Chapter 8 Lecture Pearson Physics Rotational Motion and Equilibrium Spring, 2016 © 2014 Pearson Education, Inc.

Chapter Contents Describing Angular Motion Rolling Motion and the Moment of Inertia Torque Static Equilibrium © 2014 Pearson Education, Inc.

Describing Angular Motion The bicycle wheel shown in the figure below is free to rotate about its axle. The axle is the axis of rotation for the wheel. As the wheel rotates, every point on the wheel moves in a circular path centered about the axis of rotation. © 2014 Pearson Education, Inc.

Describing Angular Motion © 2014 Pearson Education, Inc.

Describing Angular Motion While common units for measuring angles are the degree (  ) and revolution (rev), the most convenient unit for angle measurements in scientific calculations is the radian (rad). As the figure below indicates, the radian is the angle for which the length of a circular arc is equal to the radius of the circle. © 2014 Pearson Education, Inc.

Describing Angular Motion A comparison between angles measured in degrees and in radians is shown in the figure below. Angles are indicated around the circumference of the circle in both degrees and radians. © 2014 Pearson Education, Inc.

Describing Angular Motion As is illustrated in the following figure, the radian is useful because it provides a simple relationship between angles and arc lengths. © 2014 Pearson Education, Inc.

Describing Angular Motion © 2014 Pearson Education, Inc.

Describing Angular Motion As the figure below indicates, as the bicycle wheel rotates, the angular position of the spot of red paint changes. The angular displacement of the spot, Δθ, is defined as the difference between its final angle and its initial angle: Δθ = θ f − θ i © 2014 Pearson Education, Inc.

Describing Angular Motion © 2014 Pearson Education, Inc.

Describing Angular Motion © 2014 Pearson Education, Inc.

Describing Angular Motion © 2014 Pearson Education, Inc.

Describing Angular Motion The object's tangential speed is equal to the circumference of the circular path, d = 2πr, divided by the time required to complete one circuit, t = T. This gives the following: v t = d/t = 2πr/T = r(2π/T) Since 2π/T is the angular velocity ω, the tangential speed of a rotating object may be expressed as follows: © 2014 Pearson Education, Inc.

Describing Angular Motion If the angular velocity of a rotating object increases or decreases with time, then we say that the object has an angular acceleration. The average angular acceleration, α av, is the change in angular velocity, Δω, in a given time interval, Δt. © 2014 Pearson Education, Inc.

Describing Angular Motion The sign of the angular acceleration is determined by whether the change in angular velocity is positive or negative. –If ω and α have the same sign, then the speed of rotation is increasing. –If ω and α have opposite signs, then the speed of rotation is decreasing. These relationships are summarized in the following figure. © 2014 Pearson Education, Inc.

Describing Angular Motion The following example illustrates an application of the definition of average angular acceleration. Just as linear speed is equal to the radius times the angular speed, the linear acceleration is equal to the radius times the angular acceleration: tangential acceleration = radius x angular acceleration α t = rα © 2014 Pearson Education, Inc.

Rolling Motion and the Moment of Inertia As a bicycle wheel rolls freely, with no slipping between the tire and the ground, the wheel rotates about its axle, and at the same time the axle moves in a straight line. Thus, the rolling motion of a wheel is a combination of both rotational motion and linear motion. © 2014 Pearson Education, Inc.

Rolling Motion and the Moment of Inertia Because the forward motion of the axle exactly cancels out the backward motion of the bottom of the wheel, the instantaneous speed of the bottom of the wheel is zero. The top of the wheel has twice the speed of the axle. Thus, if a car's speed is v, as read on the speedometer, then the tops of its wheels have the speed 2v. © 2014 Pearson Education, Inc.

Rolling Motion and the Moment of Inertia Due to its inertia, an object at rest or moving in a straight line resists changes to that motion. A similar relationship occurs with rotational motion. Each object has a moment of inertia, I, which determines how easy or hard it is to change its rotation. An object with a large moment of inertia is difficult to start or stop rotating. For example, a merry-go-round has a large moment of inertia, whereas a baseball has a relatively small moment of inertia. © 2014 Pearson Education, Inc.

Rolling Motion and the Moment of Inertia Experiments show that an object's moment of inertia depends linearly on the mass of the object and on the square of the distance to the object from the axis of rotation. For a mass m at a distance r from the axis of rotation, the moment of inertia is I = mr 2. © 2014 Pearson Education, Inc.

Rolling Motion and the Moment of Inertia In a system with several particles with different masses and at different distances from the axis of rotation, the moment of inertia is the sum of the mr 2 terms for each object. © 2014 Pearson Education, Inc.

Rolling Motion and the Moment of Inertia The preceding results are valid for individual particles or groups of particles. What about a solid object, like a disk or sphere? The table below contains the moment of inertia for several uniform, rigid objects of various shapes and total mass m. © 2014 Pearson Education, Inc.

Rolling Motion and the Moment of Inertia For linear motion, the kinetic energy of a mass m moving with a speed v is. To find the kinetic energy of a rotating object, we replace the linear speed v with the angular speed ω and the mass m with the moment of inertia I, which we can think of as rotational mass. Thus the rotational kinetic energy of an object is one-half the product of the moment of inertia and the square of the angular speed. © 2014 Pearson Education, Inc.

