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## PowerPoint Slideshow about ' Conservative vs. Non-conservative Forces' - aline-walsh

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Conservative vs. Non-conservative Forces

The fundamental theorem of calculus can be written as follows:

Why is this important? Because from it you can derive the definition of conservative forces, which is broadly defined as:

The circle is used to denote that the limits of integration occur on a closed path, and thus the positions for evaluating the integral (i.e. at the initial and final positions) occur at the same point.

Starting from the force equation, let’s re-write force as the derivative of a scalar function, U, and evaluate the integral as the derivative of U is integrated around a closed path.

What if the force was due to gravity?

Now let’s assume that the particle’s motion occurred around a closed path (meaning that it eventually returned to its starting position). Evaluating the integral

shows that the integral of gravitational force around a closed path is zero. In other words, one can say that the state at which one can find the particle under the action of gravity is independent the path it takes between two points. Only the change in the scalar quantity U is required to compute the change in state of its initial and final positions. (i.e.)

Let’s now define the scalar quantity U as the potential energy.

Now let’s assume that the particle’s motion around a closed path is due to the force by a spring…

The integral of the force due to a spring around a closed path is zero, and thus a spring is a conservative force. If we were to evaluate the potential energy of a spring at two different points:

The scalar quantity U is now called the potential energy of the spring.

where is denoted the gradient operator

Now let’s generalize this to 3-dimensional vector fields. Assume the you have a force vector written in Cartesian coordinates.

In order to test this concept, let’s look at the 3-D gravitational field...

The force of gravity can be written as the gradient of a scalar function:

Which leads to an expression for the gravitational potential energy defined purely as a scalar of the inverse distance away from the source.

Since the integral of the force over a closed path is zero, gravity can be defined as a conservative force even in three dimensions.

What are some examples of non-conservative forces?

Friction and any processes that causes dissipation of heat.

For example, the energy dissipated in going from position 1 to position 2, assuming velocity is constant (this assumption does not matter, but just facilitates the derivation), is given by:

Likewise, the energy dissipated in going from position 2 back to position 1 is:

Thus, the total energy dissipated in a closed path is:

which is not zero! Thus friction is a non-conservative force (i.e. is path dependent).

Thus the change in potential energy, U, due to movement in a conservative force from position 1 to position 2 is re-allocated as kinetic energy, T. Another way to say this is that the sum of potential energy and kinetic energy must always remain constant. (i.e.) This is a very powerful result because now we only have to look at the initial and final states instead of integrating through the entire path of motion.

Starting from the previous derivation for potential energy

Velocity is zero, so T=0 Potential energy is U=mgy

Initial conditions at time t=0

Constraints: Bar has constant length

Solution by energy method

We must find the position at which the acceleration of slider B is zero. This may be an inflection point (i.e. where the velocity is also zero), it may be a point where the velocity is at a local maxima, or where the velocity is at a global maximum.

Solving for the velocity of slider B:

So, what conditions leads to the acceleration of slider B equal to zero?

The condition that must be satisfied for the acceleration of slider B to equal zero is shown below.

There are two solutions. One of them is: , however this is an inflection point, where the velocity is changing its direction. The other solution is:

Due to the constraints of the length of the bar, the x-position is therefore:

And therefore the maximum velocity is given as:

r = 600 mm

O

C

q

200 g

A

Problem 1

A small 200-g collar C can slide on a

semicircular rod which is made to rotate

about the vertical AB at the constant rate of

6 rad/s. Determine the minimum required

value of the coefficient of static friction

between the collar and the rod if the collar

is not to slide when (a) q = 90o, (b) q = 75o,

(c) q = 45o. Indicate in each case the

direction of the impending motion.

B

r = 600 mm

O

C

q

200 g

A

A small 200-g collar C can slide on a

semicircular rod which is made to rotate

about the vertical AB at the constant rate of

6 rad/s. Determine the minimum required

value of the coefficient of static friction

between the collar and the rod if the collar

is not to slide when (a) q = 90o, (b) q = 75o,

(c) q = 45o. Indicate in each case the

direction of the impending motion.

