m the Tcw > 0 and both the boy and theboard will fall into the water.

Rotary MotionRotary Motion

Rotary motion is an object spinning on its axis.

Examples include the hands of a clock or a bicycle tire spinning on its axis.

Rotary motion is not as intuitive as circular, projectile, or rectilinear motion.

Rotary motion uses the same terms as rectilinear motion but are defined differently.

Analogous TerminologyAnalogous Terminology

Angular Velocity (ω)

The magnitude of angular velocity is given by

ω is the angular velocity and Δθ is the angular

displacement.

ω=ΔθΔt

The angular displacement, Δθ, is usuallymeasured in radians.

A radian is defined such that length of theintercepted arc of a circle is equal to the

radiusof the circle.

When the length of s equals the

length of the radius (s = r), θ is

equal to 57.3° or 1 radian.

sr

Why 57.3°?

1 revolution = 360° = 2π radians

1 radian = 360°/2/π ≈ 57.3°

Keep in mind that a radian is a dimensionless

quantity.

The direction of the angular velocity is notintuitive either as shown on the following

slide.

Assume as you look at the wheel above that itis rotating counterclockwise in a horizontalplane.

If you curl the fingers of your right hand, yourthumb points in the direction of the angularvelocity (pointing straight, up perpendicular tothe horizontal.

A rotating object can have a constant angularvelocity of zero or be rotating at a constant

rate.

Reminiscent of Newton’s 1st Law.

In accordance with Newton’s 2nd Law, a rotatingmay accelerate by changing its rate of rotation,the direction of the rotation, or both.

Angular acceleration (α) is defined by

with units of rad/s2.α =ΔωΔt

Uniformly Accelerated RotationUniformly Accelerated Rotation

Equations for uniformly accelerated rotarymotion.

ωf = ωi + αΔt

Δθ = ωiΔt + ½αΔt2

ωf2 = ωi

2 + 2αΔθ

Rotary Motion ProblemRotary Motion Problem

A circular saw blade completes1400 revolutions in 60. s while coming to

a stop.

(a) Assuming the deceleration is constant, what

is the angular deceleration in rad/s2?ωave=ΔθΔt

ωave=

1400 rev ×

60. s

2π rad

1 rev=150 rad/s

ωave = (ωi + ωf)/2

ωf = 0; ωi = 2 × ωave = 2 × 150 rad/s = 300 rad/s

ωf = ωi + αΔt

α =0 – 300 rad/s

60. s=- 5.0 rad/s2

Its like two people on a merry-go-round. One is close to the axis of rotation and the other on the outer edge but yet they bothsweep out 360° or make one revolution in the same amount of time.

(d) Do all of the points on the circular saw blade

have the same linear speed (v)?

No because the further a point is away from the center, the faster it must travel to “keep up” with the other points.

(b) Assuming the deceleration is constant, what

was the initial angular speed?

As determined in Part (a), ωi = 300 rad/s.

(c) Do all of the points on the circular saw blade

have the same angular speed (ω)?

Yes, because every point on the blade moves through the same angle in the same time interval.

All circles consist of 360° but the larger the circle the larger the circumference.

The relationship between linear velocity and angular velocity is given by

v = ωr

and radians are dimensionless units so

m/s = 1/s × m = m/s

More Rotary Motion ProblemsMore Rotary Motion Problems

A ball rolls 5.7 m along a horizontal surface

before coming to a stop. If the ball makes12 revolutions, what is the diameter of

the ball?5.7 m =12 rev×1 rev

2 × π × r=75 r

r = 7.6 x 10-2 m

D = 2 × r = 1.5 x 10-1 m

An electric saw blade is spinning at1750 rev/min. If the blade has a diameter

of0.45 m:

(a) What is its angular velocity in rad/s?

ω =1750revmin ×

2πrad1 rev

×1 min60 s

ω =180 rad/s

(b) What is the linear speed of a point on the edge of the saw blade?

v = ωr = 180 rad/s × 0.45 m = 81 m/s

(c) What is the acceleration of a point on the edge of a saw blade?

