Chapter 9: Rotational Dynamics Tuesday, September 17, 2013

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Chapter 9: Rotational Dynamics Tuesday, September 17, 2013 10:00 PM The fundamental idea of Newtonian dynamics is that "things happen for a reason;" to be more specific, there is no need to explain

However, acceleration must be caused; that is, acceleration doesn't just happen for no reason, it must be explained, and the reason acceleration occurs is that there must be a net force to cause it.

1 Chapter 9: Rotational Dynamics Tuesday, September 17, :00 PM The fundamental idea of Newtonian dynamics is that "things happen for a reason;" to be more specific, there is no need to explain rest or motion in a straight line at a constant speed. However, acceleration must be caused; that is, acceleration doesn't just happen for no reason, it must be explained, and the reason acceleration occurs is that there must be a net force to cause it. The same is true for rotational motion, but we can be a little more specific in this case. Angular acceleration must be caused, and the cause of angular acceleration is that there must be a net force. However, the force has to be oriented appropriately to cause angular acceleration. This leads us to the concept of the torque of a force. (In other countries, the term "moment of a force" is used instead of "torque of a force;" the two terms are interchangeable and mean exactly the same thing.) Torque The torque of a force relative to a certain axis is a measure of the effectiveness of the force in producing rotation (more specifically, angular acceleration) of an object about the axis. The torque depends on the magnitude of the force, the direction of the force, and the distance between the rotation axis and the point at which the force is applied. Note that in the following three figures, which show an overhead view, the force in each case has the same magnitude, but in each case the effectiveness of the force at opening the door is different. Which force is most effective at opening the door? Definition of the torque of a force with respect to a certain axis: Ch9Preliminary Page 1

2 Various ways to calculate the torque of a force: you can use the product of the radius vector and the perpendicular component of the force, as in the hand-written figure above, or you can multiply the force by the "lever arm," as in the figure below: examples: wrench, screwdriver, crow bar, wine press, etc. Example: Determine the net torque of the tension forces about the axle of the pulley. Solution: Note that T 1 tends to turn the pulley in the CCW sense, so the torque of T 1 with respect to an axis through the centre of the pulley is positive. On the other hand, the torque of T 2 with respect to an axis through the centre of the pulley is negative, because T 2 tends to turn the pulley in the CW sense. Ch9Preliminary Page 2

3 Example: Determine the net torque about (a) point A, and (b) point B. Rigid bodies in equilibrium Ch9Preliminary Page 3

4 We learned earlier that if an object is at rest, then the total of all the external forces acting on it must be zero. If we are also considering the possibility of rotational motion, we must add a second condition for a body at rest: The net torque of all the external forces acting on the body, with respect to each axis, must also be zero. To see how this works in action, first let's review the concept of centre of mass (which is, for our purposes, the same as the concept of centre of gravity). Centre of Mass (Centre of Gravity) The centre of mass of a collection of "point" objects is a vector, and formulas for the components of this vector are given below. You can interpret the centre of mass formulas as weighted averages (double meaning very much intended). Think about how you might calculate your final grade in a course based on your grades on each graded item, including the weight of each item. The centre of mass uses the same idea, where the position of each particle is accounted for, and each particle in the formula is "weighted" by its mass. For continuous mass distributions, one can use similar formulas involving integration. Example: Determine your grade so far in a course in which you have completed the following graded items: Test 1, worth 10%: 92 Test 2, worth 10%: 80 Essay, worth 40%: 84 Ch9Preliminary Page 4

5 Group project, worth 10%: 36 Incorrect Solution: (Don't do this!) Correct solution: Example: Determine the centre of mass for the system of three identical coins. Ch9Preliminary Page 5

6 Example: (a) Determine the centre of mass of the two bars considered as one unit. (b) Determine the net torque of the bars about the far left end. Strategy: Pretend that the mass of each bar is concentrated at its geometrical centre. Ch9Preliminary Page 6

Now let's look at some examples of bodies in static equilibrium. Example: Determine the normal force on each hand and foot, assuming that the person is motionless. The weight of the person is 584 N.

