After curving (+ 8 pts)

Inelastic collision : KE is not conserved (~ thermal energy )

However, if system is closed and isolated, the total linear momentum P cannot change (whether the collision is elastic or inelastic !).

Elastic collision : TOTAL KE is conserved (~ Conservative forces )

AND if system is closed and isolated, the total linear momentum P cannot change (whether the collision is elastic or inelastic !).

€

Δ P→

= P→f − P

→i = P

→1 f + ...+ P

→nf

− P

→1i+ ...+ P

→ni

= 0

€

ΔKE = K f −Ki = K1 f + ...+Knf( )− K1i + ...+Kni( ) = 0

€

Δ P→

= P→f − P

→i = P

→1 f + ...+ P

→nf

− P

→1i+ ...+ P

→ni

= 0

€

ΔKE = K f −Ki = K1 f + ...+Knf( )− K1i + ...+Kni( ) ≠ 0

If system is closed and isolated, the total linear momentum P cannot change

Inelastic collision : KE is not conserved (~ thermal energy )

€

P before =

P after 1 - D 2 particles

m1v1i + m2v2 i( ) = m1v1 f + m2v2 f( )

v1 f =m1 − m2

m1 + m2

v1i +2m2

m1 + m2

v2i (Eqn. 9-75)

v2 f =2m1

m1 + m2

v1i +m2 − m1

m1 + m2

v2i (Eqn. 9-76)

Elastic collision : TOTAL KE is conserved (~ Conservative forces )

€

P before =

P after 1 - D

KE before= KEafter 2 particles12 m1v1i

2 + 12 m2v2i

2( ) = 1

2 m1v1 f2 + 1

2 m2v2 f2( )

Chapter 10: Rotation

Chapter 10: Rotation

In this chapter we will study the rotational motion of rigid bodies about a fixed axis. To describe this type of motion we will introduce the following new concepts:

-Angular displacement

-Average and instantaneous angular velocity (symbol: ω ) -Average and instantaneous angular acceleration (symbol: α )

-Rotational inertia, also known as moment of inertia (symbol I ) -Torque (symbol τ )

We will also calculate the kinetic energy associated with rotation, write Newton’s second law for rotational motion, and introduce the work-kinetic energy theorem for rotational motion.

Rotation Variables Here we assume (for now):

Rigid body - all parts are locked together ( e.g. not a cloud) Axis of rotation - a line about which the body rotates (fixed axis) Reference line - rotates perpendicular to rotation axis

looking down Rotation axis

Angle is in units of radians (rad) :

€

1 rev = 360 =2πrr

= 2π rad

Angular velocity of line at time t: (positive if ccw)

€

ω(t)

€

ω(t) =dθ(t)dt

Angular acceleration of line at time t: (direction: depends on ω(t) )

€

α(t)

€

α(t) =dω(t)dt

=d 2θ(t)dt 2

Angular position (displacement) of line at time t: (measured from : ccw)

€

θ(t) Δθ(t)( )

€

+ ˆ x

€

θ(t) =sr

=arc length

radius

€

Δθ = θ2 −θ1

All points move with same angular velocity Units of rad/sec

All points move with same angular acceleration Units of rad/sec2

Are angular quantities Vectors?

Some! Once coordinate system is established for rotational motion, rotational quantities of velocity and acceleration are given by right-hand rule (angular displacement is NOT a vector)

Note: the movement is NOT along the direction of the vector. Instead the body is moving around the direction of the vector

Rotation with Constant Rotational Acceleration

Recall: 1-D translational motion

€

v = v0 + atx = x0 + v0t + 1

2 at2

Rotational motion

€

ω =ω 0 +αtθ = θ0 +ω 0t + 1

2αt2

Problem 10-10: A disk, initially rotating at 120 rad/s, is slowed down with a constant acceleration of magnitude 4 rad/s2.

a)  How much time does the disk take to stop?

b)  Through what angle does the disk rotate during that time?

