Chapter 7:  Angular Kinetics

A. Center of Mass
B. Torque and Equilibrium of a Rigid Body
C. The Second Condition for Equilibrium
D. Rigid Bodies in Equilibrium
E. Levers
F. Stability of Equilibrium
G. Moment of Inertia:  The Rotational Analog to Mass
H. Angular Momentum
I. Rotational Kinetic Energy
Problems
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In Chapter 6 we saw that the rotational equations of motion were exactly the same as the corresponding equations for linear motion.  In this chapter we will continue to see that linear quantities and laws, such as Newton’s second law, kinetic energy, and momentum, have rotational counterparts.

A.     Center of Mass

The transitional and rotational motion of rigid bodies can be analyzed most easily with the help of a concept called the center of mass.  The center of mass is just the average position of the mass distribution of an object.

We make use if the center of mass because it has the following important property:

All the mass of a rigid body may be assumed to be at the center of mass of the object when considering the transitional behavior of that object under the action of external forces.

This means that the rigid body is equivalent to an equally massive point mass placed at the mass center insofar as its transitional motion under a given external force is concerned.  For example, consider the projectile motion of a bowling pin or lopsided softball.  Their center of mass follows a smooth curve, the same curve a point object or spherically symmetric object would follow if subjected to the same external forces.  As we shall see, the importance of center of mass is that it allows us to separate the transitional and rotational motion of an object.  The object will be translated as though all its mass were concentrated at its mass center.
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B.     Torque and Equilibrium of a Rigid Body

When an object is in equilibrium its state of motion is not changing.  An obvious condition for equilibrium is that the net force acting must be zero.  Then from Newton’s second law, Fnet = ma, we have that the acceleration is zero and the velocity is constant.  Although this is the only requirement necessary for a point mass to be in equilibrium it is not sufficient to guarantee the equilibrium of a rigid body.

Consider three cases below where a stick is subjected to two forces with the same magnitude.

In all three cases the net force on the stick is zero.  Yet we know that in the last case the stick does not remain at rest.  It rotates clockwise.  We are therefore forced to conclude that while a necessary condition for equilibrium is that the net force is zero, this is not a sufficient condition for a rigid body.  A second condition is required to insure rotational equilibrium.  There must be no net turning effect of forces rotating an object about a pivot point if the rigid object is to be in equilibrium.

Experience shows that the action causing a rotation depends not only on the magnitude and direction of the applied force but also on the point of application.  When you open a door, for example, you push with your hand near the doorknob.  You know that a much larger force is required to turn the door if you push near its hinges at the axis of rotation.  Two factors determine the rotational impetus:  the perpendicular component of the force and the distance from the point of application to the axis of rotation.  The turning effect is given the name torque.  The definition to torque is

Torque = (perpendicular force)(distance)

Or in symbols

                                                                                                                                                             (7.1)

Where the symbol t (Greek “tau”) has been used for torque.  Note that torque has the units of force times length (e.g., N • m, lb • ft).

Examples:

1.        Find the torque exerted by the wrench on the bolt in the diagram below.

                Using Eq. (7.1) we have

                                               

2.        A student with an arm of length 2 ft holds a weight of 20 lb in her hand.  Calculate the torque exerted on her shoulder by the weight if she holds her arm a) outstretched, b) at her side, and c) at an angle of 60˚ below horizontal.

a)       First draw a schematic diagram and then determine the torque using Eq. (7.1).

               

                               

b)       We see that when the weight is held at her side there is no component of the force perpendicular to d.  This force gives no turning effect and thus the torque is zero.

                                                                     

c)  In this case the perpendicular component of the force is (20 lb) (cos 60˚) which is equal to 10 lb.  Thus by Eq. (7.1) we have


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C.     The Second Condition for Equilibrium

We have now acquired enough insight so that we can now make a precise statement of the second condition for equilibrium.

For an object to be in equilibrium there must be no change in the rotational motion.  Thus the net torque must be zero.  This is equivalent to the requirement that the sum of all clockwise (CW) torques must equal the sum of all counterclockwise (CCW) torques.

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D.     Rigid Bodies in Equilibrium

Now that we know the second condition for equilibrium, we are able to treat all ordinary situations in which a rigid body is at equilibrium.  To analyze such situations we follow the simple procedure outlined below:

1.        Isolate the body for discussion.

2.        Draw all the forces acting on the body.

3.        Choose a pivot point about which you will determine torques.

4.        Determine the torques.

5.        Solve the equations Fnet = 0 , and  tCW = tCCW

Before we look at some examples consider the choice of the axis of rotation for solving the torque equation.  When a rigid body is in equilibrium it is not rotating about any axis.  As a result, the axis for solving the torque equation may be taken anywhere and is completely arbitrary.  We are therefore justified in taking the axis anywhere we think is convenient.

