Chapter 9 Notes Chapter 8 begins the discussion of rigid bodies, a system of particles with fixed relative positions. Previously we have dealt with translation of a particle: if a rigid body does not rotate it usually can just be treated by these earlier techniques. In Chapter 8 we discuss only rotations about a fixed axis of rotation so that the rigid body rotates in a fixed plane. We will define again the things you saw in University Physics, center of mass, moment of inertia, Newton s second law for rotation, angular momentum for this situation. We will extend the ideas from that course. After Chapter 8 we will jump to Chapter 10, but then return to Chapter 9 that discusses rigid body rotation about a general axis. In that case the moment of inertia becomes a tensor, angular momentum and angular velocity are not necessarily parallel, and we will be able to discuss gyroscopes and tops. 1 Center of Mass of a Continuous Rigid Body For point masses we defined the center of mass as x cm = xi m i mi (1) with similar expressions for y and z coordinates. For a continuous body the summation becomes an integral and we can write x dm x cm = () dm where we envision the continuous object broken into small pieces of mass dm and coordinate x. But how do we evaluate this integral? We must write dm in terms of space-variables, and to do so we use the density and the volume. The density is a function of position, ρ(x, y, z) and in Cartesian coordinates the volume is dv = dx dy dz. Then ρ x dv x cm = (3) ρ dv 1
In practice there are several things we can do to make calculation simpler. 1. Look to see if an object can be considered as a combination of simpler objects: a sledge hammer can be considered as a solid cylinder (the handle) with a second cylinder (the head) mounted perpendicular to the first.. Look for symmetry. If there is a plane of symmetry, the center of mass must be located on that plane. If there is an axis of symmetry the center of mass must be located on that axis. 3. Look for ways to convert the 3D integral into a D or 1D integral. e.g. 1 Find the center of mass of a cylinder of radius r and length L. e.g. Find the center of mass of a uniform planar triangle with a base of w and a length of l. Give answer measured from the tip. e.g. 3 Suppose we make a strange hammer using a cylinder of mass m, radius r, and length L, with a triangular head mounted at right angles to the cylinder with center of base at center of cylinder end, mass 3m, base w and length l. Find the center of mass e.g. 4 Find the center of mass of a solid hemisphere of radius R, measured from the flat base. e.g. 5 Find the center of mass of a hemispherical shell of radius R, measured from the flat base. 1.1 Numerical Integrals in Mathematica In some cases a closed integral form is not practical. If you can write the integral, you can use Mathematica (or Maple) to do the integral. For example, look at the Solid Cone of Variable Density example in the text, the center of mass of a solid unit cone with density ρ = x + y. The first task is figuring out the limits of the integrals. Once that is done, Mathematica can be used to evaluate the integral by typing NIntegrate[z Sqrt[x +y ], {x, 1, 1}, {y, Sqrt[1 x ], Sqrt[1 x ]}, {z, Sqrt[x +y ], 1}] Here the capital letters and form of the brackets, [ or {, must be used. This returns the answer 0.4188888.
Kinetic Energy and Moment of Inertia: Rotation About a Fixed Axis Consider a flat object (a laminar object) rotating about an axis perpendicular to the lamina, ˆk. A point at location r i has a speed of v i = r i ω where ω is the angular speed common to all points on the lamina. As a vector equation this is v i = ω r i (4) The system kinetic energy due to the rotation is (I could use integral signs rather than summations) T rot = 1 m ivi = 1 ( ) mi ri ω (5) We define the moment of inertia about the z-axis to be I z = m i r i = m i (x i + y i ) (6) This is a constant for the object that depends on the choice of axis Similarly we can compute the angular momentum about the axis of rotation and find that L z = I z ω (7) Since we have we can write for a rigid object N = d L dt N z = I z dω dt (8) (9) CAVEAT: Once we get to Chapter 9 life is NOT this simple! 3 Calculating the Moment of Inertia for a Rigid Object About a Fixed Axis First let us define the word moment. A moment is a weighted sum (integral) of a quantity multiplied by its position raised to some power. The first moment of mass is F irst Moment = m 1 x 1 i (10) 3
