LN-10

10-1

Note 10 Rotational Motion ISections Covered in the Text: Chapter 13

Motion is classified as being of one of three types:translational, rotational or vibrational. Translationalmotion is the motion executed by the center of mass ofan object modelled as a particle (Notes 03 and 06). Themotion of a baseball hit in a line drive is largely trans-lational. But a baseball can also rotate, so the motionof a baseball can in general possess both translationaland rotational components. A baseball does notusually vibrate. We shall study vibrational motion inNote 12.

At first thought the motion of a baseball might seemimpossibly complicated to describe mathematically.Physics shows, however, that the motion of any objectcan be separated into its translational and rotationalcomponents and those components solved forseparately. This is possible because translational androtational components of the motion of a rigid bodydo not interact with one another.1

We have already studied a particularly simple formof rotational motion in Note 05: a particle rotatingabout a point external to it. This motion is calledcircular motion. Here we survey a few aspects of therotational motion of a rigid extended body rotatingabout a point internal to itself.

We begin by laying down the kinematics of a rota-ting body as we did for a body in translationalmotion. We derive the kinematic equations of rota-tional motion and deduce the relationships betweenrotational and translational quantities. We introducethe construct of torque.

Angular Speed and Angular AccelerationFor convenience we consider a rigid extended objectwhose mass is confined to a plane and whose trans-lational motion is zero (Figure 10-1). We suppose thisobject is rotating in a counterclockwise directionabout an axis perpendicular to its plane passingthrough a point O. In the coordinate system of thefigure, this axis can be thought of as the z-axis.

We assume that the object is a rigid, extended body.By this we mean it cannot be modelled as a singleparticle. It can, however, be modelled (approximately)as a collection of particles whose positions are fixedrelative to one another. A CD is a rigid body whereasa soap bubble is not. 1 But the body must be rigid. Examples of non-rigid bodieswhose modes of translation and rotation interact are typicallystudied in a higher-level course in classical mechanics.

Figure 10-1. A particle P in a rigid body is located withrespect to O by the polar coordinates (r,θ).

Any point P in the object can be located relative to thepoint O with polar coordinates (r,θ). As the objectrotates, P follows a circle of radius r. Every other pointin the object also follows a circular path, but a pathwith a different radius. Let us suppose that in someelapsed time ∆t, P moves from a position on thepositive x-axis to the point where it is shown in thefigure. Then the subtended angle θ is called theangular displacement of P measured relative to thepositive x-axis. Because the object is rigid, the angulardisplacement of every particle in the object is the sameas the angular displacement of P.

Angular displacement is measured in the dimen-sionless unit called radian (abbreviated rad) and in thefollowing way. The arc length of the circle on which Pmoves (the distance travelled by P) is related to theradius r and angle θ by

s = rθ , …[10-1]

and therefore θ =sr

. …[10-2]

(rad). If s = r then θ = 1 radian. If the object executesone complete revolution, then the angular displace-ment of P is 2π radians. 2π radians is equivalent to360˚. If the object executes two revolutions, then itsangular displacement is 4π radians. By its nature,angular displacement is a cumulative quantity.

Note 10

10-2

We now have the tools we need to define the rota-tional equivalents of the translational quantities wedefined in Note 03. We begin by making Figure 10-1more general (Figure 10-2) by supposing that in someelapsed time ∆t = tf – ti any arbitrary point P in theobject moves from a position i to a position f. Thesepositions are shown as [A] and [B] in the figure. Thecorresponding angular positions are θ i and θ frespectively.

Figure 10-2. A more general representation of a rotatingbody than that shown in Figure 10-1. In some elapsed time∆t a particle P in the body moves counterclockwise betweentwo arbitrary angular positions.

