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**Rotation of a Rigid Object about a Fixed Axis**

Chapter 10 Rotation of a Rigid Object about a Fixed Axis

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Rigid Object Analysis models introduced so far cannot be used to analyze all motion. We can model the motion of an extended object by modeling it as a system of many particles. The analysis is simplified if the object is assumed to be a rigid object. Introduction

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**Rigid Object A rigid object is one that is non-deformable.**

The relative locations of all particles making up the object remain constant. All real objects are deformable to some extent, but the rigid object model is very useful in many situations where the deformation is negligible. In this chapter another class of analysis models based on the rigid-object model are developed. Introduction

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**Angular Position Axis of rotation is the center of the disc**

Choose a fixed reference line. Point P is at a fixed distance r from the origin. A small element of the disc can be modeled as a particle at P. Polar coordinates are convenient to use to represent the position of P (or any other point). P is located at (r, q) where r is the distance from the origin to P and q is the measured counterclockwise from the reference line. Section 10.1

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Angular Position, cont. As the particle moves, the only coordinate that changes is q. As the particle moves through q, it moves though an arc length s. The arc length and r are related: s = rq (10.1a) Section 10.1

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Radian This can also be expressed as: q is a pure number, but commonly is given the artificial unit, radian. One radian is the angle subtended by an arc length equal to the radius of the arc. Whenever using rotational equations, you must use angles expressed in radians. (10.1b) Section 10.1

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**Conversions Comparing degrees and radians**

Converting from degrees to radians Section 10.1

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**Angular Position, final**

We can associate the angle q with the entire rigid object as well as with an individual particle. Remember every particle on the object rotates through the same angle. The angular position of the rigid object is the angle q between the reference line on the object and the fixed reference line in space. The fixed reference line in space is often the x-axis. The angle q plays the same role in rotational motion that the position x does in translational motion. Section 10.1

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Angular Displacement The angular displacement is defined as the angle the object rotates through during some time interval. This is the angle that the reference line of length r sweeps out. Section 10.1

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Average Angular Speed The average angular speed, wavg, of a rotating rigid object is the ratio of the angular displacement to the time interval. (10.2) Section 10.1

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Angular Speed The instantaneous angular speed is defined as the limit of the average speed as the time interval approaches zero. This is analogous to translational speed. Units of angular speed are radians/sec. rad/s or s–1 since radians have no dimensions. Angular speed will be positive if is increasing (counterclockwise) Angular speed will be negative if is decreasing (clockwise) (10.3) Section 10.1

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Quick Quiz 10.1 A rigid object rotates in a counterclockwise sense around a fixed axis. Each of the following pairs of quantities represents an initial angular position and a final angular position of the rigid object. (i) Which of the sets can only occur if the rigid object rotates through more than 180º? (a) 3 rad, 6 rad (b) –1 rad, 1 rad (c) 1 rad, 5 rad (ii) Suppose the change in angular position for each of these pairs of values occurs in 1 s. Which choice represents the lowest average angular speed? Answer: (i) (c) (ii) (b)

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Angular Acceleration The average angular acceleration, a avg, of an object is defined as the ratio of the change in the angular speed to the time it takes for the object to undergo the change. The instantaneous angular acceleration is defined as the limit of the average angular acceleration as the time goes to 0. (10.4) (10.5) Section 10.1

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**Angular Acceleration, cont.**

Analogous to translational velocity Units of angular acceleration are rad/s² or s–2 since radians have no dimensions. Angular acceleration will be positive if an object rotating counterclockwise is speeding up. Angular acceleration will also be positive if an object rotating clockwise is slowing down. Section 10.1

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**Angular Motion, General Notes**

When a rigid object rotates about a fixed axis in a given time interval, every portion on the object rotates through the same angle in a given time interval and has the same angular speed and the same angular acceleration. So q, w, a all characterize the motion of the entire rigid object as well as the individual particles in the object. Section 10.1

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Directions, details Strictly speaking, the speed and acceleration (w, a) are the magnitudes of the velocity and acceleration vectors. The directions are actually given by the right-hand rule. Section 10.1

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**Hints for Problem-Solving**

Similar to the techniques used in linear motion problems. With constant angular acceleration, the techniques are much like those with constant linear acceleration. There are some differences to keep in mind. For rotational motion, define a rotational axis. The choice is arbitrary. Once you make the choice, it must be maintained. In some problems, the physical situation may suggest a natural axis. The object keeps returning to its original orientation, so you can find the number of revolutions made by the body. Section 10.2

