Physics 151: Lecture 25, Pg 1 Physics 151: Lecture 25 Today’s Agenda l Today’s Topics: çFinish Chapter 11 çStatics (Chapter 12)

Physics 151: Lecture 25, Pg 1

Physics 151: Lecture 25Physics 151: Lecture 25Today’s AgendaToday’s Agenda

Today’s Topics:Finish Chapter 11Statics (Chapter 12)


Lecture 25, Lecture 25, Act 3Act 3

A particle whose mass is 2 kg moves in the xy plane with a constant speed of 3 m/s along the direction r = i + j. What is its angular momentum (in kg · m2/s) relative to the origin?

a. 0 k

b. 6 (2)1/2 k

c. –6 (2)1/2 k

d. 6 k

e. –6 k


Example: Throwing ball from stoolExample: Throwing ball from stool

A student sits on a stool which is free to rotate. The moment of inertia of the student plus the stool is I. She throws a heavy ball of mass M with speed v such that its velocity vector passes a distance d from the axis of rotation. What is the angular speed F of the student-stool

system after she throws the ball ?

top view: before after

d

vM

I I

F

See example 13-9

See text: Ex. 11.11


Gyroscopic Motion...Gyroscopic Motion...

Suppose you have a spinning gyroscope in the configuration shown below:

If the left support is removed, what will happen ?The gyroscope does not fall down !

pivot

g

See text: 11.6



... instead it precessesprecesses around its pivot axis !

pivot

See text: 11.6

This rather odd phenomenon can be easily understood using the simple relation between torque and angular momentum we derived in a previous lecture.



The magnitude of the torque about the pivot is = mgd. The direction of this torque at the instant shown is out of the page

(using the right hand rule).The change in angular momentum at the instant shown must

also be out of the page!

LL pivot

d

mg

ddtL

See text: 11.6



Consider a view looking down on the gyroscope. The magnitude of the change in angular momentum in a time dt

is dL = Ld.

So

where is the “precession frequency”

top viewLL(t)

LL(t+dt)

dLL d pivot

dL

dtL

d

dtL

See text: 11.6



So

In this example = mgd and L = I:

The direction of precession is given by applying the right hand rule to find the direction of and hence of dLL/dt.

dL

dtL

LL pivot

d

mg

mgd

I

L

See text: 11.6


Lecture 24, Lecture 24, Act 1Act 1StaticsStatics

Suppose you have a gyroscope that is supported on a gymbals such that it is free to move in all angles, but the rotation axes all go through the center of mass. As pictured, looking down from the top, which way will the gyroscope precess?

(a) clockwise (b) counterclockwise (c) it won’t precess



Remember that /L.

So what is ?

= r x F

r in this case is zero. Why?

Thus is zero.

It will not precess. At All. Even if you move the base.

This is how you make a direction finder for an airplane.

Answer (c) it won’t precess


Summary:Summary:

Comparison between Rotation and Linear MotionComparison between Rotation and Linear Motion

Angular Linear

See text: 10.3

= x /R

= v / R

= a /R

x

v

a


ComparisonComparison KinematicsKinematics

Angular Linear

constant

0 t

0 021

2t t

constanta

v v at 0

x x v t at 0 021

2

2 02 2

AVE 1

2( 0)

v 2 v02 2ax

vAVE 1

2(v v0)


Comparison: Comparison: Dynamics Dynamics

Angular Linear

K 1

2I 2

K 1

2mv2

I = i mi ri2 m

F = a mr x F = I

L = r x p= I p = mv

EXT dLdt

FEXT dp

dt

W = W = F •x

K = WNET K = WNET


Statics Statics (Chapter 12)(Chapter 12)

As the name implies, “statics” is the study of systems that don’t move.Ladders, sign-posts, balanced beams, buildings,

bridges, etc...

Example: What are all ofthe forces acting on a carparked on a hill ?

xy

N

mgg

f f

See text: 12.1-3


Car on HillCar on Hill

Use Newton’s 2nd Law: FFTOT = MAACM = 0 Resolve this into x and y components:

xy

NN

mgg

f f

F 0

x: f - mg sin = 0

f = mg sin

y: N - mg cos = 0

N= mg cos

See text: 12.1-3


Example 1Example 1

The diagrams below show forces applied to an equilateral triangular block of uniform thickness. In which diagram(s) is the block in equilibrium?

a. A b. B c. C d. D e. A and

F

FF

F

F F F F

2F F

F F


Statics: Using TorqueStatics: Using Torque

Now consider a plank of mass M suspended by two strings as shown. We want to find the tension in each string:

L/2L/4

Mx cm

T1 T2

Mgy

x

F 0 First use

T1 + T2 = Mg

This is no longer enough tosolve the problem !1 equation, 2 unknowns.

We need more information !!We need more information !!

See text: 12.1-3


Using Torque...Using Torque...

We dodo have more information: We know the plank is not rotating.TOT = I = 0

The sum of all torques is zero.

This is true about any axiswe choose !

L/2L/4

Mx cm

T1 T2

Mgy

x

0

See text: 12.1-3



Choose the rotation axis to be along the z direction (out of the page) through the CM:

L/2L/4

Mx cm

T1 T2

Mgy

x

2 2 4T

L

The torque due to the string on the right about this axis is:

1 1 2T

L

The torque due to the string on the left about this axis is:

Gravity exerts notorque about CM

See text: 12.1-3



Since the sum of all torques must be 0:

L/2L/4

Mx cm

T1 T2

TL

TL

2 14 20

Mgy

x

T T2 12

We already found that

T1 + T2 = Mg

T Mg11

3

T Mg22

3


Approach to Statics:Approach to Statics:

In general, we can use the two equations

to solve any statics problems.