We used the first moment to determine the center of mass. We use the second moment Second Moment = m 1 x i (11) to find the moment of inertia. As an integral, I = r dm = ρ r dv (1) e.g. 1 Find the moment of inertia for a thin uniform rod of mass m, length a about (a) an axis at one end (b) an axis through its center of mass. The axis is perpendicular to the rod. e.g. Find the moment of inertia of a ring of mass m, radius a about an axis through center perpendicular to the plane of ring. e.g. 3 Find the moment of inertia of a solid disk of mass m, radius a about an axis through center perpendicular to the plane of disk. e.g. 4 Find the moment of inertia of a solid cylinder of mass m, radius a, height L about an axis through center perpendicular to the plane of end of the cylinder. e.g. 5 Find the moment of inertia of a solid sphere of mass m, radius a, about an axis through center. 3.1 Perpendicular Axis Theorem for Lamina For a flat object in the x y plane rotating around the z-axis, I z = m i (x i + y i ) (13) If we rotate the lamina around the x-axis instead, and use z i = 0, Therefore for a lamina I x = m i y i (14) I z = I x + I y (15) e.g. 1 Find the moment of inertia of a solid disk of radius a, mass m about an axis through its diameter. e.g. Find the moment of inertia of a rectangular slab of mass m and dimensions a, b about an axis through the center of mass and perpendicular to the slab. 4
3. Parallel Axis Theorem Consider an object rotating about a z-axis located arbitrarily. The distance from the axis to the center of mass will be denoted l. I z = m i (x i + y i ) = m i ( (xic + x cm ) + (y ic + y cm ) ) (16) Expanding this out we will get 4 terms, two of which sum to 0, leaving I z = ml + m i (x ic + y ic) (17) e.g 1 Find the moment of inertia of a thin rod about one end by using the parallel axis throrem. e.g. Find the moment of inertia of a pendulum consisting of a rod of length l and mass m 1 attached to a disk of radius a and mass m e.g. 3 Find the moment of inertia of a solid cylinder of length b, radius a, mass m, rotating about an axis through the center of mass but perpendicular to the rod. 3.3 Conservation of Energy With Rotation About a Fixed Axis Suppose I have a solid sphere of radius a and mass m supported on an ideal pivot (no friction) through a horizontal axis through its center of mass. A chain of mass m/4 is wrapped around the top half of the sphere almost perfectly in balance, but the single link of the chain will start the sphere rotating. At the instant the chain leaves the sphere, find the angular speed of the sphere. The normal force at the pivot does no work, so we can use conservation of energy. 4 Radius of Gyration and the Physical Pendulum All the moments of inertia that we have calculated can be summarized as I = mk (18) where k has units of length. Thus for a rod pivoted around one end, k = a /3, for a sphere about an axis through a diameter, k = a /5. The quantity k is called the radius of gyration and depends on the geometry of the object and the location of the axis. Table 8.3.1 has values of k. 5
We will now look at the physical pendulum and get its period (small angle oscillations first) in terms of the radius of gyration. A rigid body rotates about an axis located a distance l from the center of mass. The gravitational torque is N = r m g (19) with magnitude This is a restoring torque, so we can write Newton s Law as For small angle sin θ θ so we have with solution where and so the period is N = mgl sin θ (0) I ω = mgl sin θ θ = gl k sin θ 0 = θ + gl sin θ (1) k 0 = θ + gl k θ () θ = Θ 0 cos(πf 0 t δ) (3) πf 0 = k T 0 = π gl gl k (4) A simple pendulum consists of a small massive ball at the end of a string of lenght L = l of negligible mass so that k = L and the period is L T 0 = π (6) g Hence a physical pendulum has the same period as a simple pendulum of length L = k /l. e.g A thin meterstick is pivoted about one end. What is the length of the equivalent simple pendulum? We ll check this out experimentally. (5) 6
