today’s concepts: a) static equilibrium b) potential energy & stability physics 211 lecture 18...

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

Physics 211Lecture 18

Mechanics Lecture 18, Slide 1

ACT

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


Demo: planets mobileDemo: planets mobile

ACT

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

0sin1 LTMgL 0sin22 2 L

TL

Mg

sin1

MgT

sin2

MgT

It’s the same. Why?


M

T2 d2

Case 2

L/2

CheckPoint

In which of the static cases shown below is the tension in the supporting wire bigger? In both cases the red strut has the same mass and length.



300

L/2

T2

Case 2

300

T1

LM

Case 1

CheckPointIn which of the static cases shown below is the tension in the supporting wire bigger? In both cases the red strut has the same mass and length.

A) Case 1

B) Case 2

C) Same

A) Case 1 because the lever arm distance is the largest there B) The torque caused by gravity is equal to the torque caused by the tension. Since in Case 1, the lever arm is longer, the force does not need to be as long to equalize the torque caused by gravity. C) The angles are the same so the tensions are the same, the lengths do not matter.



Nice and clear; I like this explanation.Nice and clear; I like this explanation.Definition? Hmmm, I think I don’t know what this means.Definition? Hmmm, I think I don’t know what this means.

Good; I like this.Good; I like this.

Seems to me that a longer length means smaller tensionSeems to me that a longer length means smaller tension

It should seem VERY similar to a ladder problem.It should seem VERY similar to a ladder problem.But that is the But that is the questionquestion..

Location is important for torques.Location is important for torques.

But the question is about the force or tension.But the question is about the force or tension.

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

ACT

21

12I ML

2I Md

2 21

12I ML Md


d

M

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

ACT


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

ACT


T1

Mgy

x

d

M

ACM

Apply

Apply

Plug this into the expression for T1

ACM d

Use ACM dto find

ext CMF MA1 CMMg T MA

Iext CMA

Mgd I Id

2

CM

MdA g

I

1 CMT Mg MA


T1

Mgy

x

d

M

ACM

Conserve energy:

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?

2Mgd

I

21

2Mgd I


M

y

x

d

T

y

x

Applying

Mg

dACM 2d

Centripetal acceleration

ext CMF MA2T Mg M d

2T Mg M d


T

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 (measured about the foot):

Mg

mg

L/2

f y

xN

Fwall

daxis

cosmgd cos2

LMg sinwallF L 0

cot2

Mg

L

dmgFwall

cot2

d Mgf mg

L

wallF f


f

d

M

m

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

d Mgf mg

L


cot2

d Mgf mg

L

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

cot2

Mgf

If it’s just a ladder…


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 bigger?


CheckPoint

Case 1 Case 2Mechanics Lecture 18, Slide 22

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 bigger?


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


But what about the friction force?But what about the friction force?

I don’t know what this means.I don’t know what this means.

okayokay

A Force can not A Force can not be equal tobe equal to a torque. The frictional force will be a torque. The frictional force will be smallersmaller with a smaller with a smaller angle between the ladder and the wall.angle between the ladder and the wall.

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.


DemoDemo


But the wire is NOT perpendicular to the beam in any of the situations.But the wire is NOT perpendicular to the beam in any of the situations.

okayokayI don’t know what this means.I don’t know what this means.

GoodGoodOkayOkay

What’s a “footprint” when something is What’s a “footprint” when something is suspendedsuspended??

GoodGoodVery goodVery good

? ? ?? ? ?

What’s a “footprint” when something is What’s a “footprint” when something is suspendedsuspended??When things aren’t clear, please When things aren’t clear, please come see me — always!come see me — always!

That does That does notnot seem “intuitive” to me. But it seem “intuitive” to me. But it isistrue — from analyzing the forces and torques.true — from analyzing the forces and torques.

footprint

footprint

Stability & Potential Energy


CM DemosCM Demos

today’s concepts: a) static equilibrium b) potential energy & stability physics 211 lecture 18...

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