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**Rotational Kinematics**

Under constant angular acceleration, we can describe the motion of the rigid object using a set of kinematic equations. These are similar to the kinematic equations for linear motion. The rotational equations have the same mathematical form as the linear equations. The new model is a rigid object under constant angular acceleration. Analogous to the particle under constant acceleration model. Section 10.2

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**Rotational Kinematic Equations**

The kinematic expression for the rigid object under constant angular acceleration are of the same mathematical form as those for a particle under constant acceleration. Substitutions from translational to rotational are x → q v → w a → a Section 10.2

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**Rotational Kinematic Equations**

(10.6) (10.7) (10.8) (10.9) Section 10.2

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**Comparison Between Rotational and Linear Equations**

Section 10.2

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Quick Quiz 10.2 Consider again the pairs of angular positions for the rigid object in Quick Quiz If the object starts from rest at the initial angular position, moves counterclockwise with constant angular acceleration, and arrives at the final angular position with the same angular speed in all three cases, for which choice is the angular acceleration the highest? Answer: (b)

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Example 10.1 A wheel rotates with a constant angular acceleration of 3.50 rad/s2. (A) If the angular speed of the wheel is 2.00 rad/s at ti = 0, through what angular displacement does the wheel rotate in 2.00 s? Solution:

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Example 10.1 (B) Through how many revolutions has the wheel turned during this time interval? Solution: (C) What is the angular speed of the wheel at t = 2.00 s?

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**Relationship Between Angular and Linear Quantities**

Every point on the rotating object has the same angular motion. Every point on the rotating object does not have the same linear motion. Displacements s = q r Speeds v = w r Accelerations a = a r Section 10.3

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**Speed Comparison – Details**

The linear velocity is always tangent to the circular path. Called the tangential velocity The magnitude is defined by the tangential speed. (10.10) Section 10.3

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**Speed Comparison – Details**

Since r is not the same for all points on the object, the tangential speed of every point is not the same. The tangential speed increases as one moves outward from the center of rotation. Section 10.3

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**If you can't see the image above, please install Shockwave Flash Player.**

If this active figure can’t auto-play, please click right button, then click play. Active_Figure 10.4 10.4: Rotation of a Rigid Object About a Fixed Axis As this rigid object rotates about the fixed axis shown here, the blue point has a tangential velocity v that is always tangent to the circular path of radius r.

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**Acceleration Comparison – Details**

The tangential acceleration is the derivative of the tangential velocity. (10.11) Section 10.3

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**Speed and Acceleration Note**

All points on the rigid object will have the same angular speed, but not the same tangential speed. All points on the rigid object will have the same angular acceleration, but not the same tangential acceleration. The tangential quantities depend on r, and r is not the same for all points on the object. Section 10.3

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**Centripetal Acceleration**

An object traveling in a circle, even though it moves with a constant speed, will have an acceleration. Therefore, each point on a rotating rigid object will experience a centripetal acceleration. (10.12) Section 10.3

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**Resultant Acceleration**

The tangential component of the acceleration is due to changing speed. The centripetal component of the acceleration is due to changing direction. Total acceleration can be found from these components: (10.13) Section 10.3

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**Rotational Motion Example**

For a compact disc player to read a CD, the angular speed must vary to keep the tangential speed constant (vt = wr). Typically the constant speed of the surface at the point of the laser-lens system is 1.3 m/s. At the inner sections, the angular speed is faster than at the outer sections. Section 10.3

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Quick quiz 10.3 Ethan and Joseph are riding on a merry-go-round. Ethan rides on a horse at the outer rim of the circular platform, twice as far from the center of the circular platform as Joseph, who rides on an inner horse. (i) When the merry-go-round is rotating at a constant angular speed, what is Ethan’s angular speed? (a) twice Joseph’s (b) the same as Joseph’s (c) half of Joseph’s (d) impossible to determine (ii) When the merry-go-round is rotating at a constant angular speed, describe Ethan’s tangential speed from the same list of choices. Answer: (i) (b) (ii) (a)

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Example 10.2 On a compact disc (Fig. 10.6), audio information is stored digitally in a series of pits and flat areas on the surface of the disc. The alternations between pits and flat areas on the surface represent binary ones and zeros to be read by the CD player and converted back to sound waves. The Pits and flat areas are detected by a system consisting of a laser and lenses. The length of a string of ones and zeros representing one piece of information is the same everywhere on the disc, whether the information is near the center of the disc or near its outer edge. So that this length of ones and zeros always passes by the laser-lens system in the same time interval, the tangential speed of the disc surface at the location of the lens must be constant.