1. Kinematics:Determine the acceleration of the particle.

2. Kinetics: Draw a free body diagram showing the applied

forces and an equivalent force diagram showing the vector

ma or its components.

B

r = 600 mm

O

C

q

200 g

A

A small 200-g collar C can slide on a

semicircular rod which is made to rotate

about the vertical AB at the constant rate of

6 rad/s. Determine the minimum required

value of the coefficient of static friction

between the collar and the rod if the collar

is not to slide when (a) q = 90o, (b) q = 75o,

(c) q = 45o. Indicate in each case the

direction of the impending motion.

3. Apply Newton’s second law:The relationship between the

forces acting on the particle, its mass and acceleration is given

by SF = ma . The vectors F and a can be expressed in terms of

either their rectangular components or their tangential and normal

components. Absolute acceleration (measured with respect to

a newtonian frame of reference) should be used.

B

r = 600 mm

O

C

q

200 g

A

Kinematics.

w

B

r = 600 mm

O

C

q

an

A

an = (r sinq) w2

an = (0.6 m) sinq ( 6 rad/s )2

r sinq

an = 21.6 sinqm/s2

B

r = 600 mm

O

C

q

200 g

A

F

O

N

q

man = (0.2) 21.6 sinq

= 4.32 sinq N

(0.2 kg)(9.81 m/s2)

Kinetics; draw a free body diagram.

=

man = (0.2) 21.6 sinq

= 4.32 sinq N

+ SFt = 0: F - 0.2 (9.81) sin q = - 4.32 sinq cos q

F = 0.2 (9.81) sin q - 4.32 sinq cos q

+ SFn = man: N - 0.2 (9.81) cos q = 4.32 sinq sin q

N = 0.2 (9.81) cos q + 4.32 sin2q

F = mN

F

Apply Newton’s second law.

O

=

N

q

(0.2 kg)(9.81 m/s2)

For a given q, the values of F , N , and m can be determined

F

O

=

N

q

man = (0.2) 21.6 sinq

= 4.32 sinq N

(0.2 kg)(9.81 m/s2)

Solution:

(a) q = 90o, F = 1.962 N, N = 4.32 N, m = 0.454 (down)

(b) q = 75o, F = 0.815 N, N = 4.54 N, m = 0.1796 (down)

(c) q = 45o, F = -0.773 N, N = 3.55 N, m = 0.218 (up)

Pin B weighs 4 oz and is free to slide

in a horizontal plane along the rotating

arm OC and along the circular slot DE

of radius b = 20 in. Neglecting friction

and assuming that q = 15 rad/s and

q = 250 rad/s2 for the position q = 20o,

determine for that position (a) the

radial and transverse components of

the resultant force exerted on pin B,

(b) the forces P and Q exerted on pin

B, respectively, by rod OC and the wall

of slot DE.

r

C

B

D

.

q

O

.

.

b

A

b

E

Problem 2

r

C

B

D

q

Pin B weighs 4 oz and is free to slide

in a horizontal plane along the rotating

arm OC and along the circular slot DE

of radius b = 20 in. Neglecting friction

and assuming that q = 15 rad/s and

O

b

A

b

.

E

.

.

q = 250 rad/s2 for the position q = 20o, determine for that position

(a) the radial and transverse components of the resultant force

exerted on pin B, (b) the forces P and Q exerted on pin B,

respectively, by rod OC and the wall of slot DE.

1. Kinematics:Examine the velocity and acceleration of the

particle. In polar coordinates:

v = rer + rqeq

a = (r - r q 2 ) er + (r q + 2 r q ) eq

eq

.

er

.

r = rer

.

.

.

.

.

.

