chapter 10: rotational motion about a fixed...

Chapter 10: Rotational Motion pAbout a Fixed Axis

We will apply some of the ideas from lineari l ( i l) imotion to circular (rotational) motion.

• Displacement• VelocityVelocity• Acceleration

Angular Position, θFor circular motion, the distance (arc length) s,

the radius r and the angle θ are related by:the radius r, and the angle θ are related by:s

=θr

=θ

θ > 0 for counterclockwise i f f li

Note that θ is

rotation from reference line

π

Note that θ is measured in radians:

DEGRAD θπθ180

=1 rev = 360° = 2π rad

Consider a rotating disk:

Pr s

O P

rO

sθ

t = 0 t > 0

N ti th t th “ di ” i ll di i l tit !!Notice that the “radian” is really a dimensionless quantity!!

s=θ

r=θ

Angular Velocity ω

Notice that as the disk rotates θ changes We

Angular Velocity, ω

Notice that as the disk rotates, θ changes. We define the angular displacement, Δθ, as:

Δθ = θf - θi

which leads to the average angular speed ωav

θθθΔ

if

ifav tt

θθtθ

−

−=

ΔΔ

=ωif tttΔ

Instantaneous Angular VelocityInstantaneous Angular VelocityWe can define the instantaneous angular g

velocity as:

dΔθ θdtd

ΔtΔθ

t

θω ==→Δ

lim0

Note that the SI units of ω are: rad/s = s-1

0 f l k i iω > 0 for counterclockwise rotation

ω < 0 for clockwise rotationω 0 for clockwise rotation

Example UnitsExample - UnitsExpress the angular velocity of the second hand on a clock in the following units: (a) rev/hr (b) deg/min and (c) rad/s.

60 min( ) 1 1 60min minrev rev reva

hr hr= =

360degrees degrees( ) 1 1 360min min minrev revb

rev= =

2 min( ) 1 1min min 60sec 30 secrev rev rad radc

revπ π

= =min min 60sec 30 secrev

Angular Acceleration α

We can also define the average angular

Angular Acceleration, αWe can also define the average angular

acceleration αav:

if ωωΔω −

if

ifav tt

ωωΔtΔωα

−==

anddωωα =

Δ= lim dtt

αΔ→Δ tlim

0

The SI units of α are: rad/s2 = s-2S α

PeriodPeriodThe period, T, of rotation is the time it takes to

l l icomplete one revolution.πω 2

Tω =

Rearranging we haveg g

ωπT 2

=

What is the period of the Earth’s rotation about its own axis?

What is the angular velocity of the Earth’s rotation about itsWhat is the angular velocity of the Earth s rotation about its own axis?

Rotational KinematicsRotational motion is completely analogous to linear motion.

Li R i lLinear Rotational

x θv ωa αα

v = v0 + at ω = ω0 + αtx = x + ½(v + v)t θ = θ + ½(ω + ω)tx = x0 + ½(v0 + v)t θ = θ0 + ½(ω0 + ω)tx = x0 + v0t + ½at2 θ = θ0 + ω0t + ½αt2

2 2 2 2v2 = v02 + 2a(x-x0) ω2 = ω0

2 + 2α(θ-θ0)

ExampleExampleA ceiling fan is rotating at 0.50 rev/s. When turned off it slows uniformly to a stop in 12 s. (a) How many revolutions does it make in this time? (b) Using the result from part (a) find the number of revolutions the fan mustfrom part (a), find the number of revolutions the fan must make for its speed to decrease from 0.5 rev/s to 0.25 rev/s.

0 5 0+⎛ ⎞ ⎛ ⎞0.5 0( ) 12 3 rev2 2

i fa tω ω

θ+⎛ ⎞ +⎛ ⎞= = =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

Let's find the angular deceleration!0 0 5 1Δ

( )2 2

0 0.5 1 rev= t 12 24 s

2

ωα

ω ω α θ θ

Δ −= = −

Δ+ ( )

( ) ( )

0 0

2 2

220.25 0.5

24

ω ω α θ θ

θ

= + −

= −( ) ( )

( ) ( )( )2 2

24( ) 12 0.5 0.25 2.25 revb θ = − =

Rolling Motion (without Slipping)g ( pp g)If a wheel is rolling without slipping, the tangential speedof a point on the outer edge of the wheel is the same as the p gtranslational speed of the wheel. vt = rω

At the point of contact between the wheel and the ground the wheel is instantaneously at rest. Therefore, it is the force of static friction that allows a wheel to roll withoutforce of static friction that allows a wheel to roll without slipping. (Also the static frictional force does no work.)

E lExample

A soccer ball, which has a circumference of 70.0 cm, rolls 12 yards in 3.45 s. What was the average angular speed of the ball during this time?

3.45st;yards12;m11.0 :Given 270.0 =Δ=== m lr π

Δ=

Δ

−=

ΔΔ

= fifavg ttt

θθθθω

( )920yards12so, means slipping without Rolling

m×

= rl

lθ

( )

d/293.45s0.11m

92.0yards12 yardavg ⋅

×=

Δ⋅=

trlω

rad/s 29=avgω

Vector Nature of Angular Quantities

•Both angular velocity and angular accelerationBoth angular velocity and angular acceleration can be considered vectors. ( )αω,

• What is the direction and the sign vector?

