Rotational Kinetic Energy: Work and Energy Revisited Quiz

PDF page 452: In this module, we will learn about work and energy associated with rotational motion. Figure 10.13 shows a worker using an electric grindstone propelled by a motor. Sparks are flying, and noise and vibration are created as layers of steel are pared from the pole. The stone continues to turn even after the motor is turned off, but it is eventually brought to a stop by friction. Clearly, the motor had to work to get the stone spinning. This work went into heat, light, sound, vibration, and considerable rotational kinetic energy.

PDF page 452: sound, and vibration. (credit: U.S. Navy photo by Mass Communication Specialist Seaman Zachary David Bell) Work must be done to rotate objects such as grindstones or merry-go-rounds. Work was defined in Uniform Circular Motion and Gravitation for translational motion, and we can build on that knowledge when considering work done in rotational motion. The simplest rotational situation is one in which the net force is exerted perpendicular to the radius of a disk (as shown in Figure 10.14) and remains perpendicular as the disk starts to rotate. The force is parallel to the displacement, and so the net work done is the product of the force times the arc length traveled: 10.53 To get torque and other rotational quantities into the equation, we multiply and divide the right-hand side of the equation by , and gather terms:

PDF page 453: We recognize that and , so that 10.55 This equation is the expression for rotational work. It is very similar to the familiar definition of translational work as force multiplied by distance. Here, torque is analogous to force, and angle is analogous to distance. The equation is valid in general, even though it was derived for a special case. To get an expression for rotational kinetic energy, we must again perform some algebraic manipulations. The first step is to note that , so that 10.56

PDF page 453: The expression for rotational kinetic energy is exactly analogous to translational kinetic energy, with being

PDF page 453: analogous to and to . Rotational kinetic energy has important effects. Flywheels, for example, can be used to

PDF page 453: analogous to and to . Rotational kinetic energy has important effects. Flywheels, for example, can be used to

PDF page 453: store large amounts of rotational kinetic energy in a vehicle, as seen in Figure 10.15.

PDF page 454: flywheel. When the bus goes down a hill, its transmission converts its gravitational potential energy into . It can also convert translational kinetic energy, when the bus stops, into . The flywheel’s energy can then be used to accelerate, to go up another hill, or to keep the bus from slowing down due to friction.

PDF page 454: Consider a person who spins a large grindstone by placing her hand on its edge and exerting a force through part of a revolution as shown in Figure 10.16. In this example, we verify that the work done by the torque she exerts equals the change in rotational energy. (a) How much work is done if she exerts a force of 200 N through a rotation of ? The force is kept perpendicular to the grindstone’s 0.320-m radius at the point of application, and the effects of friction are negligible. (b) What is the final angular velocity if the grindstone has a mass of 85.0 kg? (c) What is the final rotational kinetic energy? (It should equal the work.) Strategy To find the work, we can use the equation . We have enough information to calculate the torque and are given the rotation angle. In the second part, we can find the final angular velocity using one of the kinematic relationships. In the last part, we can calculate the rotational kinetic energy from its expression in .

PDF page 454: Consider a person who spins a large grindstone by placing her hand on its edge and exerting a force through part of a revolution as shown in Figure 10.16. In this example, we verify that the work done by the torque she exerts equals the change in rotational energy. (a) How much work is done if she exerts a force of 200 N through a rotation of ? The force is kept perpendicular to the grindstone’s 0.320-m radius at the point of application, and the effects of friction are negligible. (b) What is the final angular velocity if the grindstone has a mass of 85.0 kg? (c) What is the final rotational kinetic energy? (It should equal the work.) Strategy To find the work, we can use the equation . We have enough information to calculate the torque and are given the rotation angle. In the second part, we can find the final angular velocity using one of the kinematic relationships. In the last part, we can calculate the rotational kinetic energy from its expression in .

PDF page 454: Consider a person who spins a large grindstone by placing her hand on its edge and exerting a force through part of a revolution as shown in Figure 10.16. In this example, we verify that the work done by the torque she exerts equals the change in rotational energy. (a) How much work is done if she exerts a force of 200 N through a rotation of ? The force is kept perpendicular to the grindstone’s 0.320-m radius at the point of application, and the effects of friction are negligible. (b) What is the final angular velocity if the grindstone has a mass of 85.0 kg? (c) What is the final rotational kinetic energy? (It should equal the work.) Strategy To find the work, we can use the equation . We have enough information to calculate the torque and are given the rotation angle. In the second part, we can find the final angular velocity using one of the kinematic relationships. In the last part, we can calculate the rotational kinetic energy from its expression in .

PDF page 454: Solution for (a) The net work is expressed in the equation 10.61 where net is the applied force multiplied by the radius because there is no retarding friction, and the force is perpendicular to . The angle is given. Substituting the given values in the equation above yields

PDF page 455: Discussion The final rotational kinetic energy equals the work done by the torque, which confirms that the work done went into rotational kinetic energy. We could, in fact, have used an expression for energy instead of a kinematic relation to solve part (b). We will do this in later examples.

PDF page 456: A typical small rescue helicopter, similar to the one in Figure 10.17, has four blades, each is 4.00 m long and has a mass of 50.0 kg. The blades can be approximated as thin rods that rotate about one end of an axis perpendicular to their length. The helicopter has a total loaded mass of 1000 kg. (a) Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm. (b) Calculate the translational kinetic energy of the helicopter when it flies at 20.0 m/s, and compare it with the rotational energy in the blades. (c) To what height could the helicopter be raised if all of the rotational kinetic energy could be used to lift it? Strategy Rotational and translational kinetic energies can be calculated from their definitions. The last part of the problem relates to the idea that energy can change form, in this case from rotational kinetic energy to gravitational potential energy. Solution for (a) The rotational kinetic energy is