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Example 10.2 According to equation 10.10, the angular speed must therefore vary as the laser-lens system moves radially along the disc. In a typical CD player, the constant speed of the surface at the point of the laser-lens system is 1.3 m/s. (A) Find the angular speed of the disc in revolutions per minute when information is being read from the innermost first track (r = 23 mm) and the outermost final track (r = 58 mm). Solution:

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Example 10.2 (B) The maximum playing time of a standard music disc is 74 min and 33 s. How many revolutions does the disc make during that time? Solution:

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Example 10.2 (C) What is the angular acceleration of the compact discover the 4473 s time interval? Solution:

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Torque Torque, t, is the tendency of a force to rotate an object about some axis. Torque is a vector, but we will deal with its magnitude here: t ≡ r F sin f = Fd F is the force f is the angle the force makes with the horizontal d is the moment arm (or lever arm) of the force There is no unique value of the torque on an object. Its value depends on the choice of a rotational axis. (10.14) Section 10.4

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Torque, cont The moment arm, d, is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force. d = r sin f The horizontal component of the force (F cos f) has no tendency to produce a rotation. Section 10.4

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**Torque, cont Torque will have direction.**

If the turning tendency of the force is counterclockwise, the torque will be positive. If the turning tendency is clockwise, the torque will be negative. Section 10.4

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Net Torque The force will tend to cause a counterclockwise rotation about O. The force will tend to cause a clockwise rotation about O. St = t1 + t2 = F1d1 – F2d2 Section 10.4

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**If you can't see the image above, please install Shockwave Flash Player.**

If this active figure can’t auto-play, please click right button, then click play. Active_Figure 10.8 10.8: Clockwise and Counter-Clockwise Torques For the initial conditions of this animation, one force tends to rotate the object counterclockwise about O and the other tends to rotate it clockwise. You can adjust the variables here to explore different cases.

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**Torque vs. Force Forces can cause a change in translational motion.**

Described by Newton’s Second Law Forces can cause a change in rotational motion. The effectiveness of this change depends on the force and the moment arm. The change in rotational motion depends on the torque. Section 10.4

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**Torque Units The SI units of torque are N.m.**

Although torque is a force multiplied by a distance, it is very different from work and energy. The units for torque are reported in N.m and not changed to Joules. Section 10.4

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Quick Quiz 10.4 (i) If you are trying to loosen a stubborn screw from a piece of wood with a screwdriver and fail, should you find a screwdriver for which the handle is (a) longer or (b) fatter? (ii) If you are trying to loosen a stubborn bolt from a piece of metal with a wrench and fail, should you find a wrench for which the handle is (a) longer or (b) fatter? Answer: (i) (b) (ii) (a)

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Example 10.3 A one-piece cylinder is shaped as shown in Figure 10.9, with a core section protruding from the larger drum. The cylinder is free to rotate about the central z axis shown in the drawing. A rope wrapped around the drum, which has radius R1, exerts a force to the right on the cylinder. A rope wrapped around the core, which has radius R2, exerts a force downward on the cylinder.

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Example 10.7 (A) What is the net torque acting on the cylinder about the rotation axis (which is the z axis in Fig. 10.9)? Solution: (B) Suppose T1 = 5.0 N, R1 = 1.0 m, T2 = 15 N, and R2 = 0.50 m. What is the net torque about the rotation axis, and which way does the cylinder rotate starting from rest?

