classical mechanics lecture 16 today’s concepts: a) rolling kinetic energy b) angular acceleration...

Classical Mechanics Lecture 16

Today’s Concepts:a) Rolling Kinetic Energyb) Angular Acceleration

Physics 211 Lecture 16, Slide 1

Schedule

Mechanics Lecture 14, Slide 2

One unit per lecture!

I will rely on you watching and understanding pre-lecture videos!!!!

Lectures will only contain summary, homework problems, clicker questions, Example exam problems….

Midterm 3 Wed Dec 11

Main Points


Main Points


Rolling without slipping Rv /

Total Kinetic Energy = Translational Kinetic Energy + Rotational Kinetic Energy

Energy Conservation

Rotational Kinetic Energy


H

0KEMgHPE

0PErottrans KKKE

hgv

MvhMgKU

hMgU

tot

7

10

10

70 2

Rolling without slipping

22

2

1;

2

1 IKMvK rottrans

Rv /

2

22

22

10

7

5

1

5

2

2

1

2

1

MvKKK

MvR

vMRIK

rottranstot

rot

Newton’s Second Law

Mg

f a

Acceleration of Rolling Ball


xmafmg sin

Newton’s 2nd Law for rotations

CMCMCMnet I ,

fRCMnet ,

CMCMCMnet IfR ,

CMCMI

fR

R

axCM Rolling without slipping

R

a

MR

fR x2

52

xMaf5

2

xx mamamg 5

2sin

xmamg5

7sin

sin7

5gax

Rolling Motion

Objects of different I rolling down an inclined plane:

v 0

0

K 0

K U M g hR

M

h

v = R

22

2

1

2

1MvIK


Rolling

If there is no slipping:

In the lab reference frame In the CM reference frame

v vv

Where v R


Rolling

Use v R and I cMR2 .

Doesn’t depend on M or R, just on c (the shape)

Hoop: c 1

Disk: c 1/2

Sphere: c 2/5

etc...

2222 12

1

2

1

2

1MvMvMRK c c

v

22

2

1

2

1MvIK

MghMv 212

1So: c c 1

12

ghv


Clicker QuestionA. B. C.


33% 33%33%

A) B) C)

A hula-hoop rolls along the floor without slipping. What is the ratio of its rotational kinetic energy to its translational kinetic energy?

Recall that I MR2 for a hoop about

an axis through its CM:

1trans

rot

K

K

4

3

trans

rot

K

K

2

1

trans

rot

K

K

1

2121

2

1)(

2

1

2

12

1

2

2

2222

2

Mv

Mv

K

K

MvR

vMRIK

MvK

trans

rot

rot

trans

A block and a ball have the same mass and move with the same initial velocity across a floor and then encounter identical ramps. The block slides without friction and the ball rolls without slipping. Which one makes it furthest up the ramp?

A) BlockB) BallC) Both reach the same height.

CheckPoint

v

v


The block slides without friction and the ball rolls without slipping. Which one makes it furthest up the ramp?

A) BlockB) BallC) Same

v

v

B) The ball has more total kinetic energy since it also has rotational kinetic energy. Therefore, it makes it higher up the ramp.


CheckPoint

Rolling vs Sliding


g

vh

MvhMgKU

hMgU

MvKKK

MvR

vMRIK

ball

tot

rottranstot

rot

2

2

2

22

22

10

7

10

70

10

7

5

1

5

2

2

1

2

1

Rolling Ball Sliding Block

g

vh

MvhMgKU

hMgU

MvKKK

K

MvK

block

tot

rottranstot

rot

trans

2

2

2

2

2

1

2

10

2

1

02

1

5

7

105

107

2

2

gvgv

h

h

block

ball Ball goes 40% higher!

Rolling without slipping

22

2

1;

2

1 IKMvK

KKK

rottrans

rottranstot

Rv /

CheckPoint

A cylinder and a hoop have the same mass and radius. They are released at the same time and roll down a ramp without slipping. Which one reaches the bottom first?

A) CylinderB) HoopC) Both reach the bottom

at the same time


A) CylinderB) HoopC) Both reach the bottom

at the same time

A) same PE but the hoop has a larger rotational inertia so more energy will turn into rotational kinetic energy, thus cylinder reaches it first.