Now we can cast the parallel axis theorem in terms of the radius of gyration, with k and k cm being the radii of gyration about a general axis and about an axis through the center of mass respectively. k = k cm + l (7) Thus T 0 = π (k cm + l ) gl (8) 4.1 Measuring g We can use pendula to determine g. Period can be measured accurately by measuring oscillations with an electronic timer for an extended period of time. Providing the amplitude is small we might expect an accuracy of 0.01 s during a measurement of 500 s, or parts in 10 5. For a simple pendulum we just need to measure its length accurately and trust that it meets the criteria for a simple pendulum. This is not so precise as the time measurement since the string might stretch, and we need to have a hole in the pendulum bob making the location of its center of mass more difficult. The same caveat applies to a physical pendulum, but more so. We need to measure the distance from pivot to center of mass accurately, thus must have a very precise measure of center of mass, and have the moment of inertia precisely determined, also hard. Kater s Pendulum is an approach that allows us to get very precise measurements of g without having to measure the center of mass or the moment of inertia. Consider a physical pendulum that consists of the pendulum with at least one part being adjustable in position. The pendulum has two knife edges toward either end of the pendulum. When supported from knife edge 1 the distance to the center of mass is l 1 and the period is T 1. When supported from knife edge the distance to the center of mass is l and the period is T. Now we can write the period using the radii of gyration and the parallel axis theorem, T 1 = 4π (k cm + l 1 ) gl 1 T = 4π (k cm + l ) gl (9) Rearrange these to solve for kcm and equate the two to eliminate this variable, resulting in g ( l1 4π T1 l T ) = l 1 l = (l 1 + l )(l 1 l ) (30) 7
Rearranging 8π g = (l 1T 1 l T ) (l 1 + l )(l 1 l ) Now express the numerator of the right side as (31) And factoring terms this is and substituting back, l 1 T 1 + l 1 T 1 l T l T + l T 1 l T 1 + l 1 T l 1 T (3) (l 1 + l )(T 1 T ) + (l 1 l )(T 1 + T ) (33) 8π g = T 1 + T l 1 + l + T 1 T l 1 l (34) To get g from this we would still need the values of l 1 and l, BUT if we now adjust the movable mass so that T 1 = T = T, 8π g = T l 1 + l (35) We have already indicated that we can measure period with precision. We can also measure knife edge to knife edge with precision, say to within 0.05 mm for a 500 mm separation or 1 part in 10 5. Thus combining the errors we can expect a precision of 4 parts in 10 5! Fowles and Cassiday refer to two axes on a physical pendulum that have identical periods as centers of oscillation. This seems to be a non-standard term, I have found it in no other texts, and Google returns results that say center of oscillation is synonymous with radius of gyration. 4. Large Angle Oscillations and the Elliptic Integral In the small angle approximation we have the period of oscillation of a physical pendulum about an axis a distance l from the center of mass is k T 0 = π (36) gl To get the result for large angles we start with the energy equation, E = T + V. For potential energy, V = mgh and we will choose h = 0 when the pendulum is in equilibrium. When the pendulum is rotated through an angle θ, the center of mass rises a distance h = l(1 cos θ). When the pendulum reaches maximum angle θ 0 it has no kinetic energy. We can thus write 1 I θ + mgl(1 cos θ) = mgl(1 cosθ 0 ) (37) 8
With a little algebra this becomes dθ dt = gl k (cos θ cos θ 0) (38) We choose t = 0 to be when θ = 0 and we can rearrange and integrate this equation to get k t = θ dθ (39) gl cos θ cos θ0 It is convenient to introduce a change in variables here. Let 0 sin φ = K sin θ (40) where K = sin θ 0. The text uses lower case k, but thaat can be confused with the radius of gyration, so I will use capital K. With some work (guess what will be on next problem set) the integral can be written k t = φ dφ gl 1 K sin φ k F (K, φ) (41) gl 0 and the F is the incomplete elliptic integral of the first kind. F must be compute numerically, and tables of values are available. The one I will hand out actually uses φ and θ/ (which it calls θ to confuse us!) as entries to the table. It is only evaluated for K < 1. When we reach the amplitude θ 0, we have φ = π/. The time from the above equation is 1/4 of the period, so we can write k T = 4 π/ dφ gl 1 K sin φ 4 k F (K, π/) (4) gl 0 and this is called the complete elliptic integral of the first kind. e.g. A physical pendulum is made from a solid sphere of diameter 3 cm and given the amplitude of 10. Find the period of the pendulum and compare to the result obtained from the small angle approximation. 4.3 Elliptic Integrals of second and third kind The elliptic integral of the second kind is E(K, φ) = φ 0 1 K sin φ dφ (43) 9