The angular displacement of P in the interval chosen isdefined as the difference between the angularpositions that define the interval:

∆θ = θf – θi …[10-3]

(rad).2 The angular displacement divided by thecorresponding elapsed time is defined as the averageangular velocity:

ω ≡θ f –θ it f – ti

=ΔθΔt

. …[10-4]

Average angular velocity has dimension T–1 and unitsrad.s–1 (or just s–1 since rad has no dimension). Thelimit of the average angular velocity as ∆t → 0 isdefined as the instantaneous angular velocity:

2 The alert reader will notice the word displacement hereimplying a vector quantity. Angular displacement is, in fact, avector quantity, or more correctly, a pseudo-vector. This aspect ofrotational motion is somewhat advanced and best left to a secondyear course in classical mechanics.

ω ≡ limΔt→0

ΔθΔt

=dθdt

. …[10-5]

We now extend the math to allow for changes in theinstantaneous angular velocity.3 If the instantaneousangular velocity is changing, then the object is bydefinition, undergoing an angular acceleration. Letthe instantaneous angular velocity of the point P atpositions i and f be ω i and ωf respectively. The changein the instantaneous angular velocity divided by thecorresponding elapsed time is defined as the averageangular acceleration:

α ≡ω f – ω i

tf – ti=ΔωΔt

. …[10-6]

Average angular acceleration has dimension T–2 andunits s–2. The limit of the average angular accelerationas ∆t → 0 is defined as the instantaneous angularacceleration:

α ≡ limΔt→ 0

ΔωΔt

=dωdt

. …[10-7]

ω and α are, in fact, vector quantities (pseudo-vectors)whose complete vector nature is beyond the scope ofthese notes to describe adequately.4 But you can findthe direction of the vectors with the help of the right-hand rule as was introduced for the vector crossproduct (Figure 10-3).

Figure 10-3. How to use the right hand rule to find thedirection of the vector ω of a rotating body. 3 Of course, in order for the object’s angular velocity to change,the object must be subject to a force applied to it in a special way.For the moment we ignore this force, as we did in Notes 02 and 03,and stay within the area of kinematics.4 We leave this description to a higher-level course in classicalmechanics.

Note 10

10-3

To find the direction of the ω vector, extend your righthand, curl your fingers as if you are to grip somethingand extend your thumb. Now curl your fingers in thedirection of the angular displacement of the object(the direction the object is rotating). Then your thumbpoints in the direction of the ω vector. By convention,the ω vector is placed on a diagram along the body’saxis of rotation. Later in this note we shall see otheruses of the right hand rule.

Rotational KinematicsAs implied in the previous section, a set of kinematicequations exist for rotational motion just as they dofor translational motion. They have a similar form andare derived in a similar fashion. We shall thereforejust list them (Table 10-1).

Table 10-1. Comparison of translational and rotational kine-matic equations.

Translational Motion Rotational Motionv f = vi + at ω f = ωi +αt

x f = xi + vit +12at2 θ f = θi +ωit +

12αt 2

x f = xi +12(vi + v f )t θ f = θi +

12(ω i +ω f )t

v f2 = vi

2 + 2a(x f – xi ) ω f2 = ωi

2 + 2α(θ f – θi )

Problems in rotational kinematics can be solved muchlike problems in translational kinematics. Assumingyou have memorized the translational equations andknow the rotational equivalents, you can easilyreconstruct the rotational equations. Let us consideran example.

Example Problem 10-1A Problem in Rotational Kinematics

Starting from rest a wheel is rotated with a constantangular acceleration of 2.00 rad.s–2 for 5.00 s. (a) Whatis the final angular speed of the wheel? (b) How manycomplete revolutions does the wheel execute in theelapsed time of 5.00 s?

Solution:(a) Using the first equation in Table 10-1 we have forthe final angular speed

ω f =ω i +αt = 0 + (2.00s–2 )(5.00s)

= 10.0 s–1.

The final angular speed is 10.0 s–1 or 10.0 rad.s–1.(b) Using the second equation in Table 10-1 we havefor the angular displacement

θ f =θ i +12(ω i +ω f )t

= 0 + 12(0 +10.0rad.s –1)(5.00s)

= 25.0 radians.

The number n of revolutions is this number dividedby the number of radians per revolution (i.e., 2π):

€

n =25.0(rad)

2π radrev

= 3.98 rev.