Which one reaches the bottom first?


CheckPoint

A small light cylinder and a large heavy cylinder are released at the same time and roll down a ramp without slipping. Which one reaches the bottom first?

A) Small cylinderB) Large cylinderC) Both reach the bottom

at the same time


A small light cylinder and a large heavy cylinder are released at the same time and roll down a ramp without slipping. Which one reaches the bottom first?

A) Small cylinderB) Large cylinderC) Both reach the bottom

at the same time

C) The mass is canceled out in the velocity equation and they are the same shape so they move at the same speed. Therefore, they reach the bottom at the same time.


CheckPoint


vf RM

II

I

Rf

2

52

RM

RgM

R

g

2

5


vf RM

a

M

Mg g

M

Fa MaF

v

a RM

t

Once vR it rollswithout slipping

t

v0R Rt

0v v at t


t

v

v0

atvv 0

Rv Rtatv 0

ga

R

g

2

5

tg

gtv2

50

tg

v2

70

07

2v

gt

v

a RM

Plug in a and t found in parts 2) & 3)


ga

07

2v

gt

20 2

1attvx

atvv 0

v

a RM

We can try this…

Interesting aside: how v is related to v0 :


ga

07

2v

gt

atvv 0

g

vgvv

7

2)( 0

0

00 7

2vvv

Doesn’t depend on 07

5vv

0714.0 vv

v

f RM


2

2

1MvKtran

2

2

1 IKrot 2

5

1Mv

22

5

2

2

1

R

vMR

Atwood's Machine with Massive Pulley:

A pair of masses are hung over a massive disk-shaped pulley as shown.

Find the acceleration of the blocks.

For the hanging masses use F ma

m1g T1 m1a

m2g T2 m2a

For the pulley use IR

aI

T1R T2R MRaR

aI

2

1

(Since for a disk)2

2

1MRI


y

x

m2gm1g

aT1

a

T2

R

M

m1m2

We have three equations and three unknowns (T1, T2, a).

Solve for a.

m1g T1 m1a (1)

m2g T2 m2a (2)

T1T2 (3)Ma2

1

gMmm

mma

221

21


y

x

m2gm1g

aT1

a

T2

R

M

m1m2

Atwood's Machine with Massive Pulley:

Three Masses


amTT

aRmR

aRm

R

aIIRTT

diskspherehoop

diskdiskdisk

diskdiskdisk

diskdiskdiskdiskspherehoopnetdisk

2

1

2

1

2

1 2,

amfamTamfT spherespherespherespherespherespheresphere 5

7

amf

aRmR

aIIRf

spheresphere

spherespheresphere

spherespherespherespherespherenetsphere

5

2

5

2,

)( agmTamTgm hoophoophoophoophoop

spherehoopdisk

hoop

hoopspherehoopdisk

diskspherehoop

mmm

gma

gmmmma

amamagm

57

21

5

7

2

12

1

5

7)(

sphereT

hoopT

sphereT

hoopT

spheref

Three Masses


spherehoopdisk

hoopsphere

mmm

gmaa

5

7

2

1

diskspherehoopdisk

hoop

diskdisk

diskdisk

Rmmm

gm

R

a

R

a

5

7

2

1

spherespherehoopdisk

hoop

spheresphere

spheresphere

Rmmm

gm

R

a

R

a

5

7

2

1

Three Masses


amfamTamfT spherespherespherespherespherespheresphere 5

7

)

57

21

(

)(

spherehoopdisk

hoophoophoop

hoophoophoophoophoop

mmm

gmgmT

agmTamTgm

a

yt

aty

2

2

1 2

Three Masses


yaa

yaatv

a

yt

aty

22

2

2

1 2

tR

at

R

at

spheresphere

spherespheresphere

Three Masses 2


gmmm

mma

amgmT

amTgm

amTT

2

2

1

321

21

222

111

321

Three Masses 2


pulleypulley R

a

agmT 11

agmT 22

2

2

1aty

tpulleypulley

atv

classical mechanics lecture 16 today’s concepts: a) rolling kinetic energy b) angular acceleration...

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