The elliptic integral of the third kind is π(k, n, φ) = φ 0 dφ (1 + n sin φ) 1 K sin φ In both of these, K < 1. For the third integral, n is an integer. (44) 5 Laminar Motion of A Rigid Body Thus far we have had an axis of rotation that was fixed, both in location and orientation. We now move to laminar motion, such as a ball rolling down a ramp, where the axis of rotation remains fixed in orientation, but where the axis can move and even accelerate. We start with the relation between torque and angular momentum in an inertial frame dl dt = N (45) or d ri m i v i = r i F dt i (46) Now suppose we evaluate the torques and angular momenta relative to a reference frame O that is allowed to accelerate. Refer to Figure 8.5.1 to see r i = r 0 + r i v i = v 0 + v i (47) where r 0, v 0 are the position and velocity of the non-inertial frame relative to the inertial one. In the inertial frame we have Newton s Second Law, F i = d(m i v i )/dt and hence can calculate the torque relative to the non-inertial frame. N = r 1 F i = r i d dt m i( v 0 + v = ( mi r i i ) ) d v 0 dt + = v 0 m i r i + d dt ( r ( r i d ) dt m i v i i m i v i ) (48) The time derivative can come out of the second term because in the chain rule (d r i /dt) v i = 0. The summation in the second term is just the angular momentum evaluated about O so we have N = r 0 m i r i + d L (49) dt 10
This is more complicated than Equation 45, but in three special cases the first term disappears. 1. The acceleration of O is zero. This covers rotation about a fixed axis as well as rotation about an axis having constant velocity.. O is at the center of mass. By definition m i r i = 0. This lets us write the non-inertial torque equation as N cm = d L dt cm = I cm ω (50) 3. O passes through the point of contact between a rolling object and the support, and there is rolling without slipping. If in doubt, evaluate about the center of mass for the case of laminar motion. NOTE: this discussion was for laminar motion. More general motion such as the rotation of a top is much more complicated as we shall see in Chapter 9. 6 Examples of Laminar Motion We will use Equation 50 and Newton s Second Law, to solve problems involving laminar rotation. F = m r cm (51) E.g. Body rolling on an inclined surface. Suppose a round object having mass m, radius a, and moment of inertia I cm = mk cm can roll down a surface inclined to the horizontal by θ. Choose a cartesian coordinate system with the x axis parallel to the incline (See Figure 8.6.1). The forces are the gravitational force, and the contact force between the body and the surface which we conveniently decompose into normal and frictional components. Newton s Laws then become mẍ cm = mg sin θ F P (5) mÿ cm = mg cos θ + F N (53) We constrain the object to remain on the surface, hence ÿ = 0 and F N = mg cos θ. 11
Using the Rotational Newton s Second Law, Equation 50, af P = I cm ω (54) We have two equations, but three unknowns, the two acceleration and the frictional force. Rolling Without Slipping If the coefficient of static friction is large enough, the object will roll without slipping or skidding. In this case the x velocity and angular velocity are related ẋ cm = aω (55) ẍ cm = a ω (56) The rotational equation of motion becomes af P = I cm ẍ/a and this allows us to eliminate F P in the translational equation of motion to yield mẍ cm = mg sin θ m k cm a ẍcm (57) ẍ cm = g sin θ 1 + kcm/a (58) Since the acceleration is constant, this equation can easily be integrated to get velocity and position as functions of time. Instead of choosing the center of mass for O we could have chosen the point of contact. This is done as an example in the text. Rolling and skidding If the object skids, the frictional force is a kinetic one, and we can write F P = µ k mg cos θ. The translational and rotational equations are Integrating these to get velocities, mẍ cm = mg(sin θ µ k cos θ) (59) mk cm ω = mg(µ k a cos θ) (60) ẋ cm = g(sin θ µ cos θ)t (61) ( ) µk a cos θ ω = g t (6) k cm These are both proportional to time, so we define the ratio ( ) γ = ẋcm aω = k cm tan θ a 1 µ k (63) 1
