Summary: Change in Mechanical Energy

[Pages:8]Mechanical Energy and Simple Harmonic Oscillator

8.01 Week 09D1

Summary: Change in Mechanical

Energy

Total force: Total work:

GGG

F total

=

F total c

+ Fnctotal

( ) W total

=

final

G F

total

G dr

=

final

GG

F + F total

total

c

nc

G dr

initial

initial

Change in potential energy:

Total work done is change in kinetic energy:

U

total

=

-

final

G F total

c

G dr

initial

W total = -U total + Wnc = K

Mechanical Energy Change: Conclusion:

E mechanical K + U total Wnc = K + U total

Modeling the Motion using Force and Energy Concepts

Force and Newton's Second Law:

?Draw all relevant free body force diagrams

?Identify non-conservative forces. ?Calculate non-conservative work Change in Mechanical Energy:

G final

G

Wnc = Fnc dr.

initial

?Choose initial and final states and draw energy diagrams.

?Choose zero point P for potential energy for each interaction in which

potential energy difference is well-defined.

?Identify initial and final mechanical energy.

?Apply Energy Law.

Wnc = K + U total

Mechanical Energy Accounting

Initial state:

Total initial kinetic energy Total initial potential energy Total initial mechanical energy

Kinitial = K1,initial + K2,initial +

Uinitial = U1,initial + U 2,initial +

E = K + U mechanical initial

initial

initial

Final state:

Total final kinetic energy Total final potential energy Total final mechanical energy Apply Energy Law:

Kfinal = K1,final + K2,final +

Ufinal = U1,final + U 2,final +

E = K + U mechanical final

final

final

W = E - E mechanical

mechanical

nc

final

initial

1

Example: Energy Changes

A small point like object of mass m rests on top of a sphere of radius R. The object is released from the top of the sphere with a negligible speed and it slowly starts to slide. Find an expression for the angle with respect to the vertical at which the object just loses contact with the sphere.

Energy Flow diagrams

Initial state

Final State

Example: Energy Changes

A small point like object of mass m rests on top of a sphere of radius

R. The object is released from the top of the sphere with a negligible speed and it slowly starts to slide. Find an expression for the angle with respect to the vertical at which the object just loses contact with the sphere.

Kinitial 0 Uinitial = 0

K final

=

1 2

mv2f

Ufinal = -mgR(1 - cos f )

E 0 mechanical initial

E mechanical final

=

1 2

mv

2 f

- mgR(1- cos f

)

W = 0 = E - E mechanical

mechanical

nc

final

initial

0

=

0

-

1 2

mv2f

-

mgR(1 -

cos

f

)

1 2

mv2f

=

mgR(1 -

cos f

)

Recall Modeling the Motion: Newton `s Second Law

Define system, choose coordinate system.

Draw force diagram.

Newton's Second Law for each direction.

Example: x-direction

^i :

F total x

=

m

d2x dt 2

.

Example: Circular motion

r^ :

F total r

=

-

m

v2 R

.

Example (con't): Free Body Force Diagram

Newton's Second Law

r^ : N - mg cos = - m v2 R

^ :

mg

sin

=

mR

d 2 dt 2

Constraint condition:

N = 0 at = f

Radial Equation becomes

mg cos f

=

m v2f R

1 2

mv

2 f

=

R 2

mg cos f

Energy Condition:

1 2

mv

2 f

=

mgR(1 -

cos f

)

Conclusion:

mgR(1-

cos

f

)

=

R 2

mg

cos

f

cos f

=

2 3

f

=

cos-1

2 3

2

Table Problem: Loop-the-Loop

An object of mass m is released from rest at a height h above the surface of a table. The object slides along the inside of the loop-the-loop track consisting of a ramp and a circular loop of radius R shown in the figure. Assume that the track is frictionless. When the object is at the top of the track (point a) it pushes against the track with a force equal to three times it's weight. What height was the object dropped from?

