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Chapter 10: RotationAngular position, velocity, accelerationConstant angular accelerationAngular and linear quantities Rotational kinetic energyRotational inertiaTorqueNewton’s 2nd law for rotationWork and rotational kinetic energy

10.1: What is physics?In previous chapters we have discussed the translational motionIn this chapter we will discuss the motion when object turn about an axis (Rotational Motion)Variables in rotational motion are analogous to those for translational motion with few changes We will

Discuss the quantities in angular variables (we will focus on the angle when object rotating) find the angular position, velocity , and acceleration.Apply Newton second law but instead of force and mass we will use torque and rotational inertia.Apply energy concepts to angular quantities like work kinetic energy theorem

10.2: Rotational variablesWe will focus on rotation of a rigid body about a fixed axis

Rigid body: body that can rotate with all its parts packed together without any change in its shape

Fixed axis: rotation about an axis that does not move

Figure shows a rigid body of arbitrary shape in pure rotation about the z- axis of a coordinate System every point of the body moves in a circle whose center lies on the axis of rotation (see the arbitrary reference line), and every point moves through the same angle during a particular time interval

θrs =

rs

=θ in radians (rad.) or revolution

Full circle = 360° = 2π rad. ≡ 1 revolution (دورة). where π = 22/7 = 3.14 we can convert between rad. and degree from the correlation:

(deg) 3602(rad) θπθ =

Degree rad.360° → 2πθ → ??

Consider a particle on rigid object at point p rotates through an angle θ As object rotates, the point P make an Arc length s

10.2: Rotational variables:

Angular position

Radians: ratio between the two length sand r dimensionless quantity

When r = s θ = 1 rad. ≡ 57.3°

2

All particles of a rigid object rotate at the same angular displacement, speed and acceleration.

if θθθ −=∆

ttt if

ifavg ∆

∆=

−

−=

θθθω

dtd

tt

θθω =∆∆

=→∆ 0

lim

ttt if

ifavg ∆

∆=

−

−=

ωωωα

dtd

tt

ωωα =∆∆

=→∆ 0

lim

Angular displacement:

Average angular speed:

Instantaneous angular speed:

Average angular acceleration:

Instantaneous angular

acceleration:

(Rad.)

(Rad/s)≡ s-1

(Rad/s²)≡ s-2

(Rad/s)≡ s-1

(Rad/s²)≡ s-2

As particle moves from angular position θi to θf

10.2: Rotational variables

90°R=6 cm θ1

θ2ω= 7200 rev/min

10.2: Rotational variables: Example

10.2: Rotational variables: ExampleA child’s top is spun with angular accelerationAt t = 0, the top has angular velocity 5 rad/s, and a reference line on it is at angular position (θi) θ = 2 rad. Find (a) angular velocity ω at the top at any time (b) angular position θ at any time

5245

2455

)45(

24

24

0

3

+−=⇒

−=−⇒

−=−⇒

=∆

∫

∫

tt

tt

dttt

dt

t

i

t

t

f

i

ω

ω

ωω

αω

2532

41

532

412

)5245(

35

35

0

24

++−=⇒

+−=−⇒

+−=−⇒

=∆

∫

∫

ttt

ttt

dttt

dt

t

i

t

t

f

i

θ

θ

θθ

ωθa) b)

Angular velocity ω, is a vector it can be written as

For rotation about a fixed axis, the direction of the angular velocity is along the axis of rotation.Use the right hand rule to determine direction.Also angular acceleration α is a vector quantity can be written having same rules for direction and same rules of speeding up rotation or slowing down rotation

ωr

For rotation counterclockwise ( الساعة عقارب بعكس )

+ve ω

For rotation clockwise ( الساعة عقارب مع )

-ve ω

10.3: are angular quantities vectors?

αr

3

For rotational motion with constant rotational acceleration αThe equations of motion are similar in to the equation of motion in one dimension (1D); Only do the following symbol replacement