Thus nearly 4 revolutions are required for the wheelto accelerate to the final angular speed of 10.0 s–1.

Relations exist between the angular and tangentialspeeds of a particle in a rotating rigid body, and bet-ween the angular and tangential accelerations. Sincethese relations are useful in solving rotational motionproblems we consider them next.

Relations Between Rotational andTranslational Variables

Suppose that a rigid extended body rotates about anaxis that passes through an internal point O as shownin Figure 10-4. Consider a point P in this body.

Figure 10-4. A particle P in a rotating rigid body.

Note 10

10-4

The magnitude of the tangential velocity of P is, fromeq[10-1],

v = dsdt= r dθ

dt,

since r is constant. Thus using the definition of ω ineq[10-5]

v = rω . …[10-8]

The tangential acceleration of P is, using eq[10-8],

at =dvdt

= r dωdt

,

again since r is constant, or using the definition of α ineq[10-7],

at = rα . …[10-9]

Now since P is moving in a circle it is undergoing acentripetal acceleration. The magnitude of thiscentripetal or radial component of the acceleration is

ac =v2

r= rωz

2 .

using eq[10-8].The relationship between tangential and radial com-

ponents of the acceleration of P can be seen with thehelp of Figure 10-5. The total acceleration of P is thesum of the at and ar vectors. The complete vectornature of at and ac is beyond the scope of these notesto describe. The relationships between the magnitudesof these quantities are summarized in Table 10-2.

Figure 10-5. The resultant acceleration of any particle P in arotating rigid body is the vector sum of the tangential andradial acceleration vectors.

Table 10-2. Relationships between the magnitudes of trans-lational and rotational variables.

Translational Rotational Relationshipx θ x =θrv ω v =ωra α a = αr =ω 2r

Let us consider an example.

Example Problem 10-2Tangential and Angular Speeds

A bicycle wheel of diameter 1.00 m spins freely on itsaxis at an angular speed of 2.00 rad.s–1. (a) What is thetangential speed of a point on the rim of the wheel?(b) What is the tangential speed of a point halfwaybetween the axis and the rim?

Solution:(a) Using eq[10-8] the tangential speed of a point onthe rim of the wheel is

v = rω =1.00m2

(2.00rad.s

–1) = 1.00 m.s–1.

(b) A point halfway between axis and rim will have atangential speed one half of this value, or

v = 0.50 m.s–1.

Clearly, the further a point on the wheel is from theaxis of rotation the greater is its tangential speed.Though the two points have different tangentialspeeds they have the same angular speed.

We stated in Note 07 without proof that the centre ofmass of a rigid body can be taken to be the body’sgeometric centre. We are now ready to extend the ideaof the centre of mass to a system of discrete bodies,and to define the centre of mass in a propermathematical fashion.

Note 10

10-5

Center of Mass RevisitedWe have all seen pictures of a body rotating in theweightless environment of a space capsule. If thebody is a flat plate (Figure 10-6) we would see that apoint in the body doesn’t rotate at all. That point is thebody’s centre of mass. If we model the body as acollection of particles in a coordinate system (Figure10-6b) then the coordinates of the centre of mass canbe shown to be

Figure 10-6. An unconstrained body rotating freely in spacerotates about its centre of mass.

€

xcm =1M

mixii∑

…[10-10]

€

ycm =1M

miyii∑ .

where M = m1 + m2 + … is the body’s total mass. In amoment we shall put this definition into more generalform. (In the event that the body has an appreciablethickness then we would have to add an equation in zto eqs[10-10].)

Let us consider an example.

Example Problem 10-3Finding the Centre of Mass of a System of Particles

A 0.50 kg ball and a 2.00 kg ball are connected by amassless rod of length 0.50 m. Where is the centre ofmass of the system located relative to the 2.00 kg ball?

Solution:The system is shown in Figure 10-7. We model thetwo balls as particles. For convenience we put thelarge ball at the origin. Using eqs[10-10] (and takingycm = 0) we have

Figure 10-7. Finding the centre of mass of a system.

