Chapter 10 Getting Dizzy with Rotational Dynamics In This Chapter Calculating moments of Inertia Doing rotational work Rolling with rotational work and ramps Handling angular momentum Working with rotational dynamics Inear physics translates to rotational physics in several ways - in terms of angular speed and acceleration and in terms of angular momentum and rotational kinetic energy. What's the rotational analog of force? Torque! How about F = ma? What's the analog of that? It turns out to be r = la, where / is angular Iner- tia, which is often called the moment of inertia. Just about everything In linear motion has an angular analog. This chapter covers the moment of Inertia, torque as it relates to angular acceleration, and rotational kinetic energy. Putting Newton on Wheels Figure 10-1 shows a ball at the end of stick, moving In a circle. You should remember that the force needed to accelerate the ball around a circle is given by F = ma, so the torque, which equals rf, is rf = = mar Figure 18-1: A ball moving in Circle. D John Wiley & Sons, Inc.till: Being Energetic: Work . Look at this situation in terms of a, angular acceleration. Angular acceleration is one of those Items you can multiply by the radius to get the linear equivalent, which in this case Is equal to the tangential acceleration: a = ra Substituting a = ra In the torque equation gives you: re mra This is an important result because it relates torque and angular acceleration. The quantity mr is called the moment of Inertia, I, and It represents the effort you need to get something to change its angular velocity, so this equation is usually written as: re la This equation is a general result, but the moment of inertia, /, differs depending on the situation. For example, / is different when you're spinning a solid cylinder versus a solid sphere. For a single small mass such as the ball on the end of a stick, I = mr. The units of moment of Inertia are kilogram-meters? (kg-m?) in the MKS system. Q. You're whipping a cannonball around 2. Plug in the numbers: on the end of a 0.50-m steel rod. If the cannonball has a mass of 10.0 kg, what t= mra torque do you need to apply to get an = (10.0 kg) (0.50 m)?(0.50 radians/s?) angular acceleration of 0.50 radians/s?7 = 1.25 N-m A. The correct answer is 1.25 N-m. The cannonball has so much more mass than the rod that you can 1. Use the equation ? = la = mr'a. ignore the rod's mass entirely. 1. You're rotating a 10.0-kg cannonball at the 2. end of a 1.0-m rod in a circle and want an The cannonball (10 kg) on the end of angular acceleration of 1.0 radian/s?. What a 1.0-m rod has an angular acceleration torque do you need to supply? of 2.0 radians/s'. What torque are you applying? Solve itsChapter 10: Getting Dizzy with Rotational Dynamics 3. You're shoveling snow, holding the shovel 4. A 100.0-g clock pendulum on the end of a 1.0-m rod has an angular acceleration handle In your right hand. Assuming that of 2.0 radians/s'. What torque is being you can use the equation / = mr to deter- mine the moment of Inertia of the shovel, If applied? the 2.0-kg shovel has an angular accelera- tion of 10.0 radians/s' and a length of 1.5 m, SADE TO what torque are you applying? Solve It Moments of Inertia for Everyone You know that the moment of Inertia of a small mass on the end of a thin rod is mr. What are the moments of inertia for other configurations, such as a solid sphere? By treating each mass as a collection of small masses, the moments of inertia for several other shapes have been figured out; here are some of them. In the following, m is the total mass of the object, and r is always the radius (of the disk, cylinder, sphere, or hoop). L Is the length of the rod, or the length of the rectangle in the direction perpendicular to the axis of rotation. Hoop rotating around its center (like a tire): I = mr Solid disk rotating around its center: / = (1/2)mr Hollow cylinder rotating around its center (such as a tire): / = my Solid cylinder: I - (1/2)mr Hollow sphere: I - (2/3) mr Solid sphere: / - (2/5)mr Point mass rotating at radius r. I = my Rod rotating around an axis perpendicular to it and through its center: / = (1/12)m Rod rotating around an axis perpendicular to it and through one end: / = (1/3)ml Rectangle rotating