7 Now let's look at some examples of bodies in static equilibrium. Example: Determine the normal force on each hand and foot, assuming that the person is motionless. The weight of the person is 584 N. Solution: Let N H represent the normal force that the floor exerts on each hand, and let N F represent the normal force that the floor exerts on each foot. We assume that the person is balanced so that the normal force on one hand is equal to the normal force on the other hand, and that the normal force on one foot is equal to the normal force on the other foot. Ch9Preliminary Page 7

8 Before solving the problem, guess which normal force will be greater, and roughly by how much. Draw a free-body diagram! Because the person is in static equilibrium, the net force acting on the person is zero. This gives us one equation for the two unknown normal forces. We need another equation. The other equation comes from the idea that because the person is in static equilibrium, the net torque acting on the person about any axis whatsoever is also zero. So, pick an axis and calculate the net torque of all the forces about that axis. Wait, really? We can just pick any axis? Yes, often picking any axis at all works, so just pick the axis that makes your calculation as simple as possible. (Sometimes not just any axis will do; for example, if there is an unknown force that you have to calculate, it usually doesn't work well if you choose an axis about which the torque of the required force is zero, because then the unknown force won't appear in the equation.) So, for the torque equation, I will choose an axis perpendicular to the page and passing through the red dot in the free-body diagram, as I think this will make life easier for us. (There are other choices that are equally easy; try some!) We now have two equations in two unknowns, so we can solve the problem. For the next step I will solve the first equation for 2N H and substitute the resulting expression into the second equation, and solve the resulting equation Ch9Preliminary Page 8

9 for N F. Now substitute this value into equation (1) and solve for the other normal force: Thus, the normal force at each hand is 196 N and the normal force at each foot is 96 N. Is this reasonable? Does it correspond with your experience when doing push-ups or planks? Example: A person holds a 178-N ball in his hand, with the forearm horizontal (see the drawing). The person can support the ball in this position because of the flexor muscle force M, which is applied perpendicular to the forearm. The forearm weighs 22.0 N and has a center of gravity as indicated. Determine (a) the magnitude of the flexor muscle force, M, and (b) the magnitude and direction of the force applied by the upper arm bone to the forearm at the elbow joint. Ch9Preliminary Page 9

10 Solution: Draw a free-body diagram! Focus on the arm as the body, and write down all of the forces acting on the arm. I have guessed that the upper arm exerts a force on the lower arm (at the elbow joint) that has components as shown; if I'm wrong, the actual values in our solution will come out negative, which is an indication that I guessed wrong. In other words, guessing wrong about the direction of this force does not hinder us in solving the problem. The net force acting on the arm in the x-direction is zero, because the arm is in static equilibrium: The net force acting on the arm in the y-direction is zero, because the arm is in static equilibrium: Ch9Preliminary Page 10

11 arm is in static equilibrium: The net torque (about an axis perpendicular to the page through the elbow joint) of all forces acting on the arm is zero, because the arm is in static equilibrium: Equation (1) is already solved. Equation (3) has just one unknown, so it makes sense to solve equation (3) for M, then substitute the value into equation (2) and solve for N y. Substituting the value of M into equation (2), we can solve for N y : (a) Thus, the magnitude of the flexor muscle force, M, is 1200 N. (b) The magnitude of the force applied by the upper arm bone to the forearm at the elbow joint is 1000 N, and it is directed straight down. Example: A 1220-N uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. A 1960-N crate hangs from the far end of the beam. Using the data shown in the drawing, determine Ch9Preliminary Page 11

(a) the magnitude of the tension in the wire, and (b) the magnitudes of the horizontal and vertical components of the force that the wall exerts on the left end of the beam.