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**Torque and Angular Acceleration**

Consider a particle of mass m rotating in a circle of radius r under the influence of tangential force . The tangential force provides a tangential acceleration: Ft = mat The radial force causes the particle to move in a circular path. Section 10.5

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**Torque and Angular Acceleration, Particle cont.**

The magnitude of the torque produced by on a particle about an axis through the center of the circle is St = SFt r = (mat) r The tangential acceleration is related to the angular acceleration. St = (mra) r = (mr2) a Since mr 2 is the moment of inertia of the particle, St = Ia The torque is directly proportional to the angular acceleration and the constant of proportionality is the moment of inertia. (10.15) (10.16) Section 10.5

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**Torque and Angular Acceleration, Extended**

Consider the object consists of an infinite number of mass elements dm of infinitesimal size. Each mass element rotates in a circle about the origin, O. Each mass element has a tangential acceleration. From Newton’s Second Law Fi = miai The torque associated with the force and its magnitude is given by ti = ri Fi = ri mi ai Section 10.5

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**Torque and Angular Acceleration, Extended cont.**

Finding the net torque This becomes Stext = Ia This is the same relationship that applied to a particle. This is the mathematic representation of the analysis model of a rigid body under a net torque. The result also applies when the forces have radial components. The line of action of the radial component must pass through the axis of rotation. These components will produce zero torque about the axis. (10.17) Section 10.5

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**Falling Smokestack Example**

When a tall smokestack falls over, it often breaks somewhere along its length before it hits the ground. Each higher portion of the smokestack has a larger tangential acceleration than the points below it. The shear force due to the tangential acceleration is greater than the smokestack can withstand. The smokestack breaks. Section 10.5

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**Torque and Angular Acceleration, Wheel Example**

Two analysis models need to be applied. The object is modeled as a particle under a net force. The wheel is modeled as a rigid object under a net torque. The wheel is rotating and so we apply St = Ia The tension supplies the tangential force. The mass is moving in a straight line, so apply Newton’s Second Law. SFy = may = mg – T Section 10.5

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Quick Quiz 10.5 You turn off your electric drill and find that the time interval for the rotating bit to come to rest due to frictional torque in the drill is Dt. You replace the bit with a larger one that results in a doubling of the moment of inertia of the drill’s entire rotating mechanism. When this larger bit is rotated at the same angular speed as the first and the drill is turned off, the frictional torque remains the same as that for the previous situation. What is the time interval for this second bit to come to rest? (a) 4 Dt (b) 2 Dt (c) Dt (d) 0.5 Dt (e) 0.25 Dt (f) impossible to determine Answer: (b)

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Example 10.4 A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is free to rotate about the pivot in the vertical plane as in Figure The rod is released from rest in the horizontal position. What are the initial angular acceleration of the rod and the initial translational acceleration of its right end? Solution:

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Example 10.6 A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless, horizontal axle as in Figure A light cord wrapped around the wheel supports an object of mass m. When the wheel is released, the object accelerates downward, the cord unwraps off the wheel, and the wheel rotates with an angular acceleration. Calculate the angular acceleration of the wheel, the translational acceleration of the object, and the tension in the cord.

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Example 10.6 Solution:

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Moment of Inertia The definition of moment of inertia is The dimensions of moment of inertia are ML2 and its SI units are kg‧m2. We can calculate the moment of inertia of an object more easily by assuming it is divided into many small volume elements, each of mass Dmi. Mass is an inherent property of an object, but the moment of inertia depends on the choice of rotational axis. (10.19) Section 10.6

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Moment of Inertia Moment of inertia is a measure of the resistance of an object to changes in its rotational motion, similar to mass being a measure of an object’s resistance to changes in its translational motion. The moment of inertia depends on the mass and how the mass is distributed around the rotational axis. Section 10.6

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**Moments of Inertia of Various Rigid Objects**

Section 10.5

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Moment of Inertia, cont The moment of inertia of a system of discrete particles can be calculated by applying the definition for I. For a continuous rigid object, imagine the object to be divided into many small elements, each having a mass of Dmi. We can rewrite the expression for I in terms of Dm. With the small volume segment assumption, If r is constant, the integral can be evaluated with known geometry, otherwise its variation with position must be known. (10.20) (10.21) Section 10.6

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**Notes on Various Densities**

Volumetric Mass Density → mass per unit volume: r = m / V Surface Mass Density → mass per unit thickness of a sheet of uniform thickness, t: s = r t Linear Mass Density → mass per unit length of a rod of uniform cross-sectional area: l = m / L = r A Section 10.6

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**Moment of Inertia of a Uniform Rigid Rod**

The shaded area has a mass dm = l dx Then the moment of inertia is Section 10.6

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**Moment of Inertia of a Uniform Solid Cylinder**

Divide the cylinder into concentric shells with radius r, thickness dr and length L. dm = r dV = 2p(rLr)dr Then for I Section 10.6

