pendulum (bandul)

Physics 211: Lecture 25, Pg 1

Physics 211: Lecture 25Physics 211: Lecture 25

Today’s AgendaToday’s Agenda Recap of last lecture Using “initial conditions” to solve problems The general physical pendulum The torsion pendulum Energy in SHM

Atomic Vibrations Problem: Vertical Spring Problem: Transport Tunnel SHM Review


SHM and SpringsSHM and Springs

k

s

m

0

k

m

s

0

d sdt

s2

22

km

Force:

Solution:s = A cos(t + )


Velocity and AccelerationVelocity and Acceleration

k

x

m

0

Position: x(t) = A cos(t + ) Velocity: v(t) = -A sin(t + )Acceleration: a(t) = -2A cos(t + )

by takingderivatives,since:

a t dv tdt

( ) ( )

v t dx tdt

( ) ( )

xMAX = AvMAX = AaMAX = 2A


Lecture 25, Lecture 25, Act 1Act 1Simple Harmonic MotionSimple Harmonic Motion

A mass oscillates up & down on a spring. Its position as a function of time is shown below. At which of the points shown does the mass have positive velocity and negative acceleration?

t

y(t)

(a)

(b)

(c)


Lecture 25, Lecture 25, Act 1Act 1 SolutionSolution

The slope of y(t) tells us the sign of the velocity since

t

y(t)

(a)

(b)

(c)

v dydt

y(t) and a(t) have the opposite sign since a(t) = -2 y(t)

a < 0v < 0

a > 0v > 0

a < 0v > 0

The answer is (c).


ExampleExample

A mass m = 2 kg on a spring oscillates with amplitude A = 10 cm. At t = 0 its speed is maximum, and is v = +2 m/s. What is the angular frequency of oscillation ? What is the spring constant k?

k

x

m

= 1MAX s20cm10

sm2A

v

km

Also: k = m2

So k = (2 kg) x (20 s -1) 2 = 800 kg/s2 = 800 N/m

vMAX = A


Initial ConditionsInitial Conditions

k

x

m

0

Use “initial conditions” to determine phase !

Suppose we are told x(0) = 0 , and x is initially increasing (i.e. v(0) = positive):

x(t) = A cos(t + ) v(t) = -A sin(t + )a(t) = -2A cos(t + )

sincos

x(0) = 0 = A cos() = /2 or -/2 0 < v(0) = -A sin() < 0

= -/2So


Initial Conditions...Initial Conditions...

k

x

m

0

x(t) = A cos(t - /2 ) v(t) = -A sin(t - /2 )a(t) = -2A cos(t - /2 )

So we find = -/2!!

x(t) = A sin(t) v(t) = A cos(t)a(t) = -2A sin(t)

t

x(t)A

-A


Lecture 25, Lecture 25, Act 2Act 2Initial ConditionsInitial Conditions

A mass hanging from a vertical spring is lifted a distance d above equilibrium and released at t = 0. Which of the following describes its velocity and acceleration as a function of time?

k

m

y

0

d

(a) v(t) = -vmax sin(t) a(t) = -amax cos(t)

(b) v(t) = vmax sin(t) a(t) = amax cos(t)

(c) v(t) = vmax cos(t) a(t) = -amax cos(t)

(both vmax and amax are positive numbers)

t = 0



k

m

y

0

d

Since we start with the maximum possibledisplacement at t = 0 we know that:

y = d cos(t)

t = 0

v dydt

d sin t v sin tmax

a dvdt

d cos t a cos tmax 2


Review of Simple PendulumReview of Simple Pendulum

Using = Iand sin for small

L

dm

mg

z

mgL mL ddt

22

2

I We found

ddt

2

22

Lgwhere

Which has SHM solution = 0 cos(t + )


Review of Rod PendulumReview of Rod Pendulum

Using = Iand sin for small

Ldmg

z

L/2

xCM

mg L mL ddt2

13

22

2

I

ddt

2

22 3

2gL

where

We found

Which has SHM solution = 0 cos(t + )


General Physical PendulumGeneral Physical Pendulum

Suppose we have some arbitrarily shaped solid of mass M hung on a fixed axis, and that we know where the CM is located and what the moment of inertia I about the axis is.

The torque about the rotation (z) axis for small is (sin )

= -Mgd -MgR

d

Mg

z-axis

R

xCM

ddt

2

22

MgRI

where

= 0 cos(t + )

Physical

Pendulum

2

2

dtdIMgR


Lecture 25, Lecture 25, Act 3Act 3Physical PendulumPhysical Pendulum

A pendulum is made by hanging a thin hoola-hoop of diameter D on a small nail. What is the angular frequency of oscillation of the hoop

for small displacements? (ICM = mR2 for a hoop)

(a)

(b)

(c)

gD

2gD

g2D

D

pivot (nail)



The angular frequency of oscillation of the hoop for small

displacements will be given by

pivot (nail)

Rcm x

ImgR

(see slide 13 this lect.)

Use parallel axis theorem: I = Icm + mR2

m

= mR2 + mR2 = 2mR2

mgRmR

gR

gD2 22

gDSo

Hoop

Pendulum


Torsion PendulumTorsion Pendulum

Consider an object suspended by a wire attached at its CM. The wire defines the rotation axis, and the moment of inertia I about this axis is known.

The wire acts like a “rotational spring.” When the object is rotated, the wire is twisted.