Examples:

1.        A 16 ft seesaw is pivoted in the center.  At what distance from the center would a 200 lb person sit to balance a 150 lb person on the opposite end?

A convenient point to calculate the torques would be at the fulcrum.  With this choice the upward force exerted by the fulcrum does not contribute to the torques.

                               

2.        A 10 ft, 25 lb seesaw is balanced by a little girl (50 lb) and her father (200 lb) at opposite ends as shown below.  How far from the seesaw’s center of mass must the fulcrum be placed?

Calculate the torques about the fulcrum.

                               

               

3.        A 6 ft tall 180 lb student is instructed to lie on a board as shown below.  The board is 8 ft long, weighs 10 lbs and is supported on the left end by a scale that reads 80 lbs.  Use this information to determine the distance from his feet to his center of mass.

Calculate the torques about the right pivot.

Note:  In the last example we found the 6 ft man to have a center of mass 3.33 ft from his feet.  The ratio of these two values (3.33 ft/6 ft = 0.56) tells us the position of the center of mass of the student relative to his height.  For an average man the center of mass is approximately 56% of his height.  This value drops to approximately 54% for a woman.
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E.     Levers

A lever is a rigid bar free to rotate about a fixed point called the fulcrum.  The position of the fulcrum is fixed so that it is not free to move with respect to the bar.  Levers are used to lift the loads in an advantageous way and to transfer movement from one point to another.

There are three basic parts to any lever:  the fulcrum, the applied force, and the load.  Levers are classified by which one of these parts is in the middle, as shown below.

In a class 1 lever the fulcrum is placed between the applied force and the load.  A crowbar, an automobile jack, and a pair of scissors are examples of class 1 levers.  In a class 2 lever the load is in the middle.  A wheelbarrow is an example of a class 2 lever.  A class 3 lever has the applied force in the middle.  As we shall see, some human limb movements are performed by class 3 levers.

Examples:

1.        Determine the force that must be applied to end of the class 1 lever shown below to lift the 500 lb load.

In this example we will neglect the weight of the lever.  To solve this calculate the torques about the fulcrum.

                           

2.        The bones and muscles in the arm can be examined with a very simple model of a class 3 lever.  Consider what happens when you hold a weight outstretched in your hand with your upper arm against your body.  The elbow acts like a fulcrum with the lower arm as a lever.  The load is the weight of your lower arm and the weight being held in your hand.  The biceps apply an upward force to keep the system in equilibrium.  Calculate the tension in the biceps if you hold a 15 lb weight in your hand as shown below.  The dimensions given are typical.  We assume the lower arm weighs 4 lb.

Before starting this problem you should first convince yourself that this is indeed an example of a third class lever.  To find the applied force calculate the torques about the elbow.

                      

Note:  We recognize first and second class levers as being very advantageous in lifting weights.  However, as we see in the last example, in a class 3 lever the applied force is much greater than the weight of the load.  Can you think of any particular advantage that a class 3 lever gives?
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F.     Stability of Equilibrium

The principles we have discussed in previous sections allow us to examine what happens when a body is tipped or knocked over.  Consider the motion of the glass of orange juice shown below.  Close observation shows us that the glass is rotating about point A.  The rotations is a result of an unbalanced torque on the glass.

Example:  Suppose you are given the job of moving a 100 lb refrigerator.  Being basically lazy you decide to slide the refrigerator instead of lifting it up and carrying it.  Suppose further that this task occurs in typical student ghetto housing where the linoleum is usually warped.  If the edge of the refrigerator catches a piece of linoleum as it is pushed as shown below, determine the force F that will tip it over.

This type of problem is easily solved if we recognize three important features.  First we see that the refrigerator will rotate about point A if it tips over.  Then we realize that it will only tip over if the torque about point A due to the external force exceeds the torque due to the weight of the refrigerator.  Finally remember that only the perpendicular component of the force gives a torque.  If we extend the lines of the applied force and the weight of the refrigerator, then torques are easily calculated as shown below.

If the clockwise torque is to balance the counterclockwise torque we have

Thus if the applied force exceeds 50 lb the refrigerator will be tipped over.

Problem:  If the refrigerator is 6 ft tall and you push on the top of the refrigerator, what applied force will tip it over?