around an axis along one edge: / = (1/3)ml] Rectangle rotating around an axis parallel to one edge and passing through the center: / = (1/12)ml2art Ill: Being Energetic: Work Q. A solid cylinder with a mass of 5.0 kg is A. The correct answer is 0.075 N-m. rolling down a ramp. If it has a radius of 1. Use the equation r = la. 10 cm and an angular acceleration of 3.0 radians/s'. what torque is operating 2. In this case, J = (1/2)mr. on It? 3. Plug In the numbers: 1=mr'a =(5.0 kg)(0.10 m)' (3.0 radians/s') = 0.075 N-m 5. You're spinning a 5.0-kg ball with a radius of 0.5 m. If It's accelerating at 4.0 radians/s?, what torque are you applying? 6. A tire with a radius of 0.50 m and mass of 1.0 kg is rolling down a street. If It's accelerating with an angular acceleration of 10.0 radians/s', what torque is operating on it? ServeChapter 10: Getting Dizzy with Rotational Dynamics 1 7. You're spinning a hollow sphere with a mass 8. You're throwing a 300.0-g flying disc with of 10.0 kg and radius of 1.0 m. If it has an a radius of 10 cm, accelerating it with an angular acceleration of 15 radians/s', what angular acceleration of 20.0 radians/s?. torque are you applying? What torque are you applying? Solueste 5/12 9. If you're spinning a 2.0-kg solid ball with a 10. If you're spinning a 2.0-kg hollow ball with radius of 0.5 m, starting from rest and a radius of 0.50 m, starting from rest and applying a 6.0 N-m torque, what is its applying a 12.0 N-m torque, what is Its angular speed after 60.0 seconds? angular speed after 10.0 seconds? Solve it30 Part III: Being Energetic: Work Doing Some Rotational Work If you apply a force of 500 N to the edge of a tire to get a car moving, what work do you do over 1.0 m of travel? That work looks like this equation, where s is the distance the- force was applied over: W = Fs You can also think about this force rotationally. In the case of you applying force to the edge of a tire to get a car moving, the distance s equals the radius multiplied by the angle through which the wheel turns, s = ro, so you get this equation: W - Fs - Fre But the torque, r, equals Fr in this case. So you're left with this: W = Fs - Fre = +0 Talk about a cool result - work equals torque multiplied by the angle through which that torque is applied. If you apply a torque of 500.0 N-m to a A. The correct answer is 3,140 J. ire and turn it through an angle of 2n radians, what work have you done? 1. Use the equation W= ro. 2. Plug in the numbers: W = 10 = (500.0 N-m)(2n radians) = 3,140 J 17. How much work do you do if you apply a torque of 6.0 N-m over an angle of 12. You've done 20.0 J of work turning a 200 radians? steering wheel. If you're applying 10.0 N-m of torque, what angle have you turned the steering wheel through?Chapter 10: Getting Dizzy with Rotational Dynamics 187 13. How much work do you do If you apply a 14. You've done 350 J of work turning a bicycle torque of 75 N-m through an angle of tire. If you're applying 150 N-m of torque, what angle have you turned the wheel on radians? through? Solve 14 Round and Round: Rotational Kinetic Energy In Chapter 7, you review the equation for linear kinetic energy. KE = - mu' Convert that equation to Its angular analog: KE = (mr')(;) =10' In other words, the first equation becomes the second when you're going rotational. The " angular velocity . takes the place of the linear velocity o, and the momentum of Inertia / takes the place of the mass m. 2. You have a 100-kg solid sphere with a 2. For a solid sphere, I = (2/5) mr. radius equal to 2.0 m. If It's rotating at 3. Plug in the numbers: = 10.0 radians/s, what Is its rotational kinetic energy? KE = 10 A. The correct answer is 800 J. 1. Use this equation: KE = 10' -(2)(3)(100 kg )(2.0 m)'(10.0 radians/5)' = 800 J182 Part III: Being Energetic: Work 15. How much rotational kinetic energy does 16. How much rotational kinetic energy does a spinning tire of mass 10.0 kg and a spinning tire of mass 12 kg and radius 0.50 m have if it's spinning at radius 0:80 m have if It's spinning at 40.0 rotations/s? 200.0 radians/s? Side Jo 17. How much work do you do to spin a tire, 18. How much work do you do to spin a hollow which has a mass of 5.0 kg and a radius of 0.40 m, from 0.0 radians/s to sphere, which has a mass of 10.0 kg and