Table Problem: Block-Spring

System with Friction

A block of mass m slides along a horizontal surface with speed v 0. At t =0 it hits a spring with spring constant k and begins to experience a friction force. The coefficient of friction is variable and is given by ?k= bx where b is a constant. Find how far the spring has compressed when the block has first come momentarily to rest.

Simple Harmonic Motion

Hooke's Law

Define system, choose coordinate system. Draw free-body diagram.

Hooke's Law G Fspring = -kx ^i

-kx

=

m

d2x dt 2

3

Concept Question

Which of the following functions x(t) has a second

derivative which is proportional to the negative of the

function

d2x -x? dt 2

1. x(t) = 1 at2 2

2. x(t) = Aet /T

3. x(t) = Ae-t /T

4.

x(t)

=

A cos

2 T

t

Concept Question

The first derivative

v = dx / dt x

of the sinusoidal

function is:

x

=

A cos

2 T

t

1.

vx (t)

=

A cos

2 T

t

3.

vx (t)

=

-

2 T

Asin

2 T

t

2.

v x

(t)

=

-

Asin

2 T

t

4.

v (t) x

=

2 T

2 Acos T

t

Simple Harmonic Motion

Equation of Motion:

-kx = m d 2 x dt 2

Solution: Oscillatory with Period

T = 2 m k

Position:

x

=

A cos

2 T

t

+

B sin

2 T

t

Velocity:

vx

=

dx dt

=

- 2 T

Asin

2 T

t

+

2 T

B cos

2 T

t

Initial Position at t = 0:

x x(t = 0) = A 0

Initial Velocity at t = 0:

vx,0

vx (t

=

0)

=

2 T

B

General Solution:

x

=

x0

cos

2 T

t

+

T 2

vx,0

sin

2 T

t

Period and Angular Frequency

Equation of Motion: Solution: Oscillatory with Period T

-kx = m d 2x dt 2

x

=

A cos

2 T

t

+

B sin

2 T

t

x -component of velocity:

vx

dx dt

=

- 2 T

Asin

2 T

t

+

2 T

cos

2 T

t

x -component of acceleration:

Period:

ax

d2x dt 2

=

-

2 T

2

A cos

2 T

t

-

2 T

2

B sin

2 T

t

=

-

2 T

2

x

-kx

=

m

d2x dt 2

=

-m

2 T

2

x

k

=

m

2 T

2

T

=

2

m k

Angular frequency

2 = k Tm

4

Concept Question: Simple

Harmonic Motion

A block of mass m is attached to a spring with spring constant k is free to slide along a horizontal frictionless surface.

At t = 0 the block-spring system is stretched an amount x0 > 0 from the equilibrium position and is released from rest. What is the x -component of the velocity of the block when it first comes back to the equilibrium?

1.

vx

=

- x0

T 4

2.

vx

=

x0

T 4

3.

vx = -

k m x0

4.

vx =

k m x0

Example: Block-Spring System with No Friction

A block of mass m slides along a frictionless horizontal surface with speed v x,0. At t = 0 it hits a spring with spring constant k and begins to slow down. How far is the spring compressed when the block has first come momentarily to rest?

Initial and Final Conditions

Initial state: x0 = A = 0 and

vx,0

=

2 T

B

x(t)

=

T 2

vx,0

sin

2 T

t

vx

(t)

=

vx,0

cos

2 T

t

First comes to rest when

vx (t f ) = 0

2 t

=

Tf 2

Since at time t f = T / 4

Final position

sin

2 T

t f

=

sin

2

=

1

x(t f

)

=

T 2

vx,0

=

m k vx,0

Modeling the Motion: Energy

Choose initial and final states:

( ) Change in potential energy:

U

(x

f

)

-U

(x0

)

=

1 2

k

x2f - x02

Choose zero point for potential energy: U (x = 0) = 0

Potential energy function:

