Today’s Concepts: a) Rolling Kinetic Energy b) Angular Acceleration

Physics 211Lecture 16

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

Physics 211 Lecture 16, Slide 1

Your Thoughtstranslational motion + rotational motion = DANG, BRO.The race between the hoop, the big cylinder, and the little cylinder confuse me to no end!wow. That was a LOT of information... Are you guys trying to fry us before break or something?I honestly don't know how I feel about these concepts until I start doing the homework because that's when I see what stuck and made sense and what didn't.I would like to see some examples similar to the homework that is coming up .I remember doing this stuff back in high school, my teacher took a can of pop that had liquid in it while the other was frozen solid, both having hte same mass, and he rolled them down a ramp.can we work on a practice problem or two in class...I learn best if I can see something being worked through. I can understand concepts but still be completely lost when it comes to actually doing a problem.My thoughts on this lecture are best summed up in this quote from Ozzy: "SHAAAAAROOOON."I feel like I can visualize this stuff well just from living on this earth for my whole life... Every monday night from 7-9 there is a free improv show at the Illini Union Courtyard Cafe... if you yell out physics lab as a suggestion for the group that goes up at 8, you would get to see a pretty awesome improvised MUSICAL! about physics.... i dare you to go see it march 13th is my birthday. and there's no better way to celebrate than to go to physics lecture! yay!PLEASE play Dirty Deeds Done Dirt Cheap!Stelzer > Selen. True story.

Mechanics Lecture 16, Slide 2


2nd Exam is the week after break:Covers lectures 7-13 (not the stuff we are doing now)

Sign up for the conflict if you need to before the end of this week. As before, email Dr. Jain if you have a double conflict ([email protected]).

Tuesdays class right after break will be a review(we will work out Fall 2010 exam) .

Don’t forget that there are lots of extra office hours during exam week.

mailto:[email protected]?subject=Physics%20211

Rolling

If there is no slipping:


v = R(prelecture slide 5)

Energy Conservation

H

PE MgH=0KE =

0PE =trans rotKE K K=


Two ways of looking at this

F = maa

Mg

f aI a= rr

Dynamics

Energy Conservation

Objects of different I rolling down an inclined plane:

v = 0 = 0 K = 0

RM

h

v = R


K = 0U = M g h

22

21

21 MvIK =

U = 0

0K U = 2 21 12 2

MgH I Mv=

Lets work out the general case…

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 121

21

21 MvMvMRK == c c

Ramp demo

v

22

21

21

MvIK =

MghMv = 2121So: c c 1

12

= ghv


Clicker Question

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:

A) B) C)

1rot

trans

KK

= 34

rot

trans

KK

= 12

rot

trans

KK

=


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

A) The block would have more distance as all the energy is put into moving it translationally, not rotationally.

B) Ball has more initial kinetic energy so goes up higher.C) they both have the same total kinetic energy therefore they will have the same height.


CheckPoint

CheckPointA 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) They have the same potential energy but the hoop has a larger rotational inertia so more energy converted into rotational kinetic energy

B) The moment of inertia is higher on the hoop and it will have more energy going down the ramp.

C) Same mass and radius and both rotate, no difference.

Which one reaches the bottom first?


A) CylinderB) HoopC) Both reach the bottom at the

same time

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 cylinderB) Large cylinderC) Both reach the bottom

at the same time

A) the large cylinder has a large moment of inertia.

B) greater moment of inertia.C) Both cylinders will have the same equation

for rotational inertia, so mass will end up being superfluous, so the fact that one is larger doesn't matter.


Acceleration


Acceleration depends only on the shape, not on mass or radius.

Acceleration



Example ProblemSuppose a cylinder (radius R, mass M) is used as a pulley. Two masses (m1 > m2) are attached to either end of a string that hangs over the pulley, and when the system is released it moves as shown. The string does not slip on the pulley.

What is the acceleration of the blocks?

aT1

a

T2

aR

M

m1m2

Clicker QuestionSuppose a cylinder (radius R, mass M) is used as a pulley. Two masses (m1 > m2) are attached to either end of a string that hangs over the pulley, and when the system is released it moves as shown. The string does not slip on the pulley.

Compare the tension in the string on either side of the pulley:

A) T1 > T2

B) T1 = T2

C) T1 < T2


a2T1

a1

T2

aR

M

m1m2

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

For the pulley use = IaRaI=

T1R - T2RaIR

=


m2gm1g

aT1

a

T2

aR

M

m1m2

(Since for a disk)2

21 MRI =

12

MRa=

m1g - T1 = m1aT2 - m2g = m2a

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

Solve for a.

m1g - T1 = m1a (1)

T2 - m2g = m2a (2)

T1 - T2 (3)Ma21

=

gMmm

mma

-=

221

21


m2gm1g

aT1

a

T2

aR

M

m1m2

Atwood's Machine with Massive Pulley:

I a= fRI

=22

5

MgR

MR

=

52

gR

=Ia =


vf RM

a

F Ma=FaM

= MgM

= g=


vf RM

a

v

a RM

Once v=R it rollswithout slipping

a g=

52

gRa =

v R=

0v at tRa- =

05

2gv gt t- =

07

2gv t

=

02

7t v

g=

t

= at


t

vv0

v = v0 - at

t

v0

R = a Rt

v = v0 - atv = R

v

a RM

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

20

12

x v t at= -

0v v at= -0

27

t vg

=a g=


v

a RM

We can try this…

a g=

02

7t v

g=Interesting aside: how v is related to v0 :

0v v at= -0

02( )

7vv v gg

= -

0 0

27

v v v= -

Doesn’t depend on 057

v v=

00.714v v=Mechanics Lecture 16, Slide 26

Today’s Concepts: a) Rolling Kinetic Energy b) Angular Acceleration

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