Classical Mechanics Lecture 18 Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Mechanics Lecture 18, Slide 1

Classical Mechanics Lecture 18

Today’s Concepts:a) Static Equilibriumb) Potential Energy & Stability

Mechanics Lecture 18, Slide 1

Schedule


One unit per One unit per lecture!lecture!

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

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

Midterm 3 Wed Dec 11 Midterm 3 Wed Dec 11

Lecture Thoughts


Main Points


Gravitational Potential Energy








Main Points


Main Points


footprint

footprint

Stability & Potential Energy


Unstable if CM is “outside of footprint”

CM must be lifted for the object to topple over

If CM is inside the footprint

Object is stable

Statics Problems


Statics Problems


r

r

gMr

Into the screen

Tr

Tr

TrT

Out of the screen

02

sin0

0

LMgTL

weighttensionhinge

http://hyperphysics.phy-astr.gsu.edu/hbase/torcon.html

http://hyperphysics.phy-astr.gsu.edu/hbase/torcon.html

Static Equilibrium


Clicker QuestionA. B. C. D. E.


20% 20% 20%20%20%

A (static) mobile hangs as shown below. The rods are massless and have lengths as indicated. The mass of the ball at the bottom right is 1 kg.

What is the total mass of the mobile?

A) 4 kg B) 5 kg

C) 6 kg

D) 7 kg

E) 8 kg 1 kg

1 m 3 m

1 m 2 m

Clicker QuestionA. B. C.


33% 33%33%

In which of the static cases shown below is the tension in the supporting wire bigger? In both cases M is the same, and the blue strut is massless.

A) Case 1 B) Case 2 C) Same

M

300

L/2

T2

Case 2

M

300

L

T1

Case 1

ML

T1

d1

Case 1

Balancing Torques

11dTMgL sin1LT 222dT

LMg sin

22

LT

sin1

MgT

sin2

MgT

It’s the same. Why?


M

T2 d2

Case 2

L/2

CM

Mg

T1T2 T1 T2 TSame distance from CM:

T Mg/2So:


T1 T2MgBalance forces:

Homework Problem


Homework Problem

These are the quantities we want to find:


d

Mg y

x

T1

M

ACM

What is the moment of inertia of the beam about the rotation axis shown by the blue dot?

A)

B)

C)

L

Clicker Question


d

M2

12

1MLI

2MdI

22

12

1MdMLI

The center of mass of the beam accelerates downward. Use this fact to figure out how T1 compares to weight of the beam?

A) T1 Mg

B) T1 Mg

C) T1Mg

Clicker Question


T1

Mgy

x

ACM

d

M

The center of mass of the beam accelerates downward. How is this acceleration related to the angular acceleration of the beam?

A) ACM d

B) ACMd /

C) ACM / d

Clicker Question


T1

Mgy

x

d

M

ACM

Apply

Apply

Plug this into the expression for T1

ACM d

Use ACM dto find

Iext


T1

Mgy

x

d

M

ACM

CMext MAF

CMMATMg 1

CMMAMgT 1

d

AIIMgd CM

I

dMgACM

2

After the right string is cut, the meterstick swings down to where it is vertical for an instant before it swings back up in the other direction.

What is the angular speed when the meter stick is vertical?


M

y

x

d

T

Conserve energy:2

2

1 IMgd

I

Mgd2

y

xMg

dACM 2d

Centripetal acceleration


T

Applying CMext MAF

dMMgT 2

dMMgT 2

Another HW problem:

We will now work out the general case


General Case of a Person on a Ladder

Bill (mass m) is climbing a ladder (length L, mass M) that leans against a smooth wall (no friction between wall and ladder). A frictional force f between the ladder and the floor keeps it from slipping. The angle between the ladder and the wall is .

How does f depend on the angle of the ladder and Bill’s distance up the ladder?

L

m

f

y

x

MBill


Balance forces:

x: Fwall f

y: N Mg mg

Balance torques:

Mg

mg

L/2

f y

xN

Fwall

daxis

cot2

Mg

L

dmgFwall


wallF f

cot2

Mg

L

dmgf

cosmgd cos2

LMg sinLFwall 0

f

d

M

m

Lets try it out

Climbing further up the ladder makes it more likely to slip:

Making the ladder more vertical makes it less likely to slip:

This is the General Expression:


cot2

Mg

L

dmgf

cot2

Mg

L

dmgf

Moving the bottom of the ladder further from the wall makes it more likely to slip:

If its just a ladder…


cot2

Mgf

In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the ground the biggest?


CheckPoint

Case 1 Case 2Mechanics Lecture 18, Slide 34

In the two cases shown below identical ladders are leaning against frictionless walls. In which case is the force of friction between the ladder and the ground the biggest?


Case 1 Case 2A) Because the bottom of the ladder is further away from the wall.

B) The angle is steeper, which means there is more normal force and thus more friction.

C) Both have same mass.


CheckPoint

CheckPoint

Suppose you hang one end of a beam from the ceiling by a rope and the bottom of the beam rests on a frictionless sheet of ice. The center of mass of the beam is marked with an black spot. Which of the following configurations best represents the equilibrium condition of this setup?

A) B) C)


CheckPointWhich of the following configurations best represents the equilibrium condition of this setup?

A) B) C)

If the tension has any horizontal component, the beam will accelerate in the horizontal direction.

Objects tend to attempt to minimize their potential energy as much as possible. In Case C, the center of mass is lowest.



Homework


Choose axis for torque calculation

Decompose forces into x,y components

Draw Free Body Diagram

0

0,

xTN

ixF

Sum x-component forces

Sum y-component forces

0

0,

RLy FFTf

iyFDefine torques

00

00

90sin

0180sin

90sin

T

F

WFWF

NL

fLfL

T

RF

LLF

rodN

rodrodf

R

L

Sum of Torques

0

0

Lrod

TFFNf

WFfLRL

tan

sin

cos

x

y

y

x

T

T

TT

TT

Relationship between Tx and Ty

Definition of coefficient of friction

N

f


tan/

0

0

0

xy

Lrod

RLy

x

TT

WFfL

FFTf

TN

2sign

RL

WFF

tan

02

0

0

x

y

signrod

signy

x

T

T

WW

fL

WTf

TN

tan

)2

1(

tan

)2

1(

tan/

)2

1(2

2

signrod

x

signrod

yx

signrodrod

signsigny

rod

sign

x

WL

W

TN

WL

W

TT

WL

WW

L

WWT

WL

Wf

TN

)2(

tan

)2

1(

tan2

WL

W

WLW

WL

W

N

f

rodsign

rod

rod

sign

Documents

Classical Mechanics Lecture 18 Today’s Concepts: a) Static Equilibrium b) Potential Energy & Stability Mechanics Lecture 18, Slide 1