This produces a torque that opposes the rotation.

In analogy with a spring, the torque produced is proportional to the displacement: = -k I

wire


Torsion Pendulum...Torsion Pendulum...

Since = -k = Ibecomes

k ddt

I2

2

I

wire

ddt

2

22

kIwhere

This is similar to the “mass on spring” exceptI has taken the place of m (no surprise).

Torsion

Pendulum


Energy in SHMEnergy in SHM

For both the spring and the pendulum, we can derive the SHM solution by using energy conservation.

The total energy (K + U) of a system undergoing SHM will always be constant!

This is not surprising since there are only conservative forces present, hence K+U energy is conserved.

-A A0 s

U

U

KE

U = (1/2) ks2


SHM and quadratic potentialsSHM and quadratic potentials

SHM will occur whenever the potential is quadratic. Generally, this will not be the case: For example, the potential between

H atoms in an H2 molecule lookssomething like this:

-A A0 x

U

U

KEU

x


SHM and quadratic potentials...SHM and quadratic potentials...

However, if we do a Taylor expansion of this function about the minimum, we find that for smalldisplacements, the potential IS quadratic:

U

x

U(x) = U(x0 ) + U(x0 ) (x- x0 ) + U (x0 ) (x- x0 )2+....2

1

U(x0) = 0 (since x0 is minimum of potential)

x0

U

x Define x = x - x0 and U(x0 ) = 0

Then U(x) = U (x0 ) x 2

21


SHM and quadratic potentials...SHM and quadratic potentials...

U

x

x0

U

x

U(x) = U (x0) x 2

Let k = U (x0)

Then:

U(x) = k x 2

21

21

SHM potential!!


Problem: Vertical SpringProblem: Vertical Spring

A mass m = 102 g is hung from a vertical spring. The equilibrium position is at y = 0. The mass is then pulled down a distance d = 10 cm from equilibrium and released at t = 0. The measured period of oscillation is T = 0.8 s. What is the spring constant k? Write down the equations for the position,

velocity, and acceleration of the mass as functions of time.

What is the maximum velocity? What is the maximum acceleration?

k

m

y

0

-dt = 0


Problem: Vertical Spring...Problem: Vertical Spring...

What is k ? km

2 7 85 1

Ts.

So: k s kg Nm

7 85 0102 6 291 2. . .

k m 2

k

m

y

0

-dt = 0



What are the equations of motion?

At t = 0, y = -d = -ymax

v = 0

So we conclude: y(t) = -d cos(t) v(t) = d sin(t)a(t) = 2d cos(t)

k

m

y

0

-dt = 0



y(t) = -d cos(t) v(t) = d sin(t)a(t) = 2d cos(t)

xmax = d = .1m

vmax = d = (7.85 s-1)(.1m) = 0.78 m/s

amax = 2d = (7.85 s-1)2(.1m) = 6.2 m/s2

0t

k

m

y

0

-dt = 0


Transport TunnelTransport Tunnel

A straight tunnel is dug from Urbana through the center of the Earth and out the other side. A physics 211 student jumps into the hole at noon. What time does she get back to Urbana?


Transport Tunnel...Transport Tunnel...

R

RE

F R GmMRG

R 2

where MR is themass inside radius R

MR

FG

F RF R

MR

RM

G

G E

R E

E 2

2

but 3R RM

F RF R

RR

RR

RR

G

G E

E

E E

3

2

2

3



R

RE MR

FG

EEG

G

RR

RFRF

F mg RR

kRGE

k mgRE

Like a mass ona spring with

mg)R(F EG



R

RE MR

FG

k mgRE

Like a mass ona spring with

km

gRE

So:

plug in g = 9.81 m/s2

and RE = 6.38 x 106 m

get = .00124 s-1

and so T = = 5067 s

84 min2



So she gets back to Urbana 84 minutes later, at 1:24 p.m.



Strange but true: Strange but true: The period of oscillation does not require that the tunnel be straight through the middle!! Any straight tunnel gives the same answer, as long as it is frictionless and the density of the Earth is constant.



Another strange but true Another strange but true fact: fact: An object orbiting the earth near the surface will have a period of the same length as that of the transport tunnel.

a = 2R

9.81 = 2 6.38(10)6 m = .00124 s-1

so T = = 5067 s

84 min

2


Simple Harmonic Motion: Simple Harmonic Motion: SummarySummary

d sdt

s2

22

Solution:s = A cos(t + )

Force:

km

k

s

m

0

k

m

s

0

s L

gL


An object with mass M, radius R and moment of inertia I = MR2 rolls without slipping down an incline which makes an angle with respect to the horizontal. 13. Which of the following gives the correct expression for the translational acceleration, a , of the object?

a. sin1

ga

b. sina g

c. sin1

a g

d. 1 sina g

e. sin2

ga

R

a g


Recap of today’s lectureRecap of today’s lecture

Recap of last lecture Using “initial conditions” to solve problems

(Text: 14-1) The general physical pendulum (Text: 14-3) The torsion pendulum Energy in SHM (Text: 14-2)

Atomic Vibrations Problem: Vertical Spring (Text: 14-3) Problem: Transport Tunnel SHM Review

Look at textbook problems Look at textbook problems Chapter 14: # 61, 63, 67, 68, 74, 122

pendulum (bandul)

Documents

Transcript of pendulum (bandul)