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Now, the instantaneous power is computed

P = τ ω (8.17)

= (500. dyne.cm) (419 rad/s)

P = 2.10 x 106 erg/s = 210.mW

EXAMPLE 8.34

The instantaneous power output of an electric motor is known to be 1500.W, while it is

exerting a torque of 2.50 N.m on a pump. Find the rate at which the motor is rotating (in

r/min)

Solution :

P = τ ω (8.17)ω = P/ τ

= (1500.W)/ (2.50 N.m) = 600 rad/s

= (600 rad/s) (1 r/2 π rad) (60.0s/1 min)

ω = 5730 r/min

8.6 ROTATIONAL FORM OF NEWTON’S SECOND LAW

If more than one torque is acting on the object in Figure 8.11 about the axis of rotation

and you wish to compute the total work done on the object, the torque τ in Equation 8.16

could be replaced by the resultant torque ∑τ ,thus giving *

W = (∑τ)θ (8.18)

Of course, in adding up the torques to obtain a resultant, we would have to assign a

positive ( + ) or negative (- ) sign to each torque (see Chapter 4 ), depending on the

direction (CCW or CW, respectively) in which the torque was attemping to rotate the

object. The resultant torque would than also be positive (CCW) or negative ( CW ). The

angle θ that the object rotated through would also be positive (+ ) or negative ( - ),

depending on which direction (CCW or CW, respectively) the object actually turned.

This work is equal to the change in the kinetic energy of the object, or*



(∑τ)θ = ½ I ωf 2 – I ωi

2

∑τ = I [ωf 2 - ωi

2] (8.19)

2 θ

Recall

ωf 2 = ωi

2 + 2αθ (8.3)

Solving this for α, we get

α = = I [ωf 2 - ωi2]2 θ

* A more complete discussion of vectors in rotation is given in section 8.9

This expression when substituted back into Equation 8.19 gives

(8.20)

Equation 8.20 is the rotational form of Newton’s second law. It states that the resultant

torque acting on an object equals the object’s moment of inertia times the object’s

angular acceleration. It is frequently used in conjuction with one of the rotational

kinematics equations first encountered in Section 8.1. Since Equation 8.16 was used to

derive it, the angular measure must be in radians in the angular acceleration (e.g., rad/s2)

EXAMPLE 8.35

An object whose moment of inertia is 35.0 kg.m2 about its axis of rotation has an angular

acceleration of 4.20 rad/s2 in a clockwise direction. Find the resultant torque that must be

acting on this object.

∑τ = I α



Solution :

In this example, it is assumed that the clockwise direction is the negative direction

∑τ = I α (8.20)

=(35.0 kg.m2) (- 4.20 rad/s2)

∑τ = - 147 kg.m2/s2 = -147 N.m

Therefore, the resultant torque acting on this object must be 147 N.m in the clockwise

direction.

EXAMPLE 8.36

A resultant torque of 2500.dyne.cm in the counterclockwise direction is acting on an

object whose angular acceleration is 865 mr/s2 in the counterclockwise direction. Find the

moment of inertia of this object about its axis of rotation.

Solution :

The counterclockwise direction is assumed to be the positive direction. The angular

acceleration must first changed to rad/s2

α = ( + 0.865 r/s2) (2π rad/1 r)

α = + 5.43 rad/s2



Now, the moment of inertia is computed

∑τ = I α (8.20)

(+ 2500 dyne.cm ) = I (+ 5.43 rad/s2)

I = (+ 2500 dyne.cm ) / (+ 5.43 rad/s2)

I = 460 dyne.cm.s2

Since 1 dyne = 1 g.cm/s2

EXAMPLE 8.37

A resultant torque of 18.0 N in the clockwise direction acts on a wheel whose moment of

inertia is 4.50 kg.m2 about its axis of rotation. Find the angular acceleration of this

whel(in r/min2)

Solution :

The clockwise direction is chosen to be the negative (-) direction.