We have γ 1 with γ = 1 being the case of rolling without slipping. Thus the object will roll without slipping providing µ crit = tan θ 1 + (a/k cm ) (64) To get a feel for the solution, choose a solid ball of radius 5 cm on a board inclined by 45 with coefficient of kinetic friction 0.0. We find a center of mass acceleration of 5.54 m/s and an angular acceleration of 13.9 rad/s. You can verify that the accelerations are NOT related by ẍ = a ω. A spinning billiard ball A solid billiard ball of radius a is initially spinning about a horizontal axis with angular speed ω 0 an zero cm velocity. The coefficient of kinetic friction between the ball and the table is µ k. How far does the ball travel before it begins rolling without slipping? The equations of motion are Integrate these once to get ẍ = F P /m = µ k g (65) ω = N/I = µ kga kcm (66) ẋ = µ k gt (67) ω = ω 0 µ kga kcm t (68) Notice that the center of mass speed is increasing, while the angular speed is decreasing. At the point where slipping stops, t c the relation ẋ = aω is true. ( µ k gt c = a ω 0 µ ) kga kcm t c (69) Solve for time t c = aω ( ) 0 1 µ k g 1 + (a/k cm ) and use this in the translational kinematics equation (with our initial conditions) (70) x c = 1 ẍt c = a ω0 ( ) 1 µ k g 1 + (a/k cm ) (71) Putting in some numbers, a =.8 cm, ω 0 = 4 rad/s, and µ k t = 0.196 s and x = 1.88 cm. = 0.10, we find 13
Ladder on Smooth Surfaces A uniform ladder (think thin rod) of mall m and length l is placed at an initial angle of θ 0 from the horizontal against a smooth wall the floor is also smooth. There being no friction, the ladder will begin to move. At what angle will the ladder leave the wall? The forces on the ladder are the weight, mgĵ, and normal forces at the wall and floor, N î, N 1 ĵ. We can write the translational equations of motion as mẍ = N (7) mÿ = N 1 mg (73) For an arbitrary angle θ the center of mass is located at x cm = l cos θ/, y cm = l sin θ/. Taking derivatives Writing the energy equation ẋ = l sin θ θ (74) ẍ = l sin θ θ l cos θ θ (75) ẏ = l cos θ θ (76) ÿ = l cos θ θ l sin θ θ (77) 1 mv cm + 1 I cm θ + mgy = mgy 0 (78) We can get the center of mass velocity from vcm = ẋ + ẏ = l θ /4, and using I = ml /1 and expressing all variables in terms of θ the energy equation can be manipulated to 3g θ = l (sin θ 0 sin θ) (79) Take the derivative to get θ = 1 3g l ( cosθ) θ 3g l (sin θ 0 sin θ) = 3g cos θ (80) l Put the expressions for θ, θ into Equations 75 and 7, N = 3mg ( ) sinθ cos θ + cosθ sin θ 0 sinθ cos θ The ladder loses contact with the wall when the normal force is zero, or (81) sin θ = 3 sin θ 0 (8) 14
7 Rotational Impulse and Collisions Earlier we defined linear impulse on an object, J P = F dt (83) and the net impulse equals the change in momentum. P = m v m v (84) Similarly we can define the rotational impulse P rot = N dt (85) and for the case of laminar rotation P rot = Iω Iω (86) If the linear impulse is applied so that its line of action is a distance l from the reference axis, then P rot = P l (87) Eight Ball in the Corner Pocket What must be the height of the bumper on a pool table to allow balls to bounce off the bumper without skidding? Call the radius of the ball a and the height of the bumper h. The ball has an initial velocity vî and a final velocity v î. A horizontal impulse P î is supplied by the bumper. Applying linear and rotational Impulse Momentum Theorems, The no-skid conditions require Combining, P = mv mv (88) P (h a) = Iω Iω (89) v = aω (90) v = aω (91) P + mv = mv = maω (9) ( ) P (h a) P + maω = ma + ω (93) I P = ma(h a) P I (94) ma 5 = ma(h a) (95) h = 7 5 a (96) 15
In other words, the bumper height must be 7/10 of the diameter of the ball. Check it out! Center of Percussion Consider a baseball bat, mass M, suspended so that it is free to move. A ball, mass m, strikes the bat at some point. What are the subsequent motions of the bat and the ball? Refer to figure 8.7.1. If the ball initially moves to the right along the x-axis, it bounces off and moves at a different velocity in the x-axis. The motion of the bat is more complicated, being a combination of translation and rotation. If the ball strikes the bat at its center of mass, the bat will end up in pure translation. Is it possible to find a location so that some point on the bat has no translational velocity? There is an impulse P between ball and bat. We can write From rotational impulse we can write for the bat, P = M v cm (97) P = m v 1 m v 0 (98) ω 0 = P l I cm (99) where l is the distance from the center of mass to the impact point. We can write the velocity of another point a distance l from the center of mass as v O = v cm lω = P M P ( ) l 1 l = P I cm M ll (100) I cm This velocity can be zero providing ll = I cm M (101) 16