Ch 8. Rotational Dynamics

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1 Ch 8. Rotational Dynamics Rotational W, P, K, & L (a) Translation (b) Combined translation and rotation

2 ROTATIONAL ANALOG OF NEWTON S SECOND LAW FOR A RIGID BODY ROTATING ABOUT A FIXED AXIS = τ Iα Requirement: α must be expressed in rad/s Although a rigid object possesses a unique total mass, it does not have a unique moment of inertia, for the moment of inertia depends on the location and orientation of the axis relative to the particles that make up the object.

3 Newton s nd Law and Rotational Motion: There is an equivalent to Newton s nd law for rotational motion. F net = ma τ net = Iα This can be combined with the kinematics equations from earlier in the unit to solve uniform motion problems; ω = ω o + α t ω o = ω + αδθ Δθ = ω o t + α 1 t ω + ω o Δθ = t

4 Example #: A force of 5.0 N is applied to the edge of a disk that can spin about its center. The disk has a mass of 40 kg and a diameter of 3.0 m. If the force is applied for 4.0 s, how fast will the disk be turning if it starts from rest? I disk = 1 mr τ net = Iα τ = rf sin 90 rf = τ = Iα = mr α

5 rf = mr α α = F mr ( 5 N ) ( 40kg )( 1.60 m) = 1.17 = rad s = ( rad = )( 4. 0s) ω ω o + α t s ω = 8.1 rad s

6 Rotational Work and Power

7 Rotational Work and Energy DEFINITION OF ROTATIONAL WORK The rotational work W R done by a constant torque τ in turning an object through an angle θ is W = τ θ = F s = F r θ Requirement: θ must be expressed in radians. SI Unit of Rotational Work: joule (J)

8 Work and Power in Rotational Motion l The work done by a torque on an object that undergoes an angular displacement from θ 1 to θ is given by W = τ θ Work done by a torque l If the torque is constant then the work done is given by W = τ z( θ θ1) = τ zδθ Work done by a constant torque l Note: similarity between these expressions and the equations for work done by a force (W=FS).

9 Work and Power in Rotational Motion l The rotational analog to the Work - Energy Theorem is W tot = 1 Iω 1 Iω 1 = K f K i = ΔK l The change in rotational kinetic energy of a rigid body equals the work done by forces exerted from outside the body. l The rate at which work is performed is the power P = W t! = τ θ $ # & = τ ω " t %

10 The dependence of the inertia on mass and distribution can be built up through the kinetic energy of an object moving through a circular path. The kinetic energy of a mass m moving at a speed v around the circle is: KE = 1 mv Let the circle have a radius r, and let the angular speed of the mass be ω. Write the kinetic energy in terms of the angular speed: KE = 1 mr ω Define the moment of inertia I of this point mass to be: I = mr KE = 1 Iω Then: This is now the definition.

11 Now build up to a more complicated object: The total kinetic energy becomes: KE = 1 m1v m v + v m3 3 Mass m 1 travels in a circle of radius r 1, etc. All masses have the same angular velocity, ω. Write the kinetic energy in terms of this: m1r 1 ω + m ω r ω m3r3 KE = + ( ) 1 m r + m r + m r ω I KE = = 1 ω

12 Rigid-Body Rotation about a Moving Axis l The kinetic energy of an object that is rolling without slipping is given by the sum of the rotational kinetic energy about the center of mass plus the translational kinetic energy of the center of mass: 1 K = + 1 Mv cm I cm ω Rigid body with both translation and rotation l l If a rigid body changes height as it moves, you must also consider gravitational potential energy The gravitational potential energy associated with any extended body of mass M, rigid or not, is the same as if you replace the body by a particle of mass M located at the body s center of mass: U = Mgy cm

Example #5: Four different objects are placed at the top on an incline, as shown below. A point particle can slide down without friction. The other three objects will roll down the incline.

13 Example #5: Four different objects are placed at the top on an incline, as shown below. A point particle can slide down without friction. The other three objects will roll down the incline. In what order will the objects reach the bottom, from fastest to slowest? (a) What is the speed of the sliding point particle when it reaches the bottom? E top = E bottom KE + PE = KE + top top bottom PE bottom 0 0 mgh = 1 mv v point = gh

14 (b) Solve for the speed of the sphere (solid ball) at the bottom. I cm = 5 mr ω = v cm r Energy conservation! E top = E bottom mgh = mv ω I Note that there is a fixed starting energy, and this is split between linear motion and rotation. The rolling objects are slower! mgh = 1 mv mr v r = 7 mv v = 7 gh

15 (c) Solve for the speed of the hoop at the bottom. I cm = mr Energy conservation! ω = v cm r E top = E bottom mgh = mv ω I mgh = 1 mv + 1 mr v r = mv v = gh The hoop has the slowest speed, and thus takes the longest to reach the bottom. The disk will be between the ball and the hoop. t < t < t < point ball disk t hoop

16 Angular Momentum

17 Angular Momentum We know that: p = mv Just like others: L = p r Which changes to: L = mvrsinθ L = mvr = mr ω

18 Angular Momentum DEFINITION OF ANGULAR MOMENTUM The angular momentum L of a body rotating about a fixed axis is the product of the body s moment of inertia I and its angular velocity ω with respect to that axis: L = Iω Requirement: ω must be expressed in rad/s. SI Unit of Angular Momentum: kg m /s

19 Example Four identical masses rotate about a common axis, 1. meters from the center. Each mass is.5 kg, and the system rotates at π rad/sec. The rods connecting them are assumed to be massless. Find the total angular momentum of this system. 1. m.5 kg Ans: 90.5 kg m /s

20 Example Continued The four masses are then pulled toward the center until their radii are 0.5 meters. This is done in such a manner that no external torque acts on the system. What is the new angular speed of the system? Ans: 36. rad/s

The combination rotates in the xy plane about a pivot through the center of the rod.