12 (a) the magnitude of the tension in the wire, and (b) the magnitudes of the horizontal and vertical components of the force that the wall exerts on the left end of the beam. Solution: Draw a free-body diagram! Focus on the beam as "the body," and draw on the diagram all the forces acting on the beam. (See above.) The net force acting on the beam in the x-direction is zero, because the beam is in static equilibrium: The net force acting on the beam in the y-direction is zero, because the beam is in static equilibrium: Let L represent the length of the beam. The net torque (about an axis perpendicular to the page through the point A in the free-body diagram) of all forces acting on the beam is zero, because the beam is in static equilibrium: Ch9Preliminary Page 12

13 Now substitute the value of the tension into equations (1) and (2) to solve for the components of the force that the wall exerts on the beam: Ch9Preliminary Page 13

14 Rotational Dynamics; Rotational Analogue of Newton's Second Law of Motion Newton's second law relates the total force acting on an object to its acceleration. The rotational analogue of Newton's second law of motion relates the net torque acting on an object to its angular acceleration. To understand the rotational analogue of Newton's second law of motion, we must first understand the rotational analogue of inertia. That is, for linear motion, mass is a measure of an object's "resistance to acceleration;" in other words, mass is a measure of an object's inertia. For rotational motion, it is more complex. Think about swinging a heavy mass; if you swing it close to your body it's easier to swing than if you swing it far from your body. Thus, the difficulty in rotating an object both depends on its mass and how far the mass is from the rotation axis. For an extended body (as opposed to a "point" object), different parts of the body are at different distances from the rotation axis, so how hard it is to rotate the body depends on both the total mass and how the mass is distributed within the body. This leads us to the concept of rotational inertia, or "moment of inertia," as it is called. Consider the following line of reasoning. The Concept of Moment of Inertia The heart of Newtonian mechanics is Newton's second law of motion, which tells us how forces affect motion; specifically, Newton's second law tells us how much acceleration you get when a given net force acts on an object of mass m: What is the analogue of Newton's second law of motion for rotational motion? Consider the following argument for a point particle: Ch9Preliminary Page 14

15 It appears that mr 2 plays the role for angular motion that m plays for linear motion; that is, mr 2 is a kind of rotational inertia, or resistance to angular acceleration, in the same way that mass a measure of the resistance to acceleration. The quantity mr 2 is called the moment of inertia for a point particle of mass m. For a collection of point masses, the total moment of inertia about a specified axis is For a continuous mass distribution (like a baseball bat, or rotating parts of machinery) one needs calculus to determine the moment of inertia. A table of moments of inertia for common objects is found on Page 254 of the textbook. It's worthwhile having a look at the table and getting a feel for it. A similar table, from another textbook, is reproduced below: Ch9Preliminary Page 15

16 Examples: implements attached to baseball bats for warmups; sledge hammer Example: A pulley with a frictionless bearing and mass 1.3 kg and radius 0.35 m has a rope going over it as shown in the diagram. Determine the angular Ch9Preliminary Page 16

17 acceleration of the pulley. Solution: The rope touches the pulley in such a way that the force exerted by the rope on the pulley is at right angles to the "radius vector" that starts at the centre of the pulley and touches the edge of the pulley at the point where the rope touches the pulley. Thus, the net torque on the pulley is Note that the sign of the angular acceleration is positive, which makes sense because the counterclockwise torque is greater than the clockwise torque, so the net torque is counterclockwise. Ch9Preliminary Page 17

18 Example: A wheel has a moment of inertia of 0.55 kg m 2 and an initial angular speed of 0.85 rev/s. Friction in the wheel bearings causes the wheel to stop turning after 14.5 s. Determine the magnitude of the frictional torque. Solution: First determine the angular acceleration of the wheel as it slows down, assuming that it is constant. Then use the relationship between torque and angular acceleration to determine the magnitude of the torque. Also note that because we need only the magnitude of the torque, we can ignore the signs of the torque and angular acceleration. Example: A 1.2 kg block and a 1.8 kg block are attached to the ends of a rope that hangs over a solid pulley of diameter 0.42 m and mass 1.3 kg. The axle of the pulley is frictionless and the rim of the pulley has enough friction so that the rope rotates the pulley without slipping. Determine the accelerations of the blocks, the tensions in the rope, and the angular acceleration of the pulley. Solution: As usual, the first step is to draw free-body diagrams. However, the new wrinkle in this problem is that we'll also draw a free-body diagram for the pulley. Then we'll apply Newton's second law to each block, and we'll also apply the rotational analogue of Newton's second law to the pulley. Ch9Preliminary Page 18