∑τ = Iα

α = ∑τ

I

= (-18.0 N.m)

(4.50 kg.m2) (8.20)

α = - 4.00 rad / s2



Changing these α units to r/min2, we get

α = (- 4.00 rad / s2 ) (I r/2π rad ) (60 s/1 min)2

α = -2290 r/min2

EXAMPLE 8.38

Two Torques are acting on a wheel whose moment of inertia is 8.00 x 10-4 kg.m2 about

it’s axis of rotation. One of the torques is 6.00 N.m in the clockwise direction. The wheel

is known to have an angular acceleration of 5.80 rad/s2 in the clock wise direction. Findthe second torque acting on this subject.

Solution :

The clocwise direction is assumed as the negative (-) direction.

∑τ = I

α (8.20)

Since two torques are acting,

τ1 - τ2 = I α

τ2 = I α - τ1

= (8.00 x 10-4 kg.m2) (5.80 rad/s2) – (-6.00 X 10-3 N.m)

τ2 = +1.36 mN.m



8.7 ANGULAR IMPULSE AND MOMENTUM

The symbol that is frequenly used for angular impulse is J θ. The definition of angular

impulse for a rotation in two dimensions with a constant torque acting on an object is

J θ = τ Δt

Where J θ is the angular impulse acting on the system, τ is the torque acting on the system,

and Δt is the time interval that this torque acts. Notice that angular impulse is a vector,

the means, for rotation in two dimensions, a positive (+) or negative (-) sign must be placed in front of it, depending or whether it is acting is acting in the counterclockwise

(CCW) or clockwise (CW) direction, respectively. The sign of the angular impulse must

be the same as the torque that causes it.

EXAMPLE 8.39

A (CW) torque of – 3.30 N.m acts for 4.60 s on an object. Find the angular impulse

delivered to this object by this torque.

Solution :

J θ = τ Δt

J θ = ( – 3.30 N.m ) (4.60 s)

J θ = - 15.2 N.m.s

Notice in Example 8.39 the units of angular impulse: N.m.s the SI system, dyne.cm.s in

the CGS system, and lb.ft.s in the english system. Units of angular impulse are not thesame as the unit of translational impulse first mentioned in chapter 7. Therefore, angular

impulse is a totally different quantity from tranlational impulse. These two quantities

cannot be added to or subtracted from one another since their units are different and they

are not same type of quantity.



EXAMPLE 8.40

A torque of 286 mN.m acts on an object for 865 ms in the counterclockwise directin. Find

the angular impulse exerted on the object by this torque.

Solution :

The counterclockwise direction is considered the positive direction.

J θ = τ Δt

=( + 0.286 N.m )(0.865 s)

=+ 0.247 N.m.s, or

J θ = 0.247 N.m.s in the counterclockwise direction

The symbol for angular momentum is Pθ. Angular momentum is a property of an object at

an instant and is defined by

Pθ=Iω

I in Equation 8.22 is the moment of inertia of the object at the instant the angular

momentum is being computed andω

is the instantaneous angular velocity of the object atthe instant that the angular momentum is being computed.

Angular momentum is a vector. This means (in two-dimensional rotation) that a positive

(+) or negative (-) sign must be placed in front of it, depending on whether the object is

rotating in a counterclockwise or clockwise direction, respectively.

The angular velocity ω must be in terms of radians in Equation 8.22.



EXAMPLE 8.41

An object whose moment of inertia is 360. mg.m2 has a a – 1.50 r/min angular velocity.

Find the angular momentum of this object.

Solution :

First, the angular celocity is changed to rad/s.

ω = (- 1.50 r/min) (2π rad/r) (1 min/60 s)

ω = - 0.157 rad/s

Now, the angular momentum is computed.

Pθ = Iω (8.22)

= (360.mg.m2) (- 0.157 rad/s)

` = - 56.5 mg.m2/s

Pθ = -5.65 x 10-5 kg.m2/s

The kg.m2/s is the unit of angular momentum and the gr.cm2/s is the unit in the

CGS system. These are not the same units of translational momentum first discussed in

Chapter 7. Therefore, angular momentum and translational momentum are two differentquantities and cannot be added to or subtracted from each other.