21 A light rigid rod 1.00 m in length joins two particles, with masses 4.00 kg and 3.00 kg, at its ends. The combination rotates in the xy plane about a pivot through the center of the rod. Determine the angular momentum of the system about the origin when the speed of each particle is 5.00 m/s. L = mvr i i i ( 4.00 kg)( 5.00 m s)( m) ( 3.00 kg)( 5.00 m s)( m) = + L = ( 17.5 kg m s)

22 Example. A Satellite in an Elliptical Orbit An artificial satellite is placed into an elliptical orbit about the earth. Telemetry data indicate that its point of closest approach (called the perigee) is r P = m from the center of the earth, and its point of greatest distance (called the apogee) is r A = m from the center of the earth. The speed of the satellite at the perigee is v P = 8450 m/s. Find its speed v A at the apogee. 51

23 Kepler s second law of planetary motion states that a line joining a planet to the sun sweeps out equal areas in equal time intervals. 5

24 Conservation of Angular Momentum

25 Conservation of Angular Momentum l When the net external force torque acting on a system is zero, the total angular momentum of the system is constant (or conserved) τ = Δ r L Δt = 0, r L = const Angular momentum conservation l l This principle is universal conservation law, valid at all scales from atomic and nuclear systems to the motions of galaxies Circus acrobat, diver, ice skater use this principle: l Suppose acrobat has just left a swing with arms and legs extended and rotating counterclockwise about her center of mass. When she pulls her arms and legs in, her moment of inertia I cm with respect to her center of mass changes from a large value I 1 to much smaller value I. The only external force is her weight, which has no torque (goes through center of mass). So angular momentum remains constant, and angular speed changes: I 1 ω 1 = I ω

26 If the component of the net external torque on a system along a certain axis is zero, the component of the angular momentum of the system along that axis cannot change, no matter what changes take place within the system. This conservation law holds not only within the frame of Newton s mechanics but also for relativistic particles (speeds close to light) and subatomic particles. I i ω = i I f ω f ( I i,f, ω i,f refer to rotational inertia and angular speed before and after the redistribution of mass about the rotational axis ).

27 Examples: Spinning volunteer Torque ext =0 à I i ω i = I f ω f I f < I i (mass closer to rotation axis) ω f > ω i

28 Physics of Falling Cats How does a cat land on its legs when dropped? Moment of inertia is important... ] ] ] ] To understand how a cat can land on it's feet, you must first know some concepts of rotational motion, since the cat rotates as it falls. Reminder: The moment of inertia of an object is determined by the distance it's mass is distributed from the rotational axis. Relating this to the cat, if the cat stretches out it's legs and tail, it increases it's moment of inertia; conversely, it can decrease it's moment of inertia by curling up. Remember how it was proved by extending your professor s arms while spinning around on a swivel chair? Just as a more massive object requires more force to move, an object with a greater moment of inertia requires more torque to spin. Therefore by manipulating it's moment of inertia, by extending and retracting its legs and rotating its tail, the cat can change the speed at which it rotates, giving it control over which part of it's body comes in contact with the ground.

29 Physics of Falling Cats... and the conservation of angular momentum... ] ] ] ] ] ] If a cat is dropped they almost always tend to land on their feet because they use the conservation of angular momentum to change their orientation When a cat falls, as you would expect, its centre of mass follows a parabolic path. The cat falls with a definite angular momentum about an axis through the cat s centre of mass. When the cat is in the air, no net external torque acts on it about its centre of mass, so the angular momentum about the cat s centre of mass cannot change. By pulling in its legs, cat can considerably reduce it rotational inertia about the same axis and thus considerably increase its angular speed. Stretching out its legs increases its rotational inertia and thus slows the cat s angular speed. Conservation of angular momentum allows cat to rotate its body and slow its rate of rotation enough so that it lands on its feet

as a whole is zero A free-falling cat cannot alter its total angular momentum.

30 Conservation of Angular Momentum ] ] ] Falling cat twists different parts of its body in different directions so that it lands feet first At all times during this process the angular momentum of the cat as a whole is zero A free-falling cat cannot alter its total angular momentum. Nonetheless, by swinging its tail and twisting its body to alter its moment of inertia, the cat can manage to alter its orientation

31 Falling Cats: More Information How does a cat land on its legs when dropped? ] ] ] Cats have the seemingly unique ability to orient themselves in a fall allowing them to avoid many injuries. This ability is attributed to two significant feline characteristics: righting reflex and unique skeletal structure. The righting reflex is the cat s ability to first, know up from down, and then the innate nature to rotate in mid air to orient the body so its feet face downward. Animal experts say that this instinct is observable in kittens as young as three to four weeks, and is fully developed by the age of seven weeks. A cat s righting reflex is augmented by an unusually flexible backbone and the absence of a collarbone in the skeleton. Combined, these factors allow for amazing flexibility and upper body rotation. By turning the head and forefeet, the rest of the body naturally follows and cat is able reorient itself. Like many small animals, cats are said to have a non-fatal terminal falling velocity. That is, because of their very low body volume-to-weight ratio these animals are able to slow their decent by spreading out (flying squirrel style). Animals with these characteristics are fluffy and have a high drag coefficient giving them a greater chance of surviving these falls.

32 Book Examples ] Pg ] #3, 10, 15, 8, 34, 37, 47, 51, 60, 65, 70

Chapter 8. Rotational Motion

Chapter 8. Rotational Motion Chapter 8 Rotational Motion Rotational Work and Energy W = Fs = s = rθ Frθ Consider the work done in rotating a wheel with a tangential force, F, by an angle θ. τ = Fr W =τθ Rotational Work and Energy