.

q

r

C

B

D

q

Pin B weighs 4 oz and is free to slide

in a horizontal plane along the rotating

arm OC and along the circular slot DE

of radius b = 20 in. Neglecting friction

and assuming that q = 15 rad/s and

O

b

A

b

.

E

.

.

q = 250 rad/s2 for the position q = 20o, determine for that position

(a) the radial and transverse components of the resultant force

exerted on pin B, (b) the forces P and Q exerted on pin B,

respectively, by rod OC and the wall of slot DE.

2. Kinetics: Draw a free body diagram showing the applied

forces and an equivalent force diagram showing the vector

ma or its components.

r

C

B

D

q

Pin B weighs 4 oz and is free to slide

in a horizontal plane along the rotating

arm OC and along the circular slot DE

of radius b = 20 in. Neglecting friction

and assuming that q = 15 rad/s and

O

b

A

b

.

E

.

.

q = 250 rad/s2 for the position q = 20o, determine for that position

(a) the radial and transverse components of the resultant force

exerted on pin B, (b) the forces P and Q exerted on pin B,

respectively, by rod OC and the wall of slot DE.

3. Apply Newton’s second law:The relationship between the

forces acting on the particle, its mass and acceleration is given

by SF = ma . The vectors F and a can be expressed in terms of

either their rectangular components or their radial and transverse

components. With radial and transverse components:

SFr = mar = m ( r - r q 2 ) and SFq = maq = m ( r q + 2 r q )

.

.

.

.

.

.

.

.

r

C

B

D

q

O

b

A

b

B

E

r

b

O

2q

q

A

b

q = 20o

q = 15 rad/s

q = 250 rad/s2

.

.

.

.

.

.

.

Kinematics.

r = 2 b cos q

r = - 2 b sin qq

r = - 2 b sin qq - 2 b cos qq2

.

.

.

Problem 2 Solution

r

b

O

2q

q

A

b

.

.

.

.

.

.

.

.

.

.

.

.

.

r = 2 b cos q, r = - 2 b sinqq, r = - 2 b sinqq - 2 b cosqq2

For: b = 20/12 ft, q = 20o,

q = 15 rad/s q = 250 rad/s2

r = 2 (20/12 ft) cos 20o = 3.13 ft

r = - 2 (20/12 ft) sin 20o (15 rad/s) = - 17.1 ft/s

r = -2(20/12 ft) sin 20o (250 rad/s2 ) - 2(20/12 ft) cos 20o (15 rad/s)2

r = - 989.79 ft/s2

.

.

.

r

C

B

D

q

O

b

A

b

E

maq

Fq

Fr

mar

=

r

r

B

B

q

q

A

A

O

O

(a) Radial and transverse

components of the resultant force

exerted on pin B.

Kinetics; draw a free body diagram.

maq

Fq

Fr

mar

=

r

r

B

B

q

q

A

A

O

O

.

.

(4/16)

32.2

.

.

(4/16)

32.2

Apply Newton’s second law.

+ SFr = mar: Fr = m ( r - rq2 )

Fr = [- 989.79 - ( 3.13 )(152)] = -13.16 lb Fr = 13.16 lb

+ SFq = maq: Fq = m ( r q + 2 rq )

Fq = [(3.13)(250) + 2 (-17.1)(15)] = 2.1 lb

.

.

.

Fq = 2.10 lb

r

C

B

D

q

O

b

A

b

E

q

q

40o

Fq

Q

Fr

q

=

r

r

B

B

P

q

q

A

A

O

O

Fq = - Q sin q + P

2.10 = - 14.0 sin 20o + P

P = 6.89 lb

Fr = - Q cos q

-13.16 = - Q cos 20o

Q = 14.00 lb

20o

100 mm

A

B

Problem 3

A 250-g collar can slide on a

horizontal rod which is free to rotate

about a vertical shaft. The collar is

initially held at A by a cord attached

to the shaft and compresses a spring

of constant 6 N/m, which is

undeformed when the collar is

located 500 mm from the shaft. As

the rod rotates at the rate qo = 16 rad/s, the cord is cut and the

collar moves out along the rod. Neglecting friction and the

mass of the rod, determine for the position B of the collar (a) the

transverse component of the velocity of the collar, (b) the radial

and transverse components of its acceleration, (c) the

acceleration of the collar relative to the rod.

