Lec7_WorkEnergy (2)

8/8/2019 Lec7_WorkEnergy (2)

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Lecture 7

WORK AND ENERGY

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Lecture 7

WORK AND ENERGY

1 Work Energy Theorem

2 Potential energy

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Lecture 7

WORK AND ENERGY

1 Work Energy Theorem

2 Potential energy

3 Conservation of Energy

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NEWTON’S LAWS

Fundamental: all there is to Mechanics!!

Work and Energy 2/11

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NEWTON’S LAWS


MOMENTUM WORK ENERGY


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NEWTON’S LAWS




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NEWTON’S LAWS



Tools to help solve the equations given by Newton’s Laws


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NEWTON’S LAWS




CONSERVATION LAWS


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NEWTON’S LAWS




CONSERVATION LAWS

Momentum and energy are important by themselves!


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Integration of Newton’s II Law

A point particle moves under the influence of−→F net. What is its motion?

Work and Energy Work Energy Theorem 3/11

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−→F net = ma = md2

rdt2

= mdvdt


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−→F net = ma = md

2

rdt2

= mdvdt

Objective: to obtain r(t) (motion):


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2

rdt2

= mdvdt


needs 2 integrations of the II law.


I i f N ’ II L

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2

rdt2

= mdvdt



Let’s integrate once:


I t ti f N t ’ II L

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2

rdt2

= mdvdt



Let’s integrate once:−→F net · dr =


I t ti f N t ’ II L

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2

rdt2

= mdvdt



Let’s integrate once:−→F net · dr = m

dv

dt· dr


Integration of Ne ton’s II La

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2

rdt2

= mdvdt




dv

dt· dr = mdv ·

dr

dt



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2

rdt2

= mdvdt




dv

dt· dr = mdv ·

dr

dt

= mv · dv



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Integration of Newton s II Law



2

rdt2

= mdvdt




dv

dt· dr = mdv ·

dr

dt

= mv · dv =1

2md(v2)



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2

rdt2

= mdvdt




dv

dt· dr = mdv ·

dr

dt

= mv · dv =1

2md(v2)

rb

ra

−→F net · dr

= 1

2m(v2b − v2a)



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2

rdt2 = mdv

dt




dv

dt· dr = mdv ·

dr

dt

= mv · dv =1

2md(v2)

b

a

−→F net · dr = 1

2m(v2b − v2a)

Work done by−→F net



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2

rdt2 = mdv

dt




dv

dt· dr = mdv ·

dr

dt

= mv · dv =1

2md(v2)

b

a

−→F net · dr = 1

2m(v2b − v2a)

Work done by−→F net = Change in Kinetic Energy


Work-Kinetic Energy Theorem

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Theorem

The work done by the net force on a particle,

W net = ∆KE

the change in its kinetic energy.



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Theorem


W net = ∆KE


1 Work done BY (any) force F , in translating the particle from point

a to point b:

b

a

−→F · dr



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Work Kinetic Energy Theorem

Theorem


W net = ∆KE



a to point b:

b

a

−→F · dr

2

W is a scalar, can be positive or negative



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Theorem


W net = ∆KE



a to point b:

b

a

−→F · dr

2

W is a scalar, can be positive or negative3 W depends on the path a – b.



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Theorem


W net = ∆KE



a to point b:

b

a

−→F · dr

2

W is a scalar, can be positive or negative3 W depends on the path a – b.

4 KE =1

2mv2 is a positive quantity.


Gradient Operator

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Gradient Operator

Work and Energy Potential energy 5/11

Gradient Operator

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p

Definition

−→ ≡ i

∂

∂x

+ ˆ j∂

∂y

+ k∂

∂z

:


Gradient Operator

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p

Definition

−→ ≡ i

∂

∂x

+ ˆ j∂

∂y

+ k∂

∂z

: a Derivative with vector nature


Gradient Operator

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p

Definition

−→ ≡ i

∂

∂x

+ ˆ j∂

∂y

+ k∂

∂z


Operates on scalar functions:

−→f (x,y,z) = i

∂f

∂x

+ ˆ j∂f

∂y

+ k∂f

∂z


Gradient Operator

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p

Definition

−→ ≡ i

∂

∂x

+ ˆ j∂

∂y

+ k∂

∂z



−→f (x,y,z) = i

∂f

∂x+ ˆ j

∂f

∂y+ k

∂f

∂z

Change in a function in a direction d l = idx + ˆ jdy + kdz:

(−→f ) · d l =

∂f

∂xdx +

∂f

∂ydy +

∂f

∂zdz = df


Gradient Operator

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Definition

−→ ≡ i

∂

∂x

+ ˆ j∂

∂y

+ k∂

∂z



−→f (x,y,z) = i

∂f

∂x+ ˆ j

∂f

∂y+ k

∂f

∂z


(−→f ) · d l =

∂f

∂xdx +

∂f

∂ydy +

∂f

∂zdz = df

−→f : “slope” of f in direction of max. change


Gradient Operator

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Definition

−→ ≡ i

∂

∂x

+ ˆ j∂

∂y

+ k∂

∂z



−→f (x,y,z) = i

∂f

∂x+ ˆ j

∂f

∂y+ k

∂f

∂z


(−→f ) · d l =

∂f

∂xdx +

∂f

∂ydy +

∂f

∂zdz = df

−→f : “slope” of f in direction of max. change−→f projected along a direction = rate of change of f in that

direction


Potential Energy

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For Conservative Forces,−→F ≡ −

−→U


Potential Energy

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−→U

DefinitionPotential energy function U (r): a scalar function.


Potential Energy

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−→U


Work done by a conservative force:

W C =

b

a

−(−→U · dr)


Potential Energy

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−→U



W C =

b

a

−(−→U · dr) = −

b

a

dU


Potential Energy

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−→U



W C =

b

a

−(−→U · dr) = −

b

a

dU = −[U (rb) − U (ra)]


Potential Energy

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−→U



W C =

b

a

−(−→U · dr) = −

b

a

dU = −[U (rb) − U (ra)]

W C is path-independent


Potential Energy

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−→U



W C =

b

a

−(−→U · dr) = −

b

a

dU = −[U (rb) − U (ra)]

W C is path-independentFrom

−→F one can determine U upto an additive contant: choice of

zero potential.


Examples

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Examples

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1 Spring Force

−→F (x) = −kxi


Examples

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1 Spring Force

−→F (x) = −kxi = −

d

dx

1

2

kx2 i


Examples

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1 Spring Force

−→F (x) = −kxi = −

d

dx

1

2

kx2 i

U (x) =1

2kx2


Examples

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2 Gravitational force

−→F (r) = −GMmr2

r


Examples

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−→F (r) = −GMmr2

r

Central force: depends only on |r|


Examples

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−→F (r) = −GMmr2

r


−→F (r) = −dU dr

r


Examples

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−→F (r) = −GMmr2

r


−→F (r) = −dU dr

r

GMm

r

2= −

d

dr

GMm

r

=⇒


Examples

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−→F (r) = −GMmr2

r


−→F (r) = −dU dr

r

GMm

r2= −

d

dr

GMm

r=⇒

U grav(r) = −GMm

r


Conservation of Energy

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∆KE = W net

Work and Energy Conservation of Energy 9/11


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∆KE = W net = W cons + W non-cons



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= −∆U + W non-cons



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=⇒ ∆KE + ∆U = W non-cons



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Define Mechanical Energy: E mech = KE + U =1

2mv2 + U .



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2mv2 + U .

∆E mech = W non-cons



∆

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2mv2 + U .


Conservation of mechanical energy

When only conservative forces act, total mechanical energy does not

change.



∆KE W W W

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2mv2 + U .


Conservation of mechanical energy

When only conservative forces act, total mechanical energy does not

change.

If non conservative forces act, energy is not constant.

∆E = work done by n.c. forces.


Example 1

Mass m moves initially in a circle

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with angular velocity ω0. What is r

at any later time?


Example 1


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at any later time?

Solution 1: Write down Newton’s II

Laws and solve: Difficult!


Example 1


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y


at any later time?



Solution 2: Use Energy Conservation:

∆KE + ∆U = 0


Example 1


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y


at any later time?




∆KE + ∆U = 0

KE =1

2(m + M )r2 +

1

2mr2θ2


Example 1


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y


at any later time?




∆KE + ∆U = 0

KE =1

2(m + M )r2 +

1

2mr2θ2

L = mr2θ = constant: Angular momentum


Example 1


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at any later time?




∆KE + ∆U = 0

KE =1

2(m + M )r2 +

1

2mr2θ2

L = mr2θ = constant: Angular momentum

U = −Mgy = Mg(r − l)


Example 2

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Qn: What is the work done

by the Friction force?


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Lec7_WorkEnergy (2)

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