Rolling Motion and the Moment of Inertia Thus the rotational kinetic energy of an object is one-half the product of the moment of inertia and the square of the angular speed. © 2014 Pearson Education, Inc.

Rolling Motion and the Moment of Inertia Connections between linear and rotational quantities such as these are summarized in the table below. We can apply these connections to momentum. By replacing mass with the moment of inertia and linear speed with angular speed, we obtain the angular momentum, L. © 2014 Pearson Education, Inc.

Torque A force can cause a rotation. The effectiveness of the force depends on where the force is applied and in which direction. In the figure below, we see that the force required far from the nut is much less than the force required near the nut. Less force is needed to open a revolving door if the force is applied far from the door's axis of rotation. © 2014 Pearson Education, Inc.

Torque Thus the effectiveness of a force in causing a rotation depends on both the magnitude of the force and the distance from the axis of rotation to the force. Taking this information into account, we define a physical quantity called torque. Torque,, is the product of force and distance. © 2014 Pearson Education, Inc.

Torque © 2014 Pearson Education, Inc.

Torque The figure on the left below shows that a force acting in a direction that is directly toward or away from the axis of rotation will cause no rotation. That is, radial forces produce zero torque. The figure on the right below shows that if the force is at an angle θ relative to the radial line, then only the tangential component of the force will produce a torque. © 2014 Pearson Education, Inc.

Torque Therefore, the torque in the general case is distance times the tangential component of the force, rF cosθ: © 2014 Pearson Education, Inc.

Torque The figure below shows how torque can be defined in terms of a quantity known as a moment arm. The perpendicular distance from the axis of rotation to the line of the force is defined as the moment arm. From the figure, the length of the moment arm is r  = r sinθ. Referring to the general equation for torque, © 2014 Pearson Education, Inc.

Torque The sign associated with a torque is determined by the type of angular rotation it would produce. –A torque that causes a counterclockwise rotation is defined to be positive. –A torque that causes a clockwise rotation is defined to be negative. © 2014 Pearson Education, Inc.

Torque The figure below illustrates a system with more than one torque. The net torque acting on the system is the sum of the individual torques, taking into account the appropriate sign for each. © 2014 Pearson Education, Inc.

Torque We can once again draw on the connections between linear and rotational quantities when considering the effect of torque on an object. Replacing the force with torque,, and the linear acceleration with angular acceleration, α, we obtain the rotational version of Newton's second law: © 2014 Pearson Education, Inc.

Torque Physical quantities that are conserved, such as energy and linear momentum, play an important role in physics. Angular momentum is another quantity that is conserved. Angular momentum is conserved when the net torque is zero. The motion of an ice skater may be used to illustrate how angular momentum is conserved. In the following figure, the skater pulls in her arms, reducing her momentum of inertia. However, since her angular momentum must stay the same, her angular speed must increase. © 2014 Pearson Education, Inc.

Torque In general, a large I and a small ω give the same angular momentum as a small I and a large ω: Iω = Iω © 2014 Pearson Education, Inc.

Static Equilibrium When an object is at rest and remaining at rest, it is said to be in a state of static equilibrium. The conditions for static equilibrium are the following: –The total force acting on the object must be zero. This ensures that there is no linear acceleration. –The total torque acting on the object must also be zero. This ensures that there is no angular acceleration. © 2014 Pearson Education, Inc.

Static Equilibrium An application of the these two conditions to the plank that supports the child in the figure below may be used to calculate the forces F 1 and F 2 acting on the plank. The total force acting on the plank must be zero; therefore, F 1 + F 2 – mg = 0 © 2014 Pearson Education, Inc.

Static Equilibrium The total torque acting on the plank may be found by choosing the left end of the plank to be the axis of rotation. With this choice: –Force F 1 exerts zero torque. –Force F 2 produces a torque of +F 2 L since it acts at the far end of the plank in the counterclockwise direction. –The weight of the child mg exerts a torque of –mg since it acts in the clockwise direction at a distance from the axis. © 2014 Pearson Education, Inc.

Static Equilibrium Setting the total torques equal to zero, we find 0 + F 2 L – mg(¾L) = 0 Therefore, and. In this example, the left end of the plank was arbitrarily selected as the axis of rotation. In general, you are free to chose an axis of rotation that is most convenient for a given situation. © 2014 Pearson Education, Inc.

Static Equilibrium The center of mass (CM) is the point at which an object can be balanced. In many ways, an object behaves as if all of its mass were concentrated at its center of mass. The location of the center of mass for two objects may be calculated using the following simple equation. © 2014 Pearson Education, Inc.

Static Equilibrium The figure below shows the center of mass of two objects. The center of mass is closer to the larger mass, or equidistant between the masses if they are equal. When an object is supported at its center of mass, it is in static equilibrium since the torques acting on the object are balanced. © 2014 Pearson Education, Inc.

Static Equilibrium When the center of mass of an object is directly below the suspension point, the torque due to gravity is zero. This is because the force of gravity acts through the axis of rotation. Therefore, if you allow an object with any shape to hang freely, its center of mass will be directly below its suspension point. © 2014 Pearson Education, Inc.

Static Equilibrium The center of mass of an irregularly shaped object, such as a piece of wood cut into the shape of the continental United States, may be located by suspending the model from two or more points. The center of mass lies at the intersection of vertical lines extending downward from the suspension points. © 2014 Pearson Education, Inc.