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Example 10.7 Calculate the moment of inertia of a uniform rigid rod of length L and mass M (Fig ) about an axis perpendicular to the rod (the y’ axis) and passing through its center of mass. Solution:

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Example 10.8 A uniform solid cylinder has a radius R, mass M, and length L. Calculate its moment of inertia about its central axis (the z axis in Fig.10.16). Solution:

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**Parallel-Axis Theorem**

In the previous examples, the axis of rotation coincided with the axis of symmetry of the object. For an arbitrary axis, the parallel-axis theorem often simplifies calculations. The theorem states I = ICM + MD2 I is about any axis parallel to the axis through the center of mass of the object. ICM is about the axis through the center of mass. D is the distance from the center of mass axis to the arbitrary axis. (10.22) Section 10.6

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Section 10.6

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Section 10.6

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**Moment of Inertia for a Rod Rotating Around One End – Parallel Axis Theorem Example**

The moment of inertia of the rod about its center is D is ½ L Therefore, Section 10.6

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Example 10.9 Consider once again the uniform rigid rod of mass M and length L shown in Figure Find the moment of inertia of the rod about an axis perpendicular to the rod through one end (the y axis in Fig ). Solution:

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**Rotational Kinetic Energy**

An object rotating about some axis with an angular speed, w, has rotational kinetic energy even though it may not have any translational kinetic energy. Each particle has a kinetic energy of Ki = ½ mivi2 Since the tangential velocity depends on the distance, r, from the axis of rotation, we can substitute vi = wir. Section 10.7

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**Rotational Kinetic Energy, cont**

The total rotational kinetic energy of the rigid object is the sum of the energies of all its particles. (10.23) Section 10.7

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**Rotational Kinetic Energy, final**

There is an analogy between the kinetic energies associated with linear motion (K = ½ mv2) and the kinetic energy associated with rotational motion. KR= ½ Iw2 Rotational kinetic energy is not a new type of energy, the form is different because it is applied to a rotating object. The units of rotational kinetic energy are Joules (J). (10.24) Section 10.7

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Quick Quiz 10.6 A section of hollow pipe and a solid cylinder have the same radius, mass, and length. They both rotate about their long central axes with the same angular speed. Which object has the higher rotational kinetic energy? (a) The hollow pipe does. (b) The solid cylinder does. (c) They have the same rotational kinetic energy. (d) It is impossible to determine. Answer: (a)

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Example 10.10 Four tiny spheres are fastened to the ends of two rods of negligible mass lying in the xy plane to form an unusual baton (Fig ). We shall assume the radii of the spheres are small compared with the dimensions of the rods. (A) If the system rotates about the y axis (Fig a) with an angular speed w, find the moment of inertia and the rotational kinetic energy of the system about this axis. Solution:

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Example 10.10 (B) Suppose the system rotates in the xy plane about an axis (the z axis) through the center of the baton (Fig b). Calculate the moment of inertia and rotational kinetic energy about this axis. Solution:

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**Work in Rotational Motion**

Find the work done by on the object as it rotates through an infinitesimal distance ds = r dq The radial component of the force does no work because it is perpendicular to the displacement. Section 10.8

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**Power in Rotational Motion**

The rate at which work is being done in a time interval dt is This is analogous to P = Fv in a linear system. (10.26) Section 10.8

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**Work-Kinetic Energy Theorem in Rotational Motion**

The work-kinetic energy theorem for rotational motion states that the net work done by external forces in rotating a symmetrical rigid object about a fixed axis equals the change in the object’s rotational kinetic energy. (10.27) Section 10.8

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**Work-Kinetic Energy Theorem, General**

The rotational form can be combined with the linear form which indicates the net work done by external forces on an object is the change in its total kinetic energy, which is the sum of the translational and rotational kinetic energies. Section 10.8

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**Summary of Useful Equations**

Section 10.8

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**Energy in an Atwood Machine, Example**

The system containing the two blocks, the pulley, and the Earth is an isolated system in terms of energy with no non-conservative forces acting. The mechanical energy of the system is conserved. The blocks undergo changes in translational kinetic energy and gravitational potential energy. The pulley undergoes a change in rotational kinetic energy. Section 10.8

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Example 10.11 A uniform rod of length L and mass M is free to rotate on a frictionless pin passing through one end (Fig 10.21). The rod is released from rest in the horizontal position. (A) What is its angular speed when the rod reaches its lowest position? Solution:

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Example 10.11 (B) Determine the tangential speed of the center of mass and the tangential speed of the lowest Point on the rod when it is in the vertical Position. Solution:

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Example 10.12 Two blocks having different masses m1 and m2 are connected by a string passing over a Pulley as shown in Active Figure The pulley has a radius R and moment of inertia I about its axis of rotation. The string does not slip on the pulley, and the system is released from rest. Find the translational speeds of the blocks after block 2 descends through a distance h and find the angular speed of the pulley at this time.