The axis of rotation and the right hand rule are used to define the direction of the vector.a e used to de e t e d ect o o t e vecto .

Right-Hand Rule

TorqueTorque

•Everything up to now has been rotational kinematics.

•No e ill disc ssed the effects of forces on•Now we will discussed the effects of forces on rotational motion – dynamics

• Just as every property in linear kinematics had an analogous property in rotational kinematics, so force has an analogous property called torque. (F → τ)

Consider pushing a door closed.

• Far away from the hinge the door closes without much force

• Close to the hinge the door is difficult to close

Hinge

F FF F

Th di h f i li d f h i f• The distance the force is applied from the axis of rotation is called the moment arm.

What happens if the force is applied at an angle that pp pp gis not perpendicular to the moment arm?

Ans. You only take the component of the force acting perpendicular to the moment arm.

F

θF sin θ = F┴

τ = RF┴=RF sin θHinge

θ τ RF┴ RF sin θ

Units: N mR Units: N m

Wh lti l t ti i id b d thWhen multiple torques are acting on a rigid body, the net torque is given by the sum torques (Just like Newton’s second law!)Newton s second law!)

τ τ∑net ii

τ τ= ∑Compare this to,

net ii

F F= ∑

Example24. Determine the net torque on the 2.0 m long beam shown in the figure below Calculate about (a) point C the CM and (b)the figure below. Calculate about (a) point C, the CM, and (b) point P at one end.

Rotational Dynamics;T d R t ti l I tiTorque and Rotational Inertia

Consider a particle of mass m rotates in a

R

F

m

circle. Assume a force, F, acts as shown. The torque that gives rise to the angular acceleration is given by

RF=τ

R macceleration is given by

From Newton’s second law we can write ΣF=ma. Using the relationship between tangential acceleration and angular acceleration a = Rαacceleration, at Rα.

ατα mRmRmaFR

=⇒==×

2

ατ I=

I is called the moment of inertia. If multiple torques are acting on a body, the net torque acting on the system is,

∑ I∑ = ατ I

Compare this equation to Newton’s second law,Co p e s equ o o ew o s seco d w,

∑ = maF∑ maF

What if we have something other than a point particle rotating?What if we have something other than a point particle rotating? We need to know the moment of inertia of that object.

Moment of InertiaRecall that mass (inertia) is an object’s “resistance” to acceleration.

Similarly an object’s “resistance” to rotation (angular acceleration) is known as moment of inertia. For a point mass m:

I = mr2I mrI = moment of inertiar = distance from the axis of rotationFor an extended object:

I Σ 2I =Σmiri2

Mass near the axis of rotation resists rotation less than mass far fromMass near the axis of rotation resists rotation less than mass far from the axis of rotation.

What if you have a continuous body instead of points?

Assume that the body is composed of differential mass points distributed infinitesimally close together The sum then becomesdistributed infinitesimally close together. The sum then becomes and integral and you get

2I R dm= ∫This integral is only possible if the body is geometrically simple.

Look at Example 10-10 in your book to see how this equation is used to calculate the moment of inertia of a hollow cylinderused to calculate the moment of inertia of a hollow cylinder.

Moments of Inertia for Various Objects

The Parallel-axis TheoremIf I is the moment of inertia of a body of mass M about any axis and I is the moment of inertia about an axisany axis, and ICM is the moment of inertia about an axis passing through the center of mass and parallel to the first, but a distance of h away, then, y,

2MhII CM += MhII CM +

Example: Consider a bar rotating about its center of mass

2121 MLICM = 12CM

L

What happens if I rotate it about an end?

2L ( )

21

22

21212

MLI

MMLMhII LCM +=+=

231 MLI =

Perpendicular-axis TheorempThe sum of the moments of inertia of a plane body about two perpendicular axes in the plane of the body is equal to the moment of inertia about an axis through their point of intersection perpendicular to the plane of th t bj tthat object.

yxz III += yxz

This only applies to plane figures, or bodies of uniform thickness whose thickness can be neglected compared tothickness whose thickness can be neglected compared to the other dimensions.

SEE EXAMPLE 10-12 IN BOOK!

Solving Problems in Rotational D iDynamics

1. Draw a clear and complete diagram.p g2. Draw a free-body diagram for the body under consideration

showing all the forces acting on that body. Show where and in what direction the forces actin what direction the forces act.

3. Identify the axis of rotation and calculate the torques about it. Be careful with signs!

4 C l l h d l N ’ d l f4. Calculate the torques and apply Newton’s second law for rotation Στ = Iα. If needed, calculate the moment of inertia.

5. Apply Newton’s second law for translation if needed.pp y6. Solve for the unknowns.7. Is the answer reasonable.

Example 10-8Pulley and bucket Consider a bucket with a weight ofPulley and bucket. Consider a bucket with a weight of 15.0 N (m = 1.53 kg) hanging from a cord wound around a pulley. Assume that the cord does not stretch or slip. (a) p y p ( )Calculate the angular acceleration of the pulley and the linear acceleration of the bucket. (b) Determine the angular velocity, ω, of the bucket at t = 3.0 s if the pulley and the bucket start from rest at t = 0 s.