19 In applying the rotational analogue of Newton's second law to the pulley, we will get an equation involving the angular acceleration. It will be helpful to connect this with the acceleration of the rope, which is the same as the tangential acceleration of the rim of the pulley, because we assume that the rope turns the pulley without slipping. That is, Solve equations (1) and (2) for the tensions, and then substitute the results into equation (3), then solve for the acceleration. Ch9Preliminary Page 19

20 Are these results reasonable? Is there a way to check them? Rotational work and energy When a force acts on an object in such a way that the angular speed of the object increases, then work has been done on the object, and the kinetic energy of the object also increases. The purpose of the next few paragraphs is to develop formulas for the work done when a force exerts a torque on a rotating object, and for the kinetic energy of a rotating object. First consider a rope that turns a wheel around a fixed axis or rotation, and let's determine the work done by the tension in the rope: Ch9Preliminary Page 20

To make things easy, first think about the kinetic energy of just two small "blobs" attached to the rotating object: Thus, the kinetic energy of the two blobs is one-half of their total moment of

21 As usual, make sure to use radians for the angle measure in the above formula if you wish the units for work to be reasonable. Now consider the kinetic energy of an object that rotates about a fixed centre. To make things easy, first think about the kinetic energy of just two small "blobs" attached to the rotating object: Thus, the kinetic energy of the two blobs is one-half of their total moment of inertia times the square of the angular speed. This formula is very reminiscent of the kinetic energy formula for translational motion, isn't it? It can be shown (calculus lovers!) that the same formula is valid for the kinetic energy of any continuous rigid object, where I is the moment of inertia of the rigid object. It can also be shown that the total kinetic energy for a rolling object is the sum of the translational kinetic energy (using the speed of the centre of mass of the object) and rotational kinetic energy (using the angular speed of the object about its centre of mass). This is a pleasing result, and is further justification of the usefulness of the centre of mass concept. Ch9Preliminary Page 21

22 Example: A helicopter has two blades; each blade has a mass of 240 kg and can be approximated as a thin rod of length 6.7 m. The blades are rotating at an angular speed of 44 rad/s. (a) Determine the total moment of inertia of the two blades about the axis of rotation. (b) Determine the rotational kinetic energy of the spinning blades. Solution: (a) Using the table of moments of inertia, and recognizing that there are two blades, the total moment of inertia of the two blades about the rotation axis is Example: A solid sphere is rolling on a surface without slipping. What fraction of its total kinetic energy is in the form of rotational kinetic energy about its centre of mass? Solution: Write expressions for the rotational kinetic energy and total kinetic energy, and determine their ratio. However, before taking their ratio, it would be wise to express the rotational kinetic energy in terms of the translational speed, as we expect there to be cancellation. Ch9Preliminary Page 22

23 For a solid sphere, the moment of inertial about its centre of mass is and the no-slipping condition is Thus, Thus, the ratio of rotational kinetic energy to total kinetic energy for the solid sphere that rolls without slipping is Ch9Preliminary Page 23

24 Thus, a little less than 30% of the sphere's kinetic energy is rotational, and a little more than 70% is translational. Example: A tennis ball, starting from rest, rolls (without slipping) down the hill in the drawing. At the end of the hill the ball becomes airborne, leaving at an angle of 35 degrees with respect to the horizontal. Treat the ball as a thin-walled spherical shell, and determine the range x. Discussion: Our goal is to determine the translational velocity of the ball at point 2. If we can do this, then the rest of the problem is just a projectile motion problem, of the type that we have solved back in Chapter 3. Solution: The moment of inertia of the ball is Assuming that the ball's mechanical energy is conserved between points 1 and 2 in the diagram, Ch9Preliminary Page 24

25 Because of the no-slipping condition, Thus, Now that we know the translational speed at position 2, and we know the projection angle, we can solve the projectile motion problem to determine the range. Try solving the rest of the problem for yourself; the result I obtained is x = 2.03 m. Ch9Preliminary Page 25

26 Angular momentum One perspective on Newton's second law is This led to identifying momentum as a concept of interest. Is there a rotational analogue of momentum? Principle of conservation of angular momentum: If the net external torque acting on a system is zero, then the angular momentum of the system is conserved. Example: a spinning figure skater; note how the angular speed of the skater changes as Ch9Preliminary Page 26

she brings her arms in or out. When she brings her arms in, her moment of inertia decreases, so her angular speed increases as her angular momentum stays constant.

