OpenStax-CNX module: m14307 1 Work - kinetic energy theorem for rotational motion * Sunil Kumar Singh This work is produced by OpenStax-CNX and licensed under the Creative Commons Attribution License 2.0 Abstract of a particle or rigid body about a xed axis has unique motional attributes like angular position, displacement, velocity and acceleration. There is one to one correspondence between quantities in pure translation and pure rotation. We can actually write rotational relations for work and kinetic energy by just inspecting corresponding relation of pure translational motion. Beside similarity of relations, there are other similarity governing description of pure motions in translation and rotation. We must, rst of all, realize that both pure motions are one-dimensional motions in description. It might sound a bit bizarre for rotational motion but if we look closely at the properties of vector attributes that describe pure rotational motion, then we would realize that these quantities are indeed one - dimensional. al quantities like angular displacement (θ), speed (ω) and accelerations (α) are one - dimensional quantities for rotation about a xed axis. These vector attributes are either anticlockwise (positive) or clockwise (negative). If it is anticlockwise, then the corresponding rotational vector quantity is in the direction of axis; otherwise in the opposite direction. * Version 1.5: Nov 16, 2009 8:49 am -0600 http://creativecommons.org/licenses/by/2.0/
OpenStax-CNX module: m14307 2 Figure 1: al quantities like angular displacement (θ), speed (ω) and accelerations (α) are one - dimensional quantities for rotation about xed axis. We observe yet another similarity between two pure motions. We have seen that the pure translation of a particle or a rigid body has unique (single value) motional attributes like position, displacement, velocity and acceleration. In the case of rotation, on the other hand, even if individual particles constituting the rigid body have dierent linear velocities or accelerations, but corresponding rotational attributes of the rotating body, like angular velocity and acceleration, have unique values for all particles.
OpenStax-CNX module: m14307 3 Figure 2: Individual particles constituting the rigid body have dierent linear velocities/ acceleration, but they have unique angular velocity/ acceleration. There are, as a matter of fact, more such similarities. We shall discuss them in the appropriate context as we further develop dynamics of pure rotational motion. 1 Expressions of work and energy The expressions of work in pure translation in x- direction (one dimensional linear motion) is given by : W = F x x W = F x x The corresponding expressions of work in pure rotation (one dimensional rotational motion) is given by : W = τ θ W = τ θ (2) Obviously, torque replaces component of force in the direction of displacement (Fx) and angular displacement (θ) replaces linear displacement (x). Similarly, the expression of kinetic energy as derived earlier in the course is : K = 1 2 Iω2 (2)
OpenStax-CNX module: m14307 4 Here, moment of inertia (I) replaces linear inertia (m) and angular speed ( ω) replaces linear speed (v). We, however, do not generally dene gravitational potential energy for rotation as there is no overall change in the position (elevation) of the COM of the body in pure rotation. This is particularly the case when axis of rotation passes through COM. This understanding of correspondence of expressions helps us to remember expressions of rotational motion, but it does not provide the insight into the angular quantities used to describe pure rotation. For example, we can not understand how torque accomplishes work on a rigid body, which does not have any linear displacement! For this reason, we shall develop expressions of work by considering rotation of a particle or particle like object in the next section. We need to emphasize here another important underlying concept. The description of motion in pure translation for a single particle and a rigid body dier in one very important aspect. Recall that the motion of a rigid body, in translation, is described by assigning motional quantities to the "center of mass(com)". If we look at the expression of center of mass in x-direction, then we realize that the concept of COM is actually designed to incorporate distribution of mass in the body. x COM = x m M (2) We must note here that such distinction arising due to distribution of mass between a particle and rigid body does not exist for pure rotation. It is so because the eect of the distribution of mass is incorporated in the denition of moment of inertia itself : I = r 2 m (2) For this reason, there is no corresponding "center of rotational inertia" or "center of moment of inertia" for rigid body in rotation. This is one aspect in which there is no correspondence between two motion types. It follows, then, that the expressions of various quantities of a particle or a rigid body in rotation about a xed axis should be same. Same expressions would, therefore, determine work and kinetic energy of a particle or a rigid body. The dierence in two cases will solely arise from the dierences in the values of moments of inertia. 2 Work done in rotation We consider a particle like object, attached to a "mass-less" rod, to simulate rotation of a particle. The particle like object rotates in a plane perpendicular to the plane of screen or paper as shown in the gure. Here, we shall evaluate work by a force, "F", in xy-plane, which is perpendicular to the axis of rotation. We know that it is only the tangential component of force that accelerates the particle and does the work. The component of force perpendicular to displacement does not perform work. For small linear displacement "ds", we have :
OpenStax-CNX module: m14307 5 Figure 3: A particle like object in rotation. W = F x x = F t s By geometry, = r θ Combining two equations, we have : W = F t r θ = τ θ Work done by a torque in rotating the particle like body by an angular displacement, we have : W = τ θ If work is done by a constant torque, then we can take torque out of the integral sign and : W = τ θ = τ θ As discussed earlier, this same expressions is valid for work done on a rigid body in pure rotation.