                               

Question:  In view of the previous discussion, if you were given the task of moving your own refrigerator where would you push it?
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G.     Moment of Inertia:  The Rotational Analog to Mass

The most important equation for linear motion was given to us by Newton in his second law, F = ma.  We now seek a rotational analog to the second law.  It is easy to see that for rotational motion angular acceleration must replace linear acceleration.  Furthermore, since an unbalanced torque is required to change rotational motion, just as an unbalanced force is required to change linear motion, torque will replace force.  However the rotational analog to mass is not as obvious.

To find the rotational analog to mass first consider how mass is viewed in the second law.  Mass is a measure of the inertia of an object.  It tells us how hard it is to set an object into motion or to stop it if it is already moving.  With this concept of linear inertia in mind we define the moment of inertia of an object.  The moment of inertia tells us how hard it is to set an object into rotational motion, or how difficult it is to stop a rotating object.  Just as inertia depends on the mass of an object, the moment of inertia depends on the mass of an object.  In addition, how this mass is distributed will also affect the moment of inertia.

We may now state the rotational analog to Newton’s second law as

                                                t = Ia                                                                                                                     (7.2)

Where I is the moment of inertia of the body, and α is the angular acceleration which must be in units of rad/sec2.  In this equation I is viewed as resisting angular acceleration by an external torque.  Moments of inertia for various objects are given below.  It is important to note that objects with large moments of inertia have more mass concentrated away from the axis of rotation.  This makes them more difficult to set into rotational motion.  As a result, a hollow cylinder will have a larger I than an equally massive solid cylinder.  And I is four times greater for a rod rotating about its end than about its middle.

Examples:

1.        Find the torque necessary to give a bike wheel of radius 0.4 m and mass 2 kg an angular acceleration of 10 rad/sec2.

We need to solve the Eq. 7.2.  First use the fact that a bike wheel is a full cylinder to find its moment of inertia.

                               

Now solve Eq. 7.2

                               

2.        A typical bowling ball has a mass of 5 kg and a radius of 10 cm.  What torque is necessary to increase the angular velocity of a bowling ball from rest to 40 rad/sec in a time of 2 seconds?

Once again we must solve Eq. 7.2, but in this case both I and α must be determined from the information given.  First determine I.  Neglecting the holes in a bowling ball we use the moment of inertia of a solid sphere.

                               

Now use angular kinematics to find α.

                               

                               

Noticing that the angular acceleration is in the appropriate units, we are now ready to determine the torque.

                               
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H.     Angular Momentum

In dealing with transitional motion we found it useful to introduce the quantity linear momentum (p = mv).  This allowed us to state Newton’s law F = ma in a convenient way.  Namely, that force is the rate at which linear momentum changes.  We saw that an external force was necessary to change momentum.  Furthermore, when no external force was present we found that linear momentum was conserved.  This gave us a powerful tool in treating many physical problems such as collisions.

The rotational analog to linear momentum is called angular momentum.  As one might expect from the fact that linear momentum is given by mv, the defining equation for angular momentum is

Angular momentum = Iω                                                                                   (7.3)

The angular momentum of an object or a system of objects obeys a conservations law much like the one obeyed by linear momentum.  It may be stated as follows:

If no external torques act on a system, its angular momentum will remain constant.

Example:  An ice skater with arms outstretched is spinning with an angular velocity if 0.5 rev/sec.  In this position her moment of inertia is 12 kg • m2.  Suppose she brings her arms close to her body and reduces her moment of inertia to 3 kg • m2.  What is her new angular velocity? 

To solve this we simply use the law of conservation of angular momentum.

                               
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I.      Rotational Kinetic Energy

The center of mass of a rotating wheel mounted on a stationary axle has zero velocity, and thus the wheel has no kinetic energy of translation.  Yet we know that a rotating flywheel can do work because of its rotation.  For example, it could wind up a rope with a weight hanging from it.  Therefore, it does possess kinetic energy, that is energy of motion.

Once again we can use the definition of translational kinetic energy given by ½ mv2 to find an expression for rotational kinetic energy.  We replace mass by the moment of inertia and linear velocity by angular velocity and obtain:

                                                                                                                                        (7.4)

Remembering out definition of kinetic energy, this equation tells us what amount of work must be done on a wheel to set it in motion, or what amount of work it can do by virtue of its rotation.

In many cases an object will be rotating and translating.  In such instances, the total kinetic energy is the sum of translational and rotational kinetic energies, ½ mv2 + ½ Iω2.  If the object rolls without slipping we have a further restriction that v = rω, where v is the velocity if the center of mass.  Examine the object below which rolls without slipping and summarize these equations.