If you recall that 1 dyne = 1 g.cm/s2, 1 lb = 1 slug.ft/s2, and 1 N = 1 kg.m/s2,

another interesting fact about the units of angular momentum can be shown. The units of

angular impulse are the same as the units of angular momentum when the appropriate

subtitution is made.

1 N.m.s = 1 (kg.m.s2).m.s = 1 kg.m2/s

1 dyne = 1 (g.cm/s2).cm.s = 1 g.cm2/s

1 lb.ft.s = 1 (slug.ft/s2).ft.s = 1 slug.ft/s2

Therefore, in Example 8.41, the answer was stated as Pθ = -5.65 x 10-5 kg.m2/s, but it also

could have been stated as Pθ = -5.65 x 10-5 N.m.s just as correctly.



EXAMPLE 8.42

An object has a comment of inertia of 2.60 kg.m2 when its instantaneous angular velocity

is 300 r/min in a counterclockwise direction. Find the angular momentum of this object.

Solution :

ω = (+ 300 r/min) (2π rad/1 r) (1 min/60 s) = + 31.4 rad/s

Now angular momentum is computed

Pθ = Iω (8.22)

=(2.60 kg.m2) (+ 31.4 rad/s)

= + 81.6 kg.m2/s, or

= 81.6 kg.m2/s in the counterclockwise direction, or

Pθ = 81.6 N.m.s in the counterclockwise direction

It is occasionaly interesting to compute the change in the angular momentum of an

object. The change in the angular momentum is given by

ΔPθ = Pθf - Pθi (8.23)

Often, however, the object being considered is not a rigid body and is therefore capable

of changing its shape. When its shape changes, its moment of inertia also changes. For

this reason, the previous equation is rewritten as

ΔPθ = If ωf – Ii ωi (8.23)

EXAMPLE 8.43



The rubber tire on an automobile initially has an angular velocity of +20.0 r/min and a

2.85 kg.m2 moment of inertia. As it increases its angular speed, it stretches to a larger size

until (when its final angular velocity is + 65.0 r/min) it has a 3.10 kg.m2 moment of

inertia. Find its change in angular momentum.

Solution :

First the angular velocity are converted to rad/s

ωi = (+ 20.0 r/min) (2π rad/r)(1 min/60 s)= +2.09 rad/s

ω f = (+ 65.0 r/min) (2π rad/r)(1 min/60 s)= +6.81 rad/s

Now,the change in angular momentum is computed

Δ P Ɵ= I f ω f - I i ωi

= ( 3.10 kg.m2) (+6.81 rad/s) – (2.85 kg.m2) (+ 2.09 rad/s)

=(+ 21.1 kg.m2/s) – (+ 5.96 kg.m2/s)

=+ 15.1 kg.m2/s atau

Δ P Ɵ= + 15.1 N.m.s

8.8 CONSERVATION OF ANGULAR MOMENTUM

As is the case beetween translational impulse and momentum, there is a relationship

between angular impulse and angular momentum. That relationship is

J θ = Δ PƟ

Equation 8.24 simply states that the resultant angular impulse acting on an object equals

the change in the angular momentum of the object.

EXAMPLE 8.44



A tire has an initial angular velocity of + 18.0 r/min and a 4.30 kg.m2 initial moment of

inertia. It speeds up for 5.00 s until its final angular velocity is +200. r/min when it has a

4.70 kg.m2 moment of inertia . Find the constant resultant torque that is acting on this

wheel.

Solution :

First, the angular velocities are converted to rad/s.