The tangential acceleration (aT) is given by aT = rα.

The angular velocity is constant, therefore the tangential acceleration (aT) is zero.

The centripetal (radial) acceleration is given by

aR =v2

r =(180 rad/s)2

0.45 m=7.2 x 104 m/s2

aR =v2

r =(wr)2

r = ω2r

The centripetal (radial) acceleration is given by

The total linear acceleration is equal to the

vector sum of a = aT + aR.

Moment of InertiaMoment of Inertia

The moment of inertia is an interesting conceptand a little more involved than just plain old“inertia” found in Newton’s 1st Law.

As with Newton’s 1st Law, a non-rotating wheelwill remain at rest unless a torque is applied.

A wheel rotating at a constant angular velocitywill continue to do so until a net torque isapplied.

To explain the inertia of a rotating object notonly is the mass important but also thedistribution of mass with respect to the axis ofrotation.

A familiar example is that of a figure skaterspinning or rotating in place.

While a figure skater has her arms extended,she rotates rather slowly but she rotates muchfaster when she pulls her arms into her body.

Her mass does not change but the distribution of her mass changes

when she pulls her arms into her body.

The close the mass is to the axis of

rotation, the less rotational inertia an object has.

Rotational inertia is called the moment ofinertia and is symbolized by I (capital eye).

The geometric shape and the location of theaxis of rotation determines the moment ofinertia.

Consider the following three objects all havingthe same mass and their moments of inertia:

A solid sphere, I = 2/5mr2, a solid cylinder, I = ½mr2, and a hollow cylinder, I = mr2.

What happens when they are released at the same time on the same incline from the same height?

The sphere reaches the bottom first, followed by the solid cylinder, and lastly the hollow cylinder.

The hollow cylinder because of its largest moment of inertia will be the hardest to start and stop rotating.

The units for moment of inertia are kg•m2.

Newton’s 2Newton’s 2ndnd Law Law

Newton’s 2nd Law applied to rotating objects is

given by

T = Iα

where T is the torque discussed in thebeginning slides, I is the moment of

inertia, andα is the angular acceleration.

A Moment of Inertia ProblemA Moment of Inertia Problem

A uniform cylinder has a mass of 0.680 kg and

a radius of 9.40 cm.

(a) Calculate its moment of inertia.

m = 0.680 kg r = 9.40 cm

I = ½mr2

I = ½ × 0.680 kg × (9.40 cm × 1 m/102 cm)2

I = 3.00 x 10-3 kg•m2

(b) Calculate the torque needed to accelerate it

from rest to 1700 rpm in 8.00 s.

ωi = 0 ωf = 1700 rev/min Δt = 8.00 s

ωf = ωi + αΔt1700 revmin

× 2πradrev

× 1 min60 s

=

0 + α × 8.00 s

α = 22 rad/s2

T = Iα = 3.00 x 10-3 kg•m2 × 22 rad/s2

T = 6.6 x 10-2 m•N

Angular Impulse and MomentumAngular Impulse and Momentum

Recalling impulse and the change ofmomentum for rectilinear motion:

FnetΔt = Δmv = mΔv

an analogous equation can be written for rotary

motion:

TnetΔt = IΔω

TnetΔt = IΔω

where Tnet is the net torque in N•m, Δt is thetime in s that the Tnet is acting, I is the

momentof inertia in kg•m2, and ω is the angular

velocityin rad/s.

The quantity TnetΔt is the angular impulseexpressed in N•m•s and IΔω is the change

inangular momentum expressed in kg•m2/s.

Just as there is the conservation of linearmomentum given by:

Δp = 0pi = pf

m1v1 + m2v2 = m1v1’ + m2v2’

there is a conservation of angular momentum

given by:

ΔL = 0Li = Lf

I1ω1 = I2ω2

The example of a figure skater has alreadybeen discussed.