When choosing axes about which to calculate torque, we can be clever and make the problem easy....

0F 0

See text: 12.1-3



A 1kg ball is hung at the end of a rod 1m long. The system balances at a point on the rod 0.25m from the end holding the mass. What is the mass of the rod ?

(a) 0.5 kg (b) 1 kg (c) 2 kg

1kg

1m


Lecture 25, Lecture 25, Act 2Act 2Solution ASolution A

The total torque about the pivot must be zero.

1kg

The center of mass of the rod is at its center, 0.25m to theright of the pivot.

X

CM of rod

Since this must balance the ball, which is the same distance to the left of the pivot, the masses must be the same !

same distancemROD = 1kg


Lecture 25, Lecture 25, Act 2Act 2Solution BSolution B

Since the system is not rotating, the x-coordinate of the CM must be the same as the pivot.

1kg

The center of mass of the rod is at its center, 0.25m to theright of the pivot.

X

CM of rod

.25m

x

Since the CM of the ball is 0.25m to the left of the pivot, the mass of the rod must be 1kg to make xCM = 0.

mROD = 1kg-.25m


Example Problem: Hanging LampExample Problem: Hanging Lamp

Your folks are making you help out on fixing up your house. They have always been worried that the walk around back is just too dark, so they want to hang a lamp. You go to the hardware store and try to put together a decorative light fixture. At the store you find a bunch of massless string (kind of a surprising find?), a lamp of mass 2 kg, a plank of mass 1 kg and length 2 m, and a hinge to hold the plank to the wall. Your design is for the lamp to hang off one end of the plank and the other to be held to a wall by a hinge. The lamp end is supported by a massless string that makes an angle of with the plank. (The hinge supplies a force to hold the end of the plank in place.) How strong must the string and the hinge be for this design to work ?

See text: Ex.12.3


Example: Hanging LampExample: Hanging Lamp

1. You need to solve for the forces on the string and the hinge

Use statics equations.

hinge

M

m

L

See text: Ex.12.3


Example: Hanging LampExample: Hanging Lamp

1. You need to solve for T and components of FH.

Use F = 0 in x and y.

Use = 0 in z.

m

See text: Ex.12.3

TFHy

Mg mg

L/2 L/2

FHx


Hanging Lamp...Hanging Lamp...

3. First use the fact that in both x and y directions:

M

m

L/2

Fx

Fy

TT

L/2

Mg

mg

y

x

F 0

x: T cos - Fx = 0

y: T sin + Fy - Mg - mg = 0

0 Now use in the z direction.If we choose the rotation axis tobe through the hinge then thehinge forces Fx and Fy will not enter into the torque equation:

0LTsin-mg2L

LMg

See text: Ex.12.3



3 (Cont.) So we have three equations and three unknowns:

T cos + Fx = 0

T sin + Fy - Mg - mg = 0

LMg + (L/2)mg – LTsin = 0

Which we can solve to find,

M

m

L/2

Fx

Fy

TT

L/2

Mg

mg

y

x

( )sin

g2mM

T+

=

( )tan

g2mM

Fx

+=

F mgy 1

2

See text: Ex.12.3



4. Put in numbers

M

m

L/2

Fx

Fy

TT

L/2

Mg

mg

y

x

N

smkggmMT 50

)30sin(/10~5.2

sin2

2

N

smkggmMFx 43

)30tan(/10~5.2

tan2

2

NsmkgmgFy 5)/10(~5.021 2

See text: Ex.12.3



4. Have we answered the question?

Well the string must be strong enough to exert a force of 50 N without breaking.

We don’t yet have the total force the hinge must withstand.

M

m

L/2

Fx

Fy

TT

L/2

Mg

mg

y

x

NNFFF yx 43435 2222

See text: Ex.12.3

Buy a hinge that can take more than 43 N



A box is placed on a ramp in the configurations shown below. Friction prevents it from sliding. The center of mass of the box is indicated by a white dot in each case.In which cases does the box tip over ?

(a) all (b) 2 & 3 (c) 3 only

1 2 3


Lecture 26, Lecture 26, Act 1Act 1 SolutionSolution

We have seen that the torque due to gravity acts as though all the mass of an object is concentrated at the center of mass.

Consider the bottom right corner of the box to be a pivot point.

If the box can rotate in such a way that the center of mass islowered, it will !


Lecture 26, Lecture 26, ACT 1ACT 1 SolutionSolution

We have seen that the torque due to gravity acts as though all the mass of an object is concentrated at the center of mass.

Consider the bottom right corner of the box to be a pivot point.

If the box can rotate in such a way that the center of mass islowered, it will !


Lecture 26, Lecture 26, Act 1Act 1 AddendumAddendum

What are the torques ??(where do the forces act ?)

rG

mg g

switches sign at critical point

rf

f

f

always zero

goes to zero at critical point

rN

N

N


Example 3Example 3

A square of side L/2 is removed from one corner of a square sandwich that has sides of length L. The center of mass of the remainder of the sandwich moves from C to C’. The distance from C to C’ is :


Recap of today’s lectureRecap of today’s lecture

Chapter 12 - Statics

Documents

Physics 151: Lecture 25, Pg 1 Physics 151: Lecture 25 Today’s Agenda l Today’s Topics: çFinish Chapter 11 çStatics (Chapter 12)