ω f = (+ 200 r/min) (2π rad/r)(1 min/60 s)= +20.9 rad/s

The relationship between angular impulse and angular momentum is

J θ = Δ PƟ (8.24)

and subtituting 8.21 and 8.22 into it, we get

τ (Δt) = I f ω f - I i ωi

τ = I f ω f - I i ωi

Δt

= (4.70 kg.m2)(+ 20.9 rad/s) – (430 kg.m2)(+ 1.88 rad/s)

(5.00s)

= + 98.2 kg.m2/s – 8.08 kg.m2/s

(5.00 s)

= (+ 90.1 kg.m2/s)

(5.00s)

τ = + 18.0 kg.m2/s2 = + 18.0 N.m

EXAMPLE 8.45



A solid grindstone has a constant torque of +30.0 N.m acting on it. For 2.50 s, it speeds

up from a + 4.00 r/min initial angular velocity to a +9.00 r/min finally angular velocity.

Find the constant moment of inertia of this grindstone about its axis of rotation.

Solution :

First, the angular velocities are converted to rad/s.


ωf = (+ 9.00 r/min) (2π rad/r)(1 min/60 s)= +0.942 rad/s

Jθ = Δ PƟ (8.24)

Subtituting equation as 8.21 and 8.22 into this, we get


Since the moment of inertia does not change,


(+ 30.0 Nm)(2.50 s) = I (0.942 rad/s) – I (0.419 rad/s)

75.0 N.m.s = I (0.523 rad/s)

I = 143 kg.m2

If no outside resultant torque is acting on an object ( or on a system of object ), the

angular impulse delivered is zero. Equation 8.24 then states that the angular momentum

of this object or sytem of objects does not change, or

Δ P Ɵ= 0

P f Ɵ - P iƟ = 0

P f Ɵ = P iƟ (8.25)

Equation 8.25 is only valid if no outside resultant torque is acting on the object or system

of objects.

EXAMPLE 8.46



An ice skater has a moment of inertia about this central vertical axis of 20.0 kg.m2 when

his arms are extended and 6.00 kg.m2 when his arms are held against his body. He is

initially rotating at + 80.0 r/min when he begin as a spin with his arms extended. Find his

angular velocity when he brings his arms in against his body. Neglect all frinction forces.

Solution:

Since no outside resultant torques are acting on the skater angular momentum is

conserved

P f Ɵ = P iƟ (8.25)

I f ω f = I i ωi

ω f = I i ωi

I f

= (20.0 kg.m2 ) ( + 80.0 r/min) (2π rad /r)(1 min/60 s)

(6.00 kg.m2)

ω f = (+ 267 x 2π/60) rad/s

Now, this final angular velocity can be converted back to r/minωf =[ (+ 267 x 2π/60) rad/s] (1 r/ 2π rad) (60 s/1 min)

ωf = + 267 r/min

Notice in example 8.46 how the unit where converted back and forth. We never

the units of angular velocity are given in other than rad/s, one should convert it to rad/s to

avoid errors. But in other to avoid complicated arithmetic operations, do not actually

multiply by the conversion factors separately. Instead, suibtitute the angular velocity in to

equation 8.25 as a product of several conversion factors. Quite often, the convertion

factors will cancel out before the final answer is computed as happened in the solution to

example 8.46.

8.9 VECTORS IN ROTATION



Many of the quantities used in discussing angular motion may be treated as

vectors. However, in order to more fully appreciate their usefulness, we must first

understand some other operations that are performed with vectors. One such operation is

multiplication.

In chapter 3, you learned various ways in which vectors can be added or

subtracted but no mention was made of the operation of multiplication. In regular

algebra, multiplication between two scalars can be shown by any one of a number of

symbols. For example, the multiplication of the scalar A and scalar B can be indicated by

A.B, A x B, or AB. (Note that these scalar symbols are shown in italics using standard

typeface)

In vector multiplication, there are several different multiplication methods. Each

method not only uses a different symbols but actually is a different operation. Thedifferent multiplication method arrive at different answers.