U (x) = 1 kx2, U (x = 0) = 0 2

Mechanical energy is constant (Wnc = 0)

E = E mechanical final

mechanical initial

5

Kinetic Energy vs. Potential Energy

State

Initial

x0 = 0 vx,0 > 0

Final

xf > 0 vx, f = 0

Kinetic energy

K0

=

1 2

mvx2,0

Kf =0

Potential Mechanical energy energy

U0 = 0

E = 1 mv2

2 0

x,0

U = 1 kx2 f 2f

Ef

=

1 2

kx2f

Conservation of Mechanical Energy

E f = E0

1 2

kx

2 f

=

1 2

mvx2,0

The amount the spring has compresses when the object first comes to rest is

xf =

m k vx,0

Lecture Demo: Experiment 3 Fitting the Force Curve

Position: Velocity

x = C1 cos (0t ) + C2 sin (0t ) vx = -0C1 sin (0t ) + 0C2 cos (0t )

Angular Frequency: 0 = k / mc

Initial conditions: x(t = 0) = 0 C1 = 0 vx (t = 0) = v1 C2 = v1 / 0

Lecture Demo: Experiment 3 Fitting the Force Curve

Position: Velocity:

x = (v1 / 0 ) sin (0t ) vx = v1 cos (0t )

Amplitude: xmax = 0v1

Acceleration ax = -0v1 sin (0t )

Third Law:

( ) Fsensor (t) = -Fcart (t) = -mcax = mc0v1 sin 0t

Force on sensor vs. time graph. Best Fit: Fsensor (t) = A0 sin (2 (t - A2 ) / A1 )

Fit Parameters: A2 = offset time A0 = mc0v1

A1 = 2 / 0 = 2 mc / k Spring constant: k = 4 2mc / A1

Maximum Displacement: xmax = 0v1 = A0 / mc

6

Energy Diagram

Choose zero point for potential energy:

U (x = 0) = 0

Potential energy function: U (x) = 1 kx2, U (x = 0) = 0 2

Mechanical energy is represented by a horizontal line since it is a constant

E mechanical

=

K(x) +U(x)

=

1 2

mvx2

+

1 2

kx 2

Kinetic energy is difference between mechanical energy and potential energy (independent of choice of zero point)

K = E mechanical - U

Graph of Potential energy function U(x) vs. x

Concept Question: Energy Diagram

The position of a particle is given by

x(t) = D cos(t) - D sin (t ), D > 0

Where was the particle at t = 0?

1) 1 2) 2 3) 3 4) 4 5) 5 6) 1 or 5 7) 2 or 4

Concept Question: Energy Diagram 1

A particle with total mechanical energy E has position x > 0 at t = 0 1) escapes to infinity in the ? x-direction 2) approximates simple harmonic motion 3) oscillates around a 4) oscillates around b 5) periodically revisits a and b 6) not enough information

Concept Question: Energy Diagram 2

A particle with total mechanical energy E has position x > 0 at t = 0 1) escapes to infinity 2) approximates simple harmonic motion 3) oscillates around a 4) oscillates around b 5) periodically revisits a and b 6) not enough information

7

Concept Question: Energy Diagram 3

A particle with total mechanical energy E has position x > 0 at t = 0 1) escapes to infinity 2) approximates simple harmonic motion 3) oscillates around a 4) oscillates around b 5) periodically revisits a and b 6) not enough information

Concept Question: Energy Diagram 4

A particle with total mechanical energy E has position x > 0 at t = 0 1) escapes to infinity 2) approximates simple harmonic motion 3) oscillates around a 4) oscillates around b 5) periodically revisits a and b 6) not enough information

Concept Question: Energy Diagram 5

A particle with total mechanical energy E has position x > 0 at t = 0 1) escapes to infinity 2) approximates simple harmonic motion 3) oscillates around a 4) oscillates around b 5) periodically revisits a and b 6) not enough information

8

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