Note: From example 10-7 we know I = 0.385 kg m2,r = 33 0 cm and τf = 1 10 N mr 33.0 cm and τfr 1.10 N m.

Angular Momentum and ConservationAngular Momentum and Conservation

Angular momentum is defined as,

L Iω=

g ,

The net torque on an object is equal to the time rate of q j qchange of the angular momentum

dL∑ ==dtdLIατ

If there is not net torque on the system then the angular momentum is constant.

∑ dL∑ == 0dtdLτ

Thus angular momentum will be conserved if no gexternal forces are acting on the system. Written mathematically,

fi LL = fi

ExampleExample

55. A person stands, hands at the sides, on a platform that is rotating at a rate of 1.30 rev/s. If the person now raises his g parms to a horizontal position the speed of the rotation decreases to 0.80 rev/s. (a) Why does this occur? (b) By what factor has the moment of inertia changed?factor has the moment of inertia changed?

( ) i i l l i d b(a) His rotational velocity decreases because momentum must be conserved. If I increases then ω decreases.

(b) == ffii IIL ωω

6.180030.1

=== if

II ω

80.0fiI ω

Rotational Kinetic EnergyRotational Kinetic EnergyConsider a rigid object consisting of group of particles rotating about a fixed axis The rotational kinetic energy of theseabout a fixed axis. The rotational kinetic energy of these particles is given by,

( ) ( ) ( )∑∑∑ 22222

So

( ) ( ) ( )∑∑∑ ===i

iii

iii

ii RmRmvmK 222122

212

21 ωω

So,21

2K Iω=

The work done on a rotating object is given by the work-energy theorem,

2 22 21 12 2f iW K I Iω ω= Δ = −

Rotational Plus Translational Kinetic Energy

Total Kinetic Energy = KECM +KErot

T l ti l Ki ti ETranslational Kinetic Energy:2

21

CMCM MvKE =

Rotational Kinetic Energy for an object with a fixed rotation axis:fixed rotation axis:

221 ωIKErot =

So the total kinetic energy for an object moving gy j gtranslationally with a velocity v and rotating about a fixed axis with an angular velocity ω is,

2212

21 Energy Kinetic Total ωIMvCM +=

ExampleExample

When a pitcher throws a curve ball, the ball is given a fairly rapid spin. If a 0.15-kg baseball with a radius of 3 7 i h i h li d f 46 0 / d3.7 cm is thrown with a linear speed of 46.0 m/s and an angular speed of 41 rad/s, how much of its kinetic energy is translational and how much is rotational? Assume theis translational and how much is rotational? Assume the ball is a uniform, solid sphere.

( )221 1 0.15 46 158.7 J2 2transK mv= = × × =

( )22 2

2 21 1 2 412 2 5rotK I m Rω= = × × × ×

( ) ( )2 22

2 2 51 1 2 0.15 0.037 41 0.069 J2 2 5rotK Iω= = × × × × =

ExampleExample

What will be the speed of a solid sphere of mass M andWhat will be the speed of a solid sphere of mass M and radius R when it reaches the bottom of an incline if it starts at a height h and rolls without slipping?

Conservation of energy states, Ei = Ef

( ) 00 2121 +++

+=+

IMM h

PEKEPEKE ffii

( ) 00 2212

21 ++=+ ωIMvMgh

If the ball rolls without slipping then v = rω, Substituting pp g gthis and the expression for the moment of inertia of a sphere in,

2 ⎞⎛ ⎞⎛( ) 002

252

212

21 +⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛+=+

RvMRMvMgh

⎠⎝ ⎠⎝Solving for v,

h10 ghv 710=

Example Problems7. What is the linear speed of a point (a) on the Equator, (b) on

the Arctic Circle (latitude 66.5° N), (c) at a latitude of 40.0°N due to the Earth’s rotation?N, due to the Earth s rotation?

12. A 60.0 cm diameter wheel accelerates uniformly from 210 rpm to 380 rpm in 6.5 s. How far will a point on the edge of the wheel have traveled in this time?

16. The angle through which a rotating wheel has turned in time t is given by the θ = 6.0t – 8.0t2 + 4.5t4, where θ is in radians and t is in seconds Determine an expression (a) for theand t is in seconds. Determine an expression (a) for the instantaneous angular velocity ω and (b) for the instantaneous angular acceleration α. (c) Evaluate ω and α at t = 3.0 s. (d) What is the average angular acceleration between t = 2.0 s and t = 3.0 s?

Examples (cont.)23. Calculate the net torque about the axle of a wheel shown in.

Assume that a friction torque of 0.30 N m opposes the motion.

32. A 0.84 m diameter solid sphere can be rotated about an axis through its center by a torque of 10.8 N m which accelerates it

if l f t th h t t l f 180 l ti i 15uniformly from rest though a total of 180 revolutions in 15 s. What is the mass of the sphere?