a 100.0 radians/s? radius of 0.50 m, from 0.0 radians/s to 200.0 radians/s?Chapter 10: Getting Dizzy with Rotational Dynamics 7 Working with Ramps Again Figure 10-2 shows a hollow cylinder and a solid cylinder at the top of a ramp. What happens when they roll down the ramp? Which cylinder ends up with the greater speed? Figure 10-2: A hollow cylinder and a solid cylinder on a ramp. You may think that you can simply set the original potential energy equal to the final kinetic energy and solve for the final speed that way. KE = PE = mu' = mgah Unfortunately, you can't work out the answer that way because of what you know about rotational kinetic energy. Some of the potential energy of each cylinder goes into rotational kinetic energy as the cylinders roll to the bottom of the Incline. Here's the correct equation to use: mgah = - mu' + la' Further, you can relate o and or with the equation u = ra, which means that a - v/r, so: mgan =mo" + o' -}mo"+!'("!) Given that equation, this is the final equation for v: um 2mgh So how do you evaluate this for the hollow cylinder and the solid cylinder? For a hollow gives you this: cylinder, / = mr: for a solid cylinder, / = (1/2)mr. Substituting for / for the hollow cylinder o-gh But substituting for / for the solid cylinder gives you this: The solid cylinder will be going , reach the bottom of the Incline. = 1.15 times as fast as the hollow cylinder when they84 Part III: Being Energetic: Work JAPLE If a solid sphere is at the top of a 3.0-m 3. That breaks down to: high ramp, what is Its speed when it 2gh reaches the bottom of the ramp?. D=. 1+2/5 A. The correct answer Is 6.5 m/s. 4. Plug in the numbers: 1. Use this equation: 2mgh 2(9.8 m/s' ) (3.0 m) 1+2/5 = 6.5 m/s Vm+1/r 2. For a solid sphere, I = (2/5)mr. That means o is equal to this: and Pin U= 2mgh Vm+2/5m 79. If a hollow cylinder is at the top of a 4.0-m high ramp, what is Its speed when It 20. If a solid cylinder is at the top of a 2.0-m reaches the bottom of the ramp? high ramp, what Is its speed when it reaches the bottom of the ramp?Chapter 10: Getting Dizzy with Rotational Dynamics 185 27. A tire is rolling down a ramp, starting at a 22. A basketball (that Is, a hollow sphere) is height of 3.5 m. What Is its speed when it rolling down a ramp, starting at a height of reaches the bottom of the ramp? 4.8 m. What is its speed when it reaches the bottom of the ramp? Can't Stop This: Angular Momentum In linear motion, momentum looks like this: P - mv Momentum Is conserved in collisions. In addition to linear momentum, you can have angu- lar momentum, which is represented by the symbol L. Following is the equation for angular momentum: Note that I, like ma, is actually a vector. In the simple cases, consider that L always points in the same direction as o. In physics, angular momentum is conserved. For example, If you have a skater spinning around and then she spreads her arms (giving her a different moment of Inertia), because angular momentum is conserved, you get this: With this equation, If you know the skater's original and final moments of Inertia and her original angular speed, you can calculate her final angular speed like this:186 Part III: Being Energetic: Work 4. Solve for my If a 500.0-kg merry go-round with a radius of 2.0 m is spinning at 2.0 radians/s and a boy with a mass of 40.0 kg jumps on the outer rim, what is the new angular speed mr' +my* of the merry-go-round? A. The correct answer Is 1.7 radians/s. where (1/2)mr = 1,000 kg-m* and mr = 160 kg-m'. 1. Use this equation: 5. Plug In the numbers: 2. For a solid disc like the merry go-round, 2,000 kg-m' 7 =1.7 radians/s 1 - (1/2)mr. 1,000 kg-m' +160 kg-m 3. When the boy jumps on, he adds mark to /, where m, is the mass of the boy. This means that: ( * mr ' ) , =(# mr' + mart )as 2,000 kg-m"-radians/s = ( mr* + mar" )orz 23. A merry-go-round with a mass of 500.0 kg and radius of 2.0 m Is rotating at 24. A 2,000.0-kg space station, which is a 3.0 radians/s when two children with a hollow cylinder with a radius of 2.0 m, combined mass of 70.0 kg jump on the is rotating at 1.0 radian/s when an outer rim. What is the new angular astronaut with a mass of 80.0 kg lands speed of the merry-go-round? on the outside of the station. What is the station's new angular speed