Have you ever wondered why it is so mucheasier to balance on a moving bike than astationary one?

Stationary or not, the location of your center of gravity does not change.

Stationary or not, the size of your base does not change unless your fall over.

An object spinning on its axis has a high degree

of stability.

A large torque is needed to change the angular

momentum.

Remember, the direction of the angular

momentum is given by curling the fingers

of your right hand in the direction of rotation.

Angular Momentum ProblemAngular Momentum Problem

A merry-go-round has a mass of 200. kg and aradius of 30. m. A boy whose mass is 50. kgwalks from the outer edge toward the center. Ifthe angular velocity of the merry-go-round is4.0 rad/s when the boy is at the outer edge,what will be the angular velocity of the boywhen he is 10. m from the center?

Assume I = ½mr2 for the merry-go-round andI = mr2 for the boy.

mmgr = 200. kg mb = 20. kg rb = 10. m

rmgr = 30. m ω1 = 4.0 rad/s ω2 = ?

ΔL = 0Li = Lf

I1ω1 = I2ω2

((½mr2)mgr + (mr2)b) × ω12 =

((½mr2)mgr + (mr2)b) × ω2

2

(½ × 200. kg × (30. m)2 + 20. kg × (30. m)2) ×

4.0 rad/s =(½ × 200. kg × (30. m)2 + 20. kg × (10.

m)2) ×ω2

ω2 = 4.7 rad/s

Counter Steering orCounter Steering orCounter Intuitive PhysicsCounter Intuitive Physics

A rotating wheel will resist changes in its state

of rotation when a force is applied to it

• A rotating wheel is a gyroscope

• Newton’s 2nd Law applied to rotational dynamics

• The resistance comes as a result of the I (moment of inertia)

• The applied force produces an unbalanced torque

Gyroscopic precession results when a force isapplied to a rotating mass and the force isredirected 90° in the direction of the rotation

Gyroscopic precession is best illustrated whengoing more than ≈ 20 mi/h on a motorcycle

• Not noticeable on a bicycle. Why?

Pushing the left handlebars causes the handle

bars to rotate to the right resulting in themotorcycle turning to the left

This technique is used when taking turns (leftand right) and evasive maneuvers

A gyroscope does not precess in the absenceof gravity or in a state of free fall

Is this counter intuitive or what?

Let’s explore courtesy of Prentice Hall

When the wheel rotates,the angular velocity (ωs)combined with the torqueproduced by the weight ofthe wheel (Fw) causesprecession.

All vectors lie in thesame horizontal plane. When Δωα is small, vectors α, αΔt, and Δωα are all perpendicular to Δωs

and Δωs’.

Fw

Top Viewccw

Wrap Up QuestionsWrap Up Questions

The wheel shown below has a radius, R, and

rotates counterclockwise.

•A

C

•BR

rA = 0

rB = R/2

rC = R

Do points A, B, and C have the same angularvelocity?

A, B, and C have the same angular velocity.They all trace out the same number ofrevolutions for a given time interval.

Do points A, B, and C have the same linearvelocity?

A, B, and C have different linear velocities.A has the smallest, v = 0 m/s, B has the

nextlargest, v = R/2 × ω, and C has the largest,

v =R × ω.

The formula, v = ω × r, relates the linearvelocity (m/s) to the angular velocity (rad/s)provided that ω is expressed in rad/s.

r is the distance from the axis of rotation.

What is the direction of the angular velocity?

Using the right hand rule, curl the fingers ofyour right hand in the direction of the rotation,and your thumbs points straight out(perpendicular to the plane of the page) whichis the direction of ω.

Do points A, B, and C have the same linearacceleration?

The linear acceleration equals the centripetalacceleration and is given by,

ac = ω2 × r

where ω is in units of rad/s2 and r is thedistance from the axis of rotation.

Point A has an ac = 0, Point B has anac = R/2 × ω2, and Point C has an ac = R × ω2.