x ≡ θ

v ≡ ω

a ≡ α

10.4: Rotation with constant angular acceleration

tvvxxx fiif )(21 +=∆=−

t)( 021

0 ωωθθθ +=∆=−

atvv += 0

20 2

1 attvx +=∆

xavv ∆+= 220

2

tvvxx )( 021

0 ++=

tαωω += 0

20 2

1 tt αωθ +=∆

θαωω ∆+= 220

2

t)( 021

0 ωωθθ ++=

Linear (1D) Motionwith constant linear acceleration, a

Rotational Motionwith constant rotational acceleration, α

tvvxxx )( 021

0 +=∆=−

10.4: Rotation with constant angular acceleration

α

??,?,

00.2, 00.2,0,/5.3 2

===∆

====

f

i

t

fti

srevolution

stsradstsrad

ωθ

ωα

10.4: Rotation with constant angular acceleration: Example

. 75.1. 2

. 1 11 revrad

revrad =⎟⎠⎞

⎜⎝⎛=∆π

θ

??,?,

00.2, 00.2,0,/5.3 2

===∆

====

f

i

t

fti

srevolution

stsradstsrad

ωθ

ωα

10.4: Rotation with constant angular acceleration: Example: continued from previous slide

4

10.4: Rotation with constant angular acceleration: Example

While you are operating a Rotor cylinder, the angular velocity of the cylinder from 3.4 rad/s to 2.0 rad/s in 20 rev., at constant angular acceleration. Find (a) the angular acceleration (b) time to decrease the angular speed.We have ωi= 3.4 rad/s ωf = 2 rad/s ∆θ = 20 rev.

rad/s².0304.25156.114

2

22

02

20

2

−=−

=⇒

∆−

=⇒

∆+=

α

θωωα

θαωω .7.125).1

.2.(20 radrevradrevθ ==∆

πwitha)

b)

s.

.t

t

t

7.46030

432

0

0

=−−

=⇒

−=⇒

+=

αωωαωω

θrs =Arc length s:

ωrv =Tangential speed of a point P:

αrat =Tangential acceleration of a point P:

22tr aaa +=

22

ωrrvar ==Centripetal acceleration

for rotation object:

10.5: Relating The Linear And Angular Variables

Magnitude of total acceleration

For a rotating object, both linear and angular quantities are simultaneously exist there must be a relation between them

Rotational axis -out of page

Rotational axis -out of page

A race car accelerates constantly from a speed of 40 m/s to 60 m/s in 5 s around a circular track of radius 400 m. When the car reaches a speed of 50 m/s find the (a) Angular speed (b) Centripetal acceleration, Tangential acceleration, and angular acceleration (d) The magnitude of the total acceleration.

at v = 50 m/s find ω =?, ac =?, at =?, atot=?

srad m/sv

sradrv m/sv

/15.060

/1.04004040

22

111

=⇒=

===⇒=

ω

ωWe have t = 5s,

rad/s 125.040050

===⇒=rvrv ωωa)

222

m/s 25.6400

)50(===

rvac

222 m/s 25.6)125.0(400 === ωraor cb)

10.5: Relating The Linear And Angular Variables: Example

22222 m/s 42.7425.6 =+=+= tc aaa

2/.01.0400

4 sradrara t

t ===⇒= αα

m/s 45

40601212 =

−=

−=⇒+=

tvvatavv tt

at v = 50 m/s at =?, atot=?srad m/svsrad m/sv /15.060,/1.040 2211 =⇒==⇒= ωω

We have t = 5s,

c) From linear quantities we can find the linear acceleration at

Or from angular quantities we can find angular acceleration α

2

21212

/4)01.0(400

/01.05

1.015.0

smra

sradt

t

t ===⇒

=−

=−

=⇒+=

α

ωωααωω

c) The total acceleration

10.5: Relating The Linear And Angular Variables: Example: continued from previous slide

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a collection of n particles rotating about a fixed axis has a rotational kinetic energy of:

2

21 ωIKR =

where I is the moment of inertia or rotational inertia:

2i

iirmI ∑= (for collection of particles)

(J)

(kg.m2)

Mathematically, Similar in shape to linear K with the following replacements I ≡ m, ω ≡ v

22

1

2

1 21

21 ωi

n

iii

n

ii rmvmK ∑∑

==

==

r is the distance from rotational axis

vi is the linear speed for i particle

10.6: Kinetic Energy Of Rotation

Spheres of mass m has I=0 because r=0, they lie on y-axis

10.6: Kinetic Energy Of Rotation: Example

10.6: Kinetic Energy Of Rotation: Example: continued from previous slide

For an extended, rigid object:

∫

∑=⇒

→∆⇒∆=

dmrI

mforrmI ii

ii

2

2 0

∫= dVrI 2ρdVdmIf ρ=

2i

iirmI ∑≡for collection of particles, we had:

m=ρV dm= ρdV

m=σA dm=σdA

m=λL dm=λdL

10.7: Calculating the rotational inertia

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Extended object

; λdL=λdx (L is on x-axis)

and

10.7: Calculating the rotational inertia: Example

ICM : Moment of inertia about an axis of rotation through the center of mass

10.7: Calculating the rotational inertia: Moments of inertia for various objects

If ICM is known, the moment of inertia through a parallel axis of rotation a distance h away from the center of mass is:

2MhII CM +=

22

2

2

31

2121 MLLMMLI

MhII CM

=⎟⎠⎞

⎜⎝⎛+=

+=

10.7: Calculating the rotational inertia: Parallel axis theorem

Consider a rigid object about a pivot point (نقطة ارتكاز).A force is applied to the object.This force causes the object to rotate having what is called Torque τ.