.

400 mm

100 mm

A

B

eq

er

r = rer

.

.

.

.

q

The collar is initially held at A by a

cord attached to the shaft and

compresses a spring. As the rod

rotates the cord is cut and the

collar moves out along the rod to B.

1. Kinematics: Examine the velocity and acceleration of the

particle. In polar coordinates:

v = rer + rqeq

a = (r - r q 2 ) er + (r q + 2 r q ) eq

.

.

.

.

.

400 mm

100 mm

A

B

2. Angular momentum of a particle:Determine the particle

velocity at B using conservation of angular momentum. In polar

coordinates, the angular momentum HO of a particle about O is

given by

HO = m r vq

The rate of change of the angular momentum is equal to the sum

of the moments about O of the forces acting on the particle.

SMO = HO

If the sum of the moments is zero, the angular momentum is

conserved and the velocities at A and B are related by

m ( r vq )A = m ( r vq )B

.

The collar is initially held at A by a

cord attached to the shaft and

compresses a spring. As the rod

rotates the cord is cut and the

collar moves out along the rod to B.

400 mm

100 mm

A

B

The collar is initially held at A by a

cord attached to the shaft and

compresses a spring. As the rod

rotates the cord is cut and the

collar moves out along the rod to B.

3. Kinetics: Draw a free body diagram showing the applied

forces and an equivalent force diagram showing the vector

ma or its components.

400 mm

100 mm

A

B

.

.

.

.

.

.

.

The collar is initially held at A by a

cord attached to the shaft and

compresses a spring. As the rod

rotates the cord is cut and the

collar moves out along the rod to B.

4. Apply Newton’s second law:The relationship between the

forces acting on the particle, its mass and acceleration is given

by SF = ma . The vectors F and a can be expressed in terms of

either their rectangular components or their radial and transverse

components. Absolute acceleration (measured with respect to

a Newtonian frame of reference) should be used.

With radial and transverse components:

SFr = mar = m ( r - r q 2 ) and SFq = maq = m ( r q + 2 r q )

Problem 3 Solution

100 mm

A

B

mrA(vq )A = m rB(vq )B

since (vq )A = rAq

(vq )B = q

(vq )B = (16 rad/s)

(vq )B = 0.4 m/s

.

(rA)2

q

rB

( 0.1 m )2

0.4 m

(a) The transverse component of

the velocity of the collar.

Angular momentum of a particle.

.

.

(vq)A

(vq)B

A

B

r

rA = 0.1 m

rB = 0.4 m

Problem 3 Solution

100 mm

A

B

maq

mar

F

=

(b) The radial and transverse

components of acceleration.

Apply Newton’s second law.

Only radial force F (exerted by the

spring) is applied to the collar.

For r = 0.4 m:

F = k x = (6 N/m)(0.5 m - 0.4 m)

F = 0.6 N

Kinetics; draw a free

body diagram.

+ SFr = mar: 0.6 N = (0.25 kg) ar

ar = 2.4 m/s2

+ SFq = maq: 0 = (0.25 kg) aq

aq = 0

Problem 3 Solution

100 mm

A

B

vq

r

.

q

.

.

.

.

.

.

.

.

(c) The acceleration of the

collar relative to the rod.

Kinematics.

Forr = 0.4 m:

vq = r q, q =

q = = 1 rad/s

.

.

.

0.4 m/s

0.4 m

vq

The acceleration of the collar

relative to the rod is r.

A

B

ar

r

.

ar = r - r q 2

(2.4 m/s2) = r - (0.4m)(1 rad/s)2

r = 2.8 m/s2

aq

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