27 she brings her arms in or out. When she brings her arms in, her moment of inertia decreases, so her angular speed increases as her angular momentum stays constant. When she wishes to slow down, she extends her arms again, which increases her angular momentum and decreases her angular speed, again maintaining her constant angular momentum. Behaviour of a spinning bicycle wheel when external torques are applied. demonstrations Note that angular momentum is a vector, and so is torque; using the right-hand rule we can predict how a spinning wheel behaves when various torques are applied. Example: Precession of a spinning bicycle wheel: Example: "Push Steering" (also known as counter steering) Ch9Preliminary Page 27

28 Check the following video for an explanation of how push steering works; ignore the first two minutes of physics explanation, and concentrate on the motions of the front wheel of the motorcycle. We do this unconsciously, as our bodies just do the right thing to keep our balance without us thinking Ch9Preliminary Page 28

29 consciously about it. But if you pay close attention next time you are on a bicycle (or motorcycle) then you'll become aware of the effect. The effect is more noticeable on a motorcycle because motorcycles are heavier than bicycles. Bicycles are lighter, so we can do most of the work by leaning our bodies, whereas you really have to use push steering on a motorcycle or be in danger of falling. Angular momentum of a "point" particle moving in a circle at speed v: Ch9Preliminary Page 29

30 Example: Calculate the angular momentum of the Moon in its orbit around the Earth. Solution: Treat the Moon as a point particle moving in a circular orbit around the Earth. (Ignore the spin of the Moon on its own axis.) Example: Orbital angular momentum of an electron in the ground state of a hydrogen atom The following example shows how to use conservation of angular momentum to solve a problem: Example: The figure shows a 100-g puck revolving in a 20-cm circle on a frictionless table. The string passes through a hole in the centre of the table and is tied to two 200-g masses. (a) What speed does the puck need to support the two masses? (b) The lower mass is a light bag filled with sand. Suppose a hole is poked in the bag and the sand slowly leaks out as the puck is revolving. Determine the speed of the puck and the radius of its path after all the sand is gone. Ch9Preliminary Page 30

31 Solution: (a) Draw a free-body diagram for the two hanging masses considered as one object. Then use the free-body diagram to determine the tension in the string, assuming that the speed of the puck is sufficient to keep the hanging masses still. Finally determine the speed at which the puck must move so that the tension in the string provides all the centripetal force needed to keep the puck in a circle of radius 20 cm. Ch9Preliminary Page 31

32 (b) The hanging mass decreases by a factor of 2, so the tension in the string also decreases by a factor of 2. Thus, the tension in the string is T = 0.2g. Let V be the final speed of the puck and let R be the final radius of the puck's circular path. Then Our task is to determine V and R, but we have only one equation for two unknowns. We need another independent equation. The other equation can be obtained by noting that the falling sand does not have any angular momentum with respect to the central axis of the puck's revolution. This means that if we consider the hanging mass, the sand, and the puck to be one single system, then its angular momentum is conserved. Neither the hanging weight nor the sand have any angular momentum, so this means that the angular momentum of the puck is conserved. Thus, Ch9Preliminary Page 32

33 Substituting this expression into equation (1) and solving for R, we obtain The final speed of the puck is Ch9Preliminary Page 33

34 Thus, when the sand leaks out of the bag, the puck slows down and moves in a larger circle. Is this reasonable? For example, what would happen if the other hanging mass slipped off the string? Ch9Preliminary Page 34

Chapter 9: Momentum Tuesday, September 17, :00 PM

Ch9 Page 1 Chapter 9: Momentum Tuesday, September 17, 2013 10:00 PM In this chapter and the next one, we'll explore alternative perspectives to Newton's second law. The concepts of momentum and energy