€

ω =ω 0 +αt ⇒ t =0 −120rad /s( )−4rad /s 2 = 30s

€

θ = θ0 +ω 0t + 12αt

2 ⇒ Δθ = (120rad /s)(30s) + 12 (−4rad /s 2 )(30s)2

= 1800rad = 286.5rev

Relating Translational and Rotational Variables

Rotational position and distance moved

€

s = θ r (only radian units)

Rotational and translational speed

€

v =d s→

dt=

dsdt

=d θdt

r

v = ω r

Relating period and rotational speed

€

T =2πrv

=2πω

units of time (s) €

distance = rate× time[ ]

ωT = 2π

Relating Translational and Rotational Variables

Rotational and translational acceleration

“acceleration” is a little tricky

from change in angular speed

tangential acceleration

€

at =dωdt

r = α r

b) from before we know there’s also a “radial” component

€

ar =v2

r= ω

2r radial acceleration

€

from v = ω r

dvdt

=dωdt

r

a)

€

a tot

2= a r

2+ a t

2

= ω 2r2

+ αr 2

c) must combine two distinct rotational accelerations

Relating Translational and Rotational Variables Rotational position and distance moved

€

s = θ r (only radian units)

Rotational and translational speed

€

v =d r dt

=dsdt

=d θdt

r

v = ω r

Rotational and translational acceleration

tangential acceleration

€

at =dωdt

r = α r

€

ar =v2

r= ω

2r radial acceleration

€

a tot

2= a r

2+ a t

2

= ω 2r2

+ αr 2

€

v = ω × r

€

a t = α × r

€

a r = ω ×

ω × r ( )

€

a tot = a t + a r

€

θ (t) Δ

θ = θ 2 − θ 1( )

€

ω (t) =

d θ (t)dt

€

α (t) =

d ω (t)dt

=d 2 θ (t)dt 2

€

v =dθdt

r = ω r€

s = θ r

€

a tot

2= a r

2+ a t

2

= ω 2r2

+ αr 2

€

v = ω × r

€

a t = α × r

€

a r = ω ×

ω × r ( )

€

a tot = a t + a r

Rotational Kinematics: ( ONLY IF α = constant)

€

ω =ω 0 +αtθ = θ +ω 0t + 1

2αt

Example #1

A beetle rides the rim of a rotating merry-go-round.

If the angular speed of the system is constant, does the beetle have a) radial acceleration and b) tangential acceleration?

If the angular speed is decreasing at a constant rate, does the beetle have a) radial acceleration and b) tangential acceleration?

€

If ω = constant, α = 0

€

If α = neg constant, ω =ω 0 +αt

€

v

€

a t

€

a r

€

a tot

€

ˆ r

€

a tot

2= a r

2+ a t

2

= ω 2r2

+ 0 2 ⇒ a tot =ω 2r(−ˆ r )only radial

€

a tot

2= a r

2+ a t

2

= ω 2r2

+ αr 2

atot = (ω 0 +αt)2r( )2 + αr( )2 both radial and tangential

€

a tot = a t + a r

Checkpoint #1

A ladybug sits at the outer edge of a merry-go- round, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolution once each second. The gentleman bug’s angular speed is

1. half the ladybug’s. 2. the same as the ladybug’s. 3. twice the ladybug’s. 4. impossible to determine

Checkpoint #2

A ladybug sits at the outer edge of a merry-go-round, that is turning and speeding up. At the instant shown in the figure, the tangential component of the ladybug’s (Cartesian) acceleration is:

1.  In the +x direction 2.  In the - x direction 3.  In the +y direction 4.  In the - y direction 5.  In the +z direction 6.  In the - z direction 7.  zero

Checkpoint#3 A ladybug sits at the outer edge of a merry-go- round that is turning and is slowing down. The vector expressing her angular velocity is

1.  In the +x direction 2.  In the - x direction 3.  In the +y direction 4.  In the - y direction 5.  In the +z direction 6.  In the - z direction 7.  zero

Problem10‐7:Thewheelinthepicturehasaradiusof30cmandisrota6ngat2.5rev/sec.Iwanttoshoota20cmlongarrowparalleltotheaxlewithouthi?nganspokes.(a)Whatistheminimumspeed?(b)DoesitmaGerwherebetweentheaxleandrimofthewheelyouaim?Ifsowhatisthebestposi6on.

The arrow must pass through the wheel in less time than it takes for the next spoke to rotate

Δt=

18rev

2.5rev / s= 0.05s

(a) The minimum speed is

vmn= 20cm0.05s

= 400cm / s = 4m / s

(b) No there is no dependence upon the radial position.

Problem10‐30:WheelAofradiusrA=10cmiscoupledbybeltBtowheelCofradiusrC=25cm.TheangularspeedofwheelAisincreasedfromrestataconstantrateof1.6rad/s2.Findthe6meneededforwheelCtoreachanangularspeedof100rev/min.

If the belt does not slip the tangential acceleration of each wheel is the same.rAαA = rCαC

αC =rArCαA = 0.64rad / s2

The angular velocity of C is.

ωC = αCt so t=ωC

αC

=ωCrCrAαA

with ω= 100rev/sec t=16s c