€

xcm =1M

mixii∑ =

m1x1 + m2x2m1 + m2

€

=(2.0 kg)(0.0 m) + (0.50 kg)(0.50 m)

3.0 kg + 0.50 kg

= 0.10 m

The centre of mass of the system lies between the twoballs and 0.10 m from the 2.0 kg ball. The centre ofmass of a system of particles lies at its “weighted”centre.

A solid rigid body does not consist of a collection ofdiscrete particles, but rather of a continuousdistribution of mass. We must therefore generalizeeqs[10-10] for a continuous distribution. To do this weimagine the body divided up into many small cells ofboxes, each with the small mass ∆m (Figure 10-8).Eqs[10-10] thus become

€

xcm =1M

xii∑ Δmi and ycm =

1M

yii∑ Δmi .

If we now let the number of cells become larger andlarger and the size of the cells ∆mi become smaller andsmaller these equations go over to

Note 10

10-6

Figure 10-8. The division of a solid body into cells.

€

xcm =1M

xdm and ∫ ycm =1M

ydm ∫…[10-11a and b]

Eqs[10-11] is the most general definition of the centreof mass. When evaluating eqs[10-11] you mustremember to express dm in terms of dx, dy or both andthen calculate the integrals over coordinates. Let usconsider an example.

Example Problem 10-4Finding the Centre of Mass of a Uniform Rod

Find the centre of mass of a thin, uniform rod oflength L and mass M relative to one end.

Figure 10-9. Finding the centre of mass of a long thin rod.

Solution:The rod is an example of a body with a continuousdistribution of mass. Since it is very thin we can setycm = 0 (and of course zcm = 0). We put one end of therod at the origin and imagine the length of the rod tobe divided into increments dx . Since the rod isuniform each increment has mass dm = (M/L)dx.

Eq[10-11a] becomes

€

xcm =1M

ML

xdx∫

=

1L

xdx0

L

∫ .

Evaluating the integral we get

€

xcm =1Lx 2

2

0

L

=1LL2

2− 0

=12L .

As expected, the rod’s centre of mass is locatedmidway along the rod, or at its geometric centre.

Thus far we have considered the kinematics ofrotational motion. We are now ready to broaden ourstudy to the dynamics of rotational motion.

We know that a body’s translational motion can onlybe changed by the application of a net force to thebody. By the same token, a body’s rotational motioncan only be changed by the application of a net force.But the force must be applied in a special way. Thiswarrants the definition of another tool of physicscalled torque.

TorqueWe shall see in what follows that the rotational equiv-alent of force is a construct called torque. Torque maybe thought of as the “turning action” that produces arotation in the same sense that force is the “straight-line action” that produces a translation. Torque can beunderstood with the help of Figure 10-10.

Figure 10-10. How the torque produced by a force is defined.

The figure shows a wrench whose jaws fit onto a nutcentered at the point O. We assume that the nut-bolt

Note 10

10-7

combination is a right-handed one, meaning that thenut must be rotated counterclockwise to loosen it (sothat its translational motion along the axis of thescrew is out of the plane of the page). A force musttherefore be applied to the wrench roughly as shown.

Though the ultimate cause of rotational motion is aforce, the physical “action” of loosening the nut can-not be adequately described by force alone. As anyoneknows, if you apply the force in a line through O noloosening of the nut occurs at all. Moreover, thefurther away from O you apply the force, the larger isthe turning effect for the same magnitude of force.

We shall see in what follows that this rotational“action” is best described by a new construct calledtorque. The torque about an axis O is defined to havethe magnitude

τ = Fd ,

where F is the magnitude of the force applied and d isthe perpendicular distance from O to the line alongwhich the force acts. According to Figure 10-10 we canalso write

τ ≡ rFsinφ , …[10-12]

where r is the distance from O to the point on thebody at which the force is applied and φ is thesmallest angle between the vectors r and F . Thedimension of torque is M.L2.T–2. Its units are N.m.

WARNING!Torque is not the same as Work

Torque and work have the same units, namelyN.m. But torque is not the same as work. For onething, torque is a (pseudo) vector whereas workis a scalar.