For example, if two vectors are multiplied according to the dot-product rules, as

symbolized by A . B, they yield a certain answer. If the same two vectors are multiplied

according to the cross-product rules, as symbolized by A x B, an entirely different answer

is arrived at. These two methods of vector multiplication, the dot product and the cross

product, are discussed in this section. (Note that vector symbols are shown in boldface

italic

type)When two vectors are multiplied using the dot product, the answer is a scalar . To

discuss the rules involved in the dot product, we have to imagine two vectors that are to

be multiplied together. It should be possible to imagine redrawing these two vectors such

that their tails are butted together. Two such vectors, A and B, are shown in figure 8.12.

The angle Ɵ in Figure 8.12 is the smallest angle thet can be drawn between the two

vectors A and B. Note that 00 ≤ ≤ 180Ɵ0. The dot product of two vectors defined as

Where A is the (absolute) magnitude of A , B is the (absolute) magnitude of B, and Ɵ is

the angle between them

EXAMPLE 8.47

A . B = AB cos Ɵ



A is 40.0 m, 35.00 south of east and B is 60.0 m, 10.00 west of north. Find A . B

Solution:

Drawing the two vectors, we obtain a figure very similar to figure 8.12 in which Ɵ is

135o;, therefore

A . B = AB cos Ɵ (8.26)

= (40.0 m)(60.0 m) cos 135.00

= (40.0 m)(60.0 m)(-0.707)

A . B = - 1700 m2

Notice in Example 8.47 that the dot product can be a negative number (scalars can

be either positive ir negative), depending on the size of the angle Ɵ , but also notice that

it does not have a direction associated with it; it has only a magnitude. Of course, unitsmust be carried through and appear in the dot product.

One other important item to notice is that the dot product of two vectors is

commutative; that is, A . B = B . A

B Ɵ

A

Several quantities in physics are defined using the dot product. One such quantity

that has already been studied is work that was defined (chapter 6) as

W = FX cos Ɵ (6.1)

Work can now be redefined as



(8.27)

Where F is the force exerted on an object and X is the displacement of the object.

EXAMPLE 8.48

An object is displaced 2.00 m while a constant force of magnitude 28.0 N acts on

it. If the force maintains an angle of 30.00 with the displacement of the object, compute

the work done by this force.

Solution :

(Refer to example 6.1) W = F . X = FD cos Ɵ

= (28.0 N) (2.00 m) cos 300

= (28.0 N) (2.00 m)(0.866)

W = 48.5 J

When two vectors are mulitiplied using the cross product , the answer is a vector .

Therefore, in order to find the cross product of two vectors, we have to compute twothings; the direction and magnitude of the cross-product vector.

We will learn first how to compute the magnitude of the cross product. The two

vectors always must be drawn so that their tails are together (as A and B have been drawn

in Figure 8.12). The angle Ɵ ia again the smallest angle between the two vectors. The

magkitude of the cross product is defined as

(8.28)

Where A and B are the magnitudes of A and B, respectively, and │ A x B│is the

magnitude of the cross-product vector. You may have recognized the two vertical bars

from your math studies as meaning “ the absolute value” in algebra. When two vertical

bars are used with quantity, they mean something else, namely, the magnitude of the

vector.

W = F . X

│A x B│= AB sin Ɵ



The direction of the cross-product vector is an easy idea to grasp if you think in

three dimensions and remember to use your right hand. Whenever two vectors are drawn

so that their tails are together, the two vectors then lie in the same single plane.* The

cross product of these two vectors are always in a direction perpendicular to this plane

(and therefore perpendicular to each of the two vectors). A and B together with a segment

of the plane in which they lie are shown in Figure 8.13. Also notice in Figure 8.13 that

the cross-product vector A x B has been drawn perpendicular to the plane. The only

Problem that has not yet been explained here is how the cross product in Figure 8.13 was

chosen to come up and out of the plane instead of down and into the plane. After all, both

of these directions would be perpendicular to the plane. This is where your right hand is

useful.

Since you already know that the cross-product vector is directed in one of twodirections, the right-hand rule is used to determine which of these directions is the correct

one (and therefore which one is incorrect). After placing the tails of the two vectors

together, choose one of the two possible directions that the cross-product vector could

take and orient your right hand so that the thumb is pointing in this direction (see Figure

8.14). If the fingers are curling from the first vector to the second, through the smallest

angle , the thumb is pointed in the direction of the cross product. If he fingers are not

directed from the first vector to the second, the cross product vector is in the oppositedirection to the direction of thumb.