φr

F

Fcosφ

Fsinφ

φτ sinrF=

φ Is angle between F and r directions

10.8: Torque

7

Torque and Angular Acceleration

tt maF =

αrat =

rFt=τ

ατ I=⇒

αττ Inet ==∑

(sin 90° = 1)

but

and

ατ 2mrrFt ==⇒

If more than one force applied to the object

consider a particle of mass m rotating a bout a fixed axis underan influence of applied force FThe component Fr does no torque since (anti-parallel to r) the tangent component Ft has a torque

Newton’s second law in rotation

10.8: Torque

Two forces T1 and T2 are applied as shown

For rotation counterclockwise (بعكس عقارب الساعة)

+ve α

For rotation clockwise ( مع عقاربve- (الساعة α

10.8: Torque: Example

a

α=?

a=?

T=?

10.8: Example: a uniform disk, with mass M = 2.5 kg and radius R = 20 cm, mounted on a fixed horizontal axle. A block with mass m = 1.2 kg hangs from a massless cord that is wrapped around the rim of the disk. Find the acceleration of the falling block, the angular acceleration of the disk, and the tension in the cord. The cord does not slip, and there is no friction at the axle.

22

222

/24,6,/8.4

.05.0)2.0)(5.2(21

21

sradandNTsma

mkgMRI

===⇒

===

α

with

* The rode will move like pendulumunder the effect of Fg=Mg

•Extended object look at the CM

10.8: Example: A uniform rod of length L and mass M is attached as shown. The rod is released from rest in the horizontal position. What are the initial angular acceleration of the rod and the initial translational acceleration of its right end?

MgLrFrF )2

(sin === φτat

Lg

MLMgL

I

MgLIbut

23

3/1)2/(

)2

(

2 ===⇒

==

τα

ατ

solution

α=? and at=?

The translationalacceleration is

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Work in rotational motion

Work in linear motion

vFdt

dWW

ddW

rr

rr

.

sF

sF

==

⋅=

⋅=

P

τωθτ

=∆==

⋅=

PW

τdθdWddW sF

10.8: Work and Rotational Kinetic Energy

22

21

21

if mvmvW∑ −=

Rif KIIW ∆=−=∑ 22

21

21 ωω

The work-kinetic energy theorem for linear motion:

… and for rotational motion:

External work done on an object changes its kinetic energy

External rotational work done on an object changes its rotational kinetic energy

10.8: Work and Rotational Kinetic Energy: Work Kinetic Energy theorem

10.8: Work and Rotational Kinetic Energy: Example

In previous example of disk, if the disk start from rest at time t = 0 . What is its rotational kinetic energy KR at t = 2.5 s?

22

222

/24,6,/8.4

.05.0)2.0)(5.2(21

21

sradandNTsma

mkgMRI

===

===

α

2

21 ωIKR =

rad/s06)5.2(2400 =+=+= tαωω

We need to find ω at t = 2.5s

0J90.05)(60)(21

21 22 ===⇒ ωIKR

From previous example we have

M = 2.5 kg ,radius R = 20 cm, and m = 1.2 kg

10.8: Work and Rotational Kinetic Energy: Example: continued from previous slide

)(0 θτ ∆=−=−=∆

KKKWK

if

R

rad.αttω∆θ

N.m.))(.(RTτ

75)52)(24(210

21

21620

220 =+=+=

===

We need to find τ and ∆θ

0J9)(75)2.1( ==∆=⇒ θτRK

or

M = 2.5 kg ,radius R = 20 cm, and m = 1.2 kg

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ReviewLinear quantities have analogous angular counterparts.

ReviewObject rotating make both linear and angular quantities at same instant there is a relation with angular and linear quantitiesTorque is the tendency of a force to rotate an object.The total kinetic energy of a rotating object has to include its rotational kinetic energy.