                               

For all objects that rotate, we see that a rotational kinetic energy exists due to the rotational motion of the object.  This rotational kinetic energy is interconvertible with work, potential energy, and translational energy.  So we can again use the law of conservation of energy.

Example:  A flywheel of radius 0.2 meters and mass 40 kg is rotating at 50 rad/sec.  Find the moment of inertia of the wheel and its rotational kinetic energy.

We assume the flywheel is a solid cylinder so we first calculate the moment of inertia.

                               

Now use Eq 7.4 to find the rotational kinetic energy

                               

Question:  In the previous example how much work is required to set the flywheel in motion?
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Summary

Center of Mass Moment of Inertia
Equilibrium  Angular Momentum
Torque  Conservation of Angular Momentum
Levers  Rotational Kinetic Energy
Stability of Equilibrium  

Problems

1.        Calculate the torque exerted by the applied force in the situations described below:

a.        A 200 lb force acting perpendicularly at a distance of 3 ft.

b.       A 70 kg person 2 m from a seesaw fulcrum.

c.        A uniform 12 ft ladder of 50 lb suspended at one end on a horizontal position.

d.       A uniform 12 ft ladder of 50 lb suspended at one end on a vertical position.

2.        A 40 lb little boy would like to ride on the seesaw but there is no one else to play with.  The seesaw is 12 ft long and weighs 20 lb.  Where must the fulcrum be placed to balance the boy alone at one end?

3.        A 20 ft long, 200 lb plant rests on two points as shown below.  The plank is not fastened at either point.  What is the furthest distance x that the 100 lb girl can walk before the plank tips over?

4.        The position of the foot when standing with the tip of the feet on the edge of a step is shown below.  The floor pushes upward at the toes with a force equal to the weight of the person.  Calculate the tension on the Achilles tendon and the compressional force on the tibia when a 200 lb person stands in this position.

 

5.        A dolly is very useful instrument to have when you need to move massive objects.  It is another example of a class 2 lever.  Consider the task of moving a 4 ft, 300 lb sleeper sofa with a 6 foot dolly as diagrammed below.  What force must you apply to lift the sofa?

6.        What weights must be placed on a 4 foot hurdle if it is to be just knocked over with a horizontal force of 7 lb?  Assume the bottoms of the hurdle are each 2 ft long and weigh 2 lb each.

7.        In the two situations shown below identical objects are shown in different positions.  In which position are the objects more stable against a toppling force?

8.        A 200 lb student stands 6 ft tall.  What horizontal force is required to knock him over in the following circumstances?

a.        The force is applied 3 ft from the ground and his feet are spread 1 ft apart.

b.       The force is applied 3 ft from the ground and his feet are spread 2 ft apart.

c.        The force is applied 5 ft from the ground and his feet are spread 1 ft apart.

9.        A beer keg has a radius of 20 cm (.2 m) and a mass of 25 kg.  What is its moment of inertia?  What torque is necessary to give it an angular acceleration of 8 rad/sec2?

10.     A high diver is rotating in the tuck position with an angular velocity of 2 rev/sec.  His mass is 80 kg.

a.        Assume that in the tuck position he has a moment of inertia similar to a solid cylinder of radius 0.5 meters.  Find his moment of inertia.

b.       Before entering the water a diver stretches out of the tuck position and assumes the shape of a long rigid rod.  Assuming he stretches out to a length of 2 meters calculate his moment of inertia in this position.

c.        What is his angular velocity in this new position?

11.     Why is it necessary to have a second propeller of a helicopter?

12.     A merry-go-round of radius 2 m and moment of inertia 400 kg•m2 is rotating with an angular velocity of 0.30 rev/sec.  If a 50 kg student steps onto the edge of the merry-go-round find:

a.        The final moment of inertia of the system.

b.       The final angular velocity if the system.

13.     A bowling ball has a mass of 5 kg and a radius of 0.15 m.  As it rolls down the bowling alley without slipping its center of mass travels with a velocity of 4 m/sec.

a.        What is the moment of inertia of the bowling ball?

b.       What is the angular velocity of the bowling ball?

c.        What is the rotational kinetic energy of the bowling ball?

d.       What is the translational kinetic energy of the bowling ball?

e.        What is the total kinetic energy of the bowling ball?

14.     A hollow cylinder is rolling with a speed of 4 m/sec on a horizontal surface as it approaches an incline.  What height above the horizontal will it rise?


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