Therefore, the cross product of two vectors can be computed using Equation 8.28

to obtain the magnitude and the right-hand rule to acquire the direction.

Notice that the cross product of two vectors is not commutative. That is, A x B does not

equal B x A since their directions are opposite.

To discuss the cross product correctly, we have to think in three dimensions. It is

sometimes helpful to have standard symbols to represent vectors either pointed into or

out of a page. Figure 8.15a shows a vector directed out of the page. The central dot is

meant to represent the point of an arrow coming at you. Figure 8.15b is a vector directed

into the page. Here, the crossed lines are supposed to be the tail feathers (on the back of

an arrow) as it travels away from you.



One example of a cross product is a torque, which was introduced in Chapters 4.3 Recall

that torque was defined there as

(4.3)*

*Two other definitions of torque were also given in chapter 4, but since they are all

equivalent, only this one is discussed here.

Where F is the magnitude of the force, D is the magnitude of the displacement vector

drawn from the axis of rotation to the force, and θ is the angle between F and B. please

recall that a sign convention [(+) or (-)] was also used. The torque can now be redefined

as

(8.29)

Where D is the displacement vector drawn from the axis of rotation to the point of

application of the force F that is causing the torque. Note that torque actually is vector

and has both a magnitude and a direction.

EXAMPLE 8.49

The board in Figure 8.16a is balanced horizontally when a weight W1 is placed on its left

edge ang a weight W2 is at its right edge. If W1= 50.0 N and D1 is 4.00 m, find the torque

caused by this force about the fulcrum.

Solution:

; therefore, the magnitude of the torque is



The direction can be found by first positioning the tails of the D1 and W1 vectors together,

as in Figure 8.16b, and applying the right-hand rule. The direction in this case is out of

page. Therefore, we can finally say

; out of the page

Other variables in rotational physics may also be considered to be vectors. To start, an

angle itself may be considered a vector. The magnitude of such an angular displacement

vector is measured in terms of one of the units of angular measure (e.g., rad or degrees).

The direction of the angular displacement vector is given by a slightly different version

of the right-hand rule. If the fingers of the right hand are oriented in such a way that they

curl in the direction in which the object has rotated, the thumb is pointing in the directionof the angular displacement vector. Try it yourself with the angle shown in Figure 8.17.

Angular velocity is also a vector. To the find the direction of an angular velocity vector,

curl the fingers of your right hand in the direction in which the object is actually rotating.

The thumb points in the direction of the angular velocity vector. Try it with Figure 8.18

to verify the direction of ω.

Angular acceleration is a vector also, but to find its direction, one must go back to thedefinition of angular acceleration, Equation 8.2, where

Therefore, to find the direction of an angular acceleration vector, we must subtract the

initial angular velocity vector from the final angular velocity vector to determine the

change in angular velocity. The direction of this change angular velocity is the same asthe direction of the angular acceleration vector.



Quite a few other angular quantities may also be considered vector quantities. Angular

momentum Pθ is a vector that has the same direction as the direction of the angular

velocity of the object since.

(8.22)

Several equations are written as vector equations; for example, the resultant torque acting

on an object is

(8.20)

Which states that the resultant torque is in the same direction as the angular acceleration.

Combining this with Equation 8.2, we get

But since Iω is angular momentum, this could be written as

(8.30)

Notice that the torque is in the same direction as the change in the angular momentum of

the abject.

8.10 THE TOP AND THE GYROSCOPE

Consider the device shown in Figure 8.19. It consist of a rotating disk of a fairly high

moment of inertia supported by a light but rigid axle that runs from point A to point B. As

it is shown, the right- hand rule would tell us that the initial angular momentum of this

device, P θi , is to the right.