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Example 10.12 Solution:

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Rolling Object The red curve shows the path moved by a point on the rim of the object. This path is called a cycloid. Section 10.9

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Rolling Object The green line shows the path of the center of mass of the object. In pure rolling motion, an object rolls without slipping. In such a case, there is a simple relationship between its rotational and translational motions. Section 10.9

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**Pure Rolling Motion, Object’s Center of Mass**

The translational speed of the center of mass is The linear acceleration of the center of mass is (10.28) (10.29) Section 10.9

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Rolling Motion Cont. Rolling motion can be modeled as a combination of pure translational motion and pure rotational motion. The contact point between the surface and the cylinder has a translational speed of zero (c). Section 10.9

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**Total Kinetic Energy of a Rolling Object**

The total kinetic energy of a rolling object is the sum of the translational energy of its center of mass and the rotational kinetic energy about its center of mass K = ½ ICM w2 + ½ MvCM2 The ½ ICMw2 represents the rotational kinetic energy of the cylinder about its center of mass. The ½ Mv2 represents the translational kinetic energy of the cylinder about its center of mass. (10.31) Section 10.9

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**Total Kinetic Energy, Example**

Accelerated rolling motion is possible only if friction is present between the sphere and the incline. The friction produces the net torque required for rotation. No loss of mechanical energy occurs because the contact point is at rest relative to the surface at any instant. Section 10.9

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**Total Kinetic Energy, Example**

In reality, rolling friction causes mechanical energy to transform to internal energy. Rolling friction is due to deformations of the surface and the rolling object. Section 10.9

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**If you can't see the image above, please install Shockwave Flash Player.**

If this active figure can’t auto-play, please click right button, then click play. Active_Figure 10.26 10.26: Conservation of Mechanical Energy for Rolling Objects In this animation, you can roll several objects down the incline and see how the final speed depends on the type of object.

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**Total Kinetic Energy, Example cont.**

Apply Conservation of Mechanical Energy: Let U = 0 at the bottom of the plane Kf + U f = Ki + Ui Kf = ½ (ICM / R2) vCM2 + ½ MvCM2 Ui = Mgh Uf = Ki = 0 Solving for v (10.33) Section 10.9

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**Sphere Rolling Down an Incline, Example**

Conceptualize A sphere is rolling down an incline. Categorize Model the sphere and the Earth as an isolated system in terms of energy. No non-conservative forces are acting. Section 10.9

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**Sphere Rolling Down an Incline, Example cont.**

Analyze Use Conservation of Mechanical Energy to find v. Solve for the acceleration of the center of mass. Finalize Both the speed and the acceleration of the center of mass are independent of the mass and the radius of the sphere. Generalization All homogeneous solid spheres experience the same speed and acceleration on a given incline. Similar results could be obtained for other shapes. Section 10.9

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Quick Quiz 10.7 A ball rolls without slipping down incline A, starting from rest. At the same time, a box starts from rest and slides down incline B, which is identical to incline A except that it is frictionless. Which arrives at the bottom first? (a) The ball arrives first. (b) The box arrives first. (c) Both arrive at the same time. (d) It is impossible to determine. Answer: (b)

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Example 10.13 For the solid sphere shown in Active Figure 10.26, calculate the translational speed of the center of mass at the bottom of the incline and the magnitude of the translational acceleration of the center of mass. Solution:

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Example 10.14 A cylindrically symmetric spool of mass m and radius R sits at rest on a horizontal table with friction (Fig ). With your hand on a massless string wrapped around the axle of radius r, you pull on the spool with a constant horizontal force of magnitude T to the right. As a result, the spool rolls without slipping a distance L along the table with no rolling friction. (A) Find the final translational speed of the center of mass of the spool.

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Example 10.14 Solution:

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Example 10.14 (B) Find the value of the friction force f. Solution:

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