Any number of forces may act on a body at the sametime. For example, Figure 10-11 shows two forceswith magnitudes, F1 and F2, acting on the same body.The force F1 tends to produce a rotation of the bodycounterclockwise about O. For this reason the torqueproduced by this force is arbitrarily given a positivesign. The force F2 tends to produce a rotation of thebody clockwise about O. For this reason the torqueproduced by this force is given a negative sign. Theresultant torque has the magnitude

τ∑ = F1d1 – F2d2 .

According to our sign convention, if this result ispositive, then the body will tend to rotate counter-

clockwise about O. If the result is negative, then thebody will tend to rotate clockwise about O.

Thus far we have considered the magnitude oftorque. To understand the direction of torque wemust work with its vector definition. This brings us tothe major reason for introducing the vector crossproduct:

Figure 10-11. Two forces applied to a body in arbitrarydirections.

τ = r × F , …[10-13]

where r is the vector locating the point at which theforce F is applied relative to the axis of rotation. Thedirections of the three vectors in relation to oneanother can be seen with the help of Figure 10-12. Letus suppose for convenience that a force is applied to apoint P in the xy-plane of the body. The vector rlocates the point P relative to the point O about whichthe object rotates. The torque vector τ is then givenby the right hand rule. With reference to the figure,the magnitude of the torque as given by eq[10-13] canbe seen to be equivalent to eq[10-12].

We now consider an example of two forces appliedto a body.

Note 10

10-8

Figure 10-12. The relationahip between the vectors τ , r andF .

Example Problem 10-5The Net Torque on a Cylinder

A one-piece cylinder is shaped as in Figure 10-13 witha core section protruding from a larger drum. Thecylinder is free to rotate around the central (z-) axisshown in the drawing. Two ropes are wound, onearound the core, the other around the drum, and exertforces T1 and T2 on the cylinder in the directionsshown. Calculate the net torque acting on the cylinderabout the rotation axis.

Solution:The magnitude of the torque produced by T1 is –R1T1(given a negative sign because it tends to produce aclockwise rotation about the axis through O). Themagnitude of the torque produced by T2 is +R 2T2

(given a positive sign because it tends to produce acounterclockwise rotation). The magnitude of the nettorque about the rotation axis is the sum

Figure 10-13. Two torques acting on a cylinder.

τnet = τ1 +τ 2 = R2T2 – R1T1 ,

This result can be positive or negative depending onthe relative magnitudes of the forces and radii.Clearly, if the sum is zero, then the body undergoesno rotation at all.

Question. Does the body in Figure 10-13 tend to movewith translational motion? Explain.

10-9

To Be Mastered

• Definitions: angular displacement, radian, average angular velocity, average angular speed, instantaneous angularvelocity, instantaneous angular acceleration, average angular acceleration

• equations of rotational kinematics (to be memorized):

ω f = ωi +αt θ f = θi +ωit +12αt 2 θ f = θi +

12(ω i +ω f )t ω f

2 = ωi2 + 2α(θ f – θi )

• relations between the magnitudes of translational and rotational quantities

x =θr v =ωr a = αr =ω 2r

Typical Quiz/Test/Exam Questions

1. (a) Define torque(b) What are the units of torque?

2. State the following as being a vector or a scalar. Give the units of each.(a) coefficient of friction(b) potential energy(c) linear momentum(d) angular velocity

3. A body is subject to non-concurrent forces. Write down the conditions for the body to be in a state ofmechanical equilibrium. Explain the meaning of each quantity in your expressions.

4. A uniform plank of length 2.00 m and weight 100.0 N is to be balanced on a fulcrum or support point (see thefigure). A 500.0 N weight is suspended from the right end of the plank and a 200.0 N weight suspended fromthe left end. Answer the following questions.

left right

fulcrum

(a) Describe the state of the plank when it is balanced.(b) What are the conditions for this state?(c) How far from the left end of the plank must the fulcrum be placed?(d) What is the reaction force exerted by the fulcrum on the plank?