While the top was being started, forces were probably being exerted at several points

along the axis to keep it in its present horizontal position. However, we assume that only



the upward force F and the weight of the disk itself are acting on the system. The

question is: What will happen to the device; that is, will is fall? Will it remain where it is?

If you attempt to compute the resultant torque acting on the system about point A, you

will first find that the only torque about this point will be the torque due to the weight of

the disk itself. This torque could be computed using the methods givens in Section 8.9, in

which

The D and W vectors have been drawn in Figure 8.20 and the resultant torque acting on

the top about point A is shown to be directed into the page.

Now recall that

This could be rewritten as

τ . ΔƩ t = Δ PƟ

Equation 8.24 tells us that the resultant torque acting on an object is in the same direction

as the change in the angular momentum of the object. Therefore, the change in angular

momentum of the top that was first shown in Figure 8.19 is directed into the page. But,

again, how does this affect the top? The new angular momentum must be equal to the old

angular momentum plus the change in angular momentum,or

P f Ɵ = P i Ɵ + Δ P Ɵ

Since these vectors, vector addition must be used, However, since Δ P Ɵ is directed into

the page and P i Ɵ is directed to the right. It is difficult to the draw the vector diagram from

the same view point as the previous figures. Therefore, the diagram showing the addition



of the angular momentum vectors has been drawn from above the device (looking down)

as in Figure 8.21

Picture 8.19.20.21

Notice in Figure 8.21 that the final angular momentum vector is turned slightly toward

the back. The right-hand rule then tells us that this means the axis of rotation of the top

has itself rotated or precessed and the top has not fallen down. Tops do indeed precess

like this. Remember that the precession phenomenon is predicteble from the vector nature

of angular quantities.

Now, we find the rate at which the top preccesses, sometimes called the rate of

precession. The symbol that is commonly used for the rate of precession is Ω, anuppercase Greek omega.

If you recall that in Figure 8.21 the change in angular momentum Δ P Ɵ took place over a

small time interval Δt and the angle through which the top precessed is shown as ΔØ ,

the rate of precession of the top is given by

Ω = ΔØ

Δt

In the absence of friction, the final angular momentum has the same magnitude as the

initial angular momentum. Since the solid disk itself keeps rotating about its axis at a

constant angular velocity ω, we could call this P ϴ, or

Therefore, the triangle shown in Figure 8.21 has equal legs and we approximate the

relationship between P ,ϴ △ P ϴ, and △ø using

(8.7)



(8.33)

Combining Equations 8.30, 8.32, and 8.33, we get

(8.34)

This is the basic equation for computing the rate of precession of a top. The only problem

with this equation is that all angular quantities must be in radians (as is the case with

Equation 8.7).

EXAMPLE 8.50

A top is constructed as in Figure 8.19 with the disk having 0.0500 kg*m2 moment of

inertia. The weight of the disk is 8.00 N and its center of gravity is 60.0 mm from the

support point. If the disk is rotating about is horizontal axis at 300 r/min, at what rate will

the top precess? Assume friction forces and the weight of the axle are negligible.

Solution:

The angular velocity of the disk will first be changed to rad/s as follows:



An interesting point to notice is that as the angular velocity ω of the top increases, the

magnitude of the angular momentum P ϴ of the top also increases. Therefore, the rate of

precession of the top Ω decreases.

Bearings are sometimes used to mount a top in such a manner that the top’s center of

gravity always remains in the same position. Such a device is called a gyroscope. The

rotor in gyroscope is often brought to speed by means of a small motor mounted with the

gyroscope. When the rotor is spinning, the gyroscope does not precess unless an outside

torque acts on it.

Gyroscope are frequently mounted on gimbal rings as shown in Figure 8.22. If the

friction in the mounting bearings is negligible, the gimbals are not capable of exertingany net torque on the rotor (with the exception of the case in which all gimbal rings are

oriented exactly in the same plane). Such a gyroscope then maintains the same (original)

axis of rotation (it does not precess). This type of device has proved to be quite useful in

the area of navigation.

Tugas Fisika Yakin Full

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