OpenStax-CNX module: m14307 6 2.1 Power in rotation Power measures the rate at which external torque does work on a particle like object or a rigid body. It is equal to the rst time dierential of work done, P = W t = τ θ t = τω (3) We must note the correspondence of this expression with its linear counterpart, where : P = F v Clearly, torque (τ) replaces force (F) and angular speed (ω) replaces linear speed (v) in the expression of power. Example 1 Problem : A uniform disk of mass 5 kg and radius 0.2 m is mounted on a horizontal axle to rotate freely. One end of a rope, attached to the disk, is wrapped around the rim of the disk. If a force of 10 N is applied vertically as shown, then nd the work done by the force in rotating the disk for 2 s, starting from rest at t = 0 s. Figure 4: A uniform disk is mounted on a horizontal axle. Solution : Here, torque due to vertical force is constant. Thus, in order to determine the work on the disk in rotation, we need to use the expression of work done by a constant torque as :
OpenStax-CNX module: m14307 7 W = τ θ = τ ( θ 2 θ 1 ) It means that we need to calculate torque and angular displacement. Here, torque (positive as rotation is anti-clockwise) is given by : τ = F t r = 10 x 0.2 = 2 N m This is a rotational motion with constant acceleration (due to constant torque). Thus, we can use the equation of motion for constant angular acceleration to nd the displacement : θ f = θ i t + 1 2 αt2 Here, disk is at rest in the beginning, θi = 0. Hence, θ f = 1 2 αt2 But, we do not know the angular acceleration. We can, however, use corresponding Newton's second law for rotation. Since, disk rotates freely, we can assume that there is no other torque involved : α = τ I On the other hand, the moment of inertia for the disk about an axis passing through the center of mass and perpendicular to its surface is given by : I = MR2 2 = 5 x 0.22 2 = 0.1 kg m 2 Putting this value of moment of inertia, we get the value of angular acceleration : Now nal angular position is : α = τ I = 2 0.1 = 20 rad s 2 θ f = 1 2 x 20 x 22 = 40 rad Hence, work done is : W = τ ( θ 2 θ 1 ) = τθ 2 = 2 x 40 = 80 J 3 Work - kinetic energy theorem for rotational motion We have seen earlier that net work on a body is equal to change in its kinetic energy. In translation, we have : W = K = 1 2 mv f 2 1 2 mv i 2 A corresponding work - kinetic energy theorem for pure rotation, therefore, is : W = K = 1 2 Iω f 2 1 2 Iω i 2 (4) where "W" represents the net work on the rigid body in rotation.
OpenStax-CNX module: m14307 8 Example 2 Problem : A uniform disk of mass 5 kg and radius 0.2 m is mounted on a horizontal axle to rotate freely. One end of a rope, attached to the disk, is wrapped around the rim of the disk. If a force of 10 N is applied vertically as shown, then nd the work done by the force in rotating the disk for 2 s, starting from rest at t = 0 s. Use work - kinetic energy theorem to nd the result. Figure 5: A uniform disk is mounted on a horizontal axle. Solution : This is the same question as the previous one except that we are required to solve the question, using work-kinetic energy theorem : W = K = 1 2 Iω f 2 1 2 Iω i 2 From the previous example, we have : and I = MR2 2 = 5 x 0.22 2 = 0.1 kg m 2 α = τ I = 20 rad s 2 Now, as the disk starts from rest, ω i = 0. In order to nd the nal angular speed, we use equation of motion for constant angular acceleration : ω f = ω i + αt = 0 = 2 x 20 = 40 rad / s
OpenStax-CNX module: m14307 9 Thus, W = K = 1 2 Iω f 2 0 = 1 2 Iω f 2 W = 1 2 x 0.1 x 402 = 80 J As expected, the result is same as calculated in earlier example. 4 Summary 1: al quantities describing the motion are linear (one-dimensional) vectors. 2: Work done by a torque is given by : W = τ θ 3: Work done by a constant torque is given by : W = τ θ 4: We generally do not consider change in gravitational potential energy in the case of rotation about an axis that passes through its center of mass. Since there is no change in the position of center of mass, there is no corresponding change in its potential energy.