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.

Now let’s generalize this to 3-dimensional vector fields Now let’s generalize this to 3-dimensional vector fields. Assume the you have a force vector written in Cartesian coordinates. where is denoted the gradient operator 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).

What about kinetic energy? Starting from the previous derivation for potential energy 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.

Solution by energy method 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:

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. q r = 600 mm C A B 200 g O

A small 200-g collar C can slide on 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. q r = 600 mm C A B 200 g O 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.

A small 200-g collar C can slide on 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. q r = 600 mm C A B 200 g O 3. Apply Newton’s second law: The relationship between the forces acting on the particle, its mass and acceleration is given by S F = m a . 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.

an = (r sinq) w2 an = (0.6 m) sinq ( 6 rad/s )2 an = 21.6 sinq m/s2 Problem 1 Solution Kinematics. q r = 600 mm C A B 200 g O 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 sinq m/s2

= man = (0.2) 21.6 sinq = 4.32 sinq N Problem 1 Solution B r = 600 mm C A B 200 g O Kinetics; draw a free body diagram. q (0.2 kg)(9.81 m/s2) O N F = man = (0.2) 21.6 sinq = 4.32 sinq N

Problem 1 Solution F Apply Newton’s second law. O = man = (0.2) 21.6 sinq = 4.32 sinq N q N (0.2 kg)(9.81 m/s2) + 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 For a given q, the values of F , N , and m can be determined

= man = (0.2) 21.6 sinq = 4.32 sinq N Problem 1 Solution F O N q (0.2 kg)(9.81 m/s2) O N F man = (0.2) 21.6 sinq = 4.32 sinq N = 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)

Problem 2 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. . q b r O A B C D E

. eq v = r er + r q eq er a = (r - r q 2 ) er + (r q + 2 r q ) eq b r O A B C D E Problem 2 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. 1. Kinematics: Examine the velocity and acceleration of the particle. In polar coordinates: v = r er + r q eq a = (r - r q 2 ) er + (r q + 2 r q ) eq . r = r er eq er q

q b r O A B C D E Problem 2 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. 2. Kinetics: Draw a free body diagram showing the applied forces and an equivalent force diagram showing the vector ma or its components.

q b r O A B C D E Problem 2 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. 3. Apply Newton’s second law: The relationship between the forces acting on the particle, its mass and acceleration is given by S F = m a . 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: S Fr = m ar = m ( r - r q 2 ) and S Fq = m aq = m ( r q + 2 r q ) . .

r = 2 b cos q q = 20o r = - 2 b sin q q q = 15 rad/s E Problem 2 Solution Kinematics. 2q b r O A B q r = 2 b cos q r = - 2 b sin q q r = - 2 b sin q q - 2 b cos q q 2 q = 20o q = 15 rad/s q = 250 rad/s2 . . . . . .

r = 2 b cos q, r = - 2 b sinq q, r = - 2 b sinq q - 2 b cosq q 2 A B q Problem 2 Solution . . . r = 2 b cos q, r = - 2 b sinq q, r = - 2 b sinq q - 2 b cosq q 2 . . 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 . . . . . .

= Problem 2 Solution r C (a) Radial and transverse B D q b r O A B C D E (a) Radial and transverse components of the resultant force exerted on pin B. Kinetics; draw a free body diagram. r O A B q Fr Fq mar maq =

= + SFr = mar: Fr = m ( r - r q2 ) Apply Newton’s second law. Problem 2 Solution r O A B q Fr Fq mar maq = + SFr = mar: Fr = m ( r - r q2 ) Fr = [- 989.79 - ( 3.13 )(152 )] = -13.16 lb Fr = 13.16 lb + SFq = maq: Fq = m ( r q + 2 r q ) Fq = [(3.13)(250) + 2 (-17.1)(15)] = 2.1 lb . . (4/16) 32.2 . . . (4/16) 32.2 Fq = 2.10 lb

= Fr = - Q cos q Fq = - Q sin q + P -13.16 = - Q cos 20o b r O A B C D E Problem 2 Solution Fq Fr Q q = r B r q B q q q P A O A O Fr = - Q cos q -13.16 = - Q cos 20o Q = 14.00 lb Fq = - Q sin q + P 2.10 = - 14.0 sin 20o + P P = 6.89 lb 20o 40o

Problem 3 A 250-g collar can slide on a 400 mm 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 Problem 3 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 = r er + r q eq a = (r - r q 2 ) er + (r q + 2 r q ) eq r = r er eq er q . . . . . . .

. Problem 3 400 mm 100 mm The collar is initially held at A by a A B 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. 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. S MO = 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 .

400 mm 100 mm A B Problem 3 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.

. . Problem 3 400 mm 100 mm A The collar is initially held at A by a B 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 S F = m a . 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: S Fr = m ar = m ( r - r q 2 ) and S Fq = m aq = m ( r q + 2 r q ) . .

. . . . . . . . vq = r q ar = r - r q 2 aq = r q + 2 r q 400 mm 100 mm B Problem 3 Solution Kinematics. q . vq . vq = r q ar = r - r q 2 aq = r q + 2 r q A B ar r . . aq . . . .

. . . m rA (vq )A = m rB (vq )B since (vq )A = rA q (vq )B = q 400 mm 100 mm A B Problem 3 Solution (a) The transverse component of the velocity of the collar. Angular momentum of a particle. m rA (vq )A = m rB (vq )B since (vq )A = rA q (vq )B = q (vq )B = (16 rad/s) (vq )B = 0.4 m/s (rA)2 rB ( 0.1 m )2 0.4 m . q . . (vq)A (vq)B A B r rA = 0.1 m rB = 0.4 m

400 mm 100 mm A B Problem 3 Solution (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. F m ar m aq = + SFr = mar: 0.6 N = (0.25 kg) ar ar = 2.4 m/s2 + SFq = maq: 0 = (0.25 kg) aq aq = 0

. . . . . . . . . vq = r q, q = vq q = = 1 rad/s r 0.4 m/s 0.4 m 400 mm 100 mm A B Problem 3 Solution (c) The acceleration of the collar relative to the rod. Kinematics. For r = 0.4 m: vq = r q, q = q = = 1 rad/s . . vq r . 0.4 m/s q . 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 .