axis of rotation

7/28/2019 Axis of Rotation

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UNIT 6 MOMENT OF INERTIA ANDFIXED AXIS ROTATIONStructure

6.1 IntroductionObjectives

6.2 Angular Momentum of a Rigid Body6.2.1 Moments and Products of Inertia6.2.2 Moments of Inertia of Simple Rigid Bodies6.2.3 Axis Transfer Theorems6.2.4 Calculation of Products of Inertia

6.3 Rotation of a Rigid Body about a Fixed Axis6.3.1 Equations of Motion6.3.2 Rotational Kinetic Energy and Power6.3.3 Impulse Momentum Principle

6.4 Oscillations of Rigid Bodies6.4.1 Compound (Physical) Pendulum6.4.2 Torsional Pendulum

6.5 Summary6.6 Answers to SAQs

6.1 INTRODUCTIONIn the preceding unit, you learnt about the motion of a system of particles. Itwas found that many of the physical principles which were earlier found fordescribing the mechanics of a single particle may be generalized and used indescribing the motion of a system of particles. Towards the end of the lastunit we defined a special system of particles called rigid bodies. In a rigidbody the relative position of any two mass points of the body does not changewith time. This means that the motion of the rigid body is completelydescribed when we find : (a) the trajectory of any one particle (e.g. the centreof mass of the body) associated with the body, and (b) the relative orientationof the body in course of its motion.It was pointed out that a rigid body is capable of two distinct types ofmotion, viz., a pure translation and a rotation. You have further seen that anarbitrary motion of a rigid body can be looked upon as a translation of thecentre of mass ( or any other point in the body ) followed by a rotation of thebody about an axis passing through the centre of mass (or the pointmentioned above). In the earlier units, we have discussed in detail thetranslational motion of a particle. The motion of the centre of mass istherefore well understood by us. In this unit we will consider a particularlysimple case of rigid body rotation, viz. the rotation of the body about an axisfixed in space. In this case, if we consider any particle in the body, itsperpendicular distance from the axis of rotation remains constant in time.The trajectory of the particle is a circle and its instantaneous position isdescribed by specifying its angular position, with respect to a reference line.We will begin by a review of the single particle rotation and recall therelationship between the angular momentum of the particle and the appliedtorque. One can define the angular momentum of the rigid body in a waysimilar to the way the momentum was defined in the last unit for a system of


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Consider a set of axes (z, y, z) attached to the body with its origin located atthe centre of mass of the body (Fig. 6.2). The set of space fixed axes aretaken to be ( X , Y ,2)with its origin at 0 . The body rotates about an axisthrough the centre of mass with.an angular velocity w. Since we are,interestedin fixed axis rotation, we will take the position vector T', of the centre of masswith respect to the fixed set of axes as constant.

MomentdFIX& Axb

Figure 6.2 : Rotation of a rigid body about an arbitrary fixed a xisConsider a mass element dm of the body located at the position r'withrespect to the body fixed axes (z , y, z). The absolute velocity of the masselement is v'= 3x r'. The angular momentum d c of the mass element aboutthe centre of mass may therefore be written as r 'x (dmv'). Summing over allthe mass elements, the total angular momentum of the rigid body aboutcentre of mass is given by

The double cross product in the above expression may be expanded in thecomponent form using r' = z i + y j+ zk and 3= w,i +w,j +w,k

so thatF x ( 3 x i ) = [ w * ( ~ ~z2) - W,ZY - W,ZZ] i + [-W,YZ+wy(z2+ z2)- ,YZ] j

+ [-w,zz - w,zy +w,(z2 + y2)1kThe cartesian components of the angular momentum may therefore be writtenas

L, = -% / ydm +w, ( 2 z2)dm- q, / zdmJL, = -a, / zdm +w / zdm+w / (z2+ Z)dm (6-4)


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~otatienolDynamia We define the moments of inedia as

Note that the integrand in each case contains the square of the .distance fromthe cartesian axes and hence the moments of inertia are positive. I,,, I,, andI,, are usually written as I,, I, and I, respectively. The products of inertiaare defined through

I,, = / rd mI, = / r z d m (6.6)

Unlike the moments of inertia, the products of inertia contains the cartesiancoordinates and may therefore need not be positive. We therefore get,

You may observe that the moments and the products of inertia are definedwith respect to a set of axes fixed with the body. One could have definedthese in terms of axes fixed in space, but if we did so, these quantities wouldchange as the rigid body rotates in space. If you are familiar with matrixmultiplication you may rewrite Eqn. (6,7) above using the elementary rules ofmatrix multiplication

Calculation of moments of inertia of bodies is a task very similar to what wehave done for finding the centres of mass. In the following section we willobtain the moments of inertia of a few simple rigid bodies.6.2.2 Moments of Inertia of Simple Rigid BodiesWe will generally be concerned with bodies of uniform density p. In this caseEqn.(6.2) can be written as

where T is the distance of the volume element du from the axis. We will find itconvenient to introduce the concept of a radius of gyration k . Note that themoment of inertia has the dimension of mass times distance squared. Theradius of gyration k is defined tbrough

If we imagine the entire mass of the rotating body to be concentrated a t apoint, the angular momentum of the rotating body will be the same as that ofthis imagined mass point, if the point mass were rotating about the axis at adistance k.


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Example 1Find the moment of inertia of a uniform flat disk or a uniform solid cylinderof radius R and mass M about the axis of symmetry.Solution :A solid cylinder can be regarded as consisting of a very large number of flatdisks stacked one over another. Thus the radius of gyration of a solid cylinderand a flat disk are the same. To find the moment of inertia of a flat diskabout an axis which is perpendicular to its plane and passing through itscentre C, consider the contribution to the moment of inertia by an annularring at radius r and width dr (Fig. 6.3). The moment of inertia of this ring isr2dm, where dm, the mass element is given by

The moment of inertia of the disk therefore is

Example 2Find the moment of inertia of a uniform rectangular bar of length L, width band mass M about an axis perpendicular to its plane and passing through itscentre.Solution :Consider (Fig. 6.4) a mass element dm at (z, y). The contribution of thiselement to the moment of inertia is

-2where Lb is the total area of the bar. The total moment of inertia is foudd byintegrating over z and y. I

0Example 3Find the moment of inertia of a solid sphere of uniform density about one ofits diameters.Solution :The sphere may be built up of a number of circular disks of varying radii.Consider (Fig. 6.5) one such disk at a distance r from the centre andthicknessdr . The radius of this disk is d ( n . he moment of inertia of this disk is

Momentof InertladEh e d AxisRotation

Figure 6 .3

x .

Figure 6. 4

Figure 6.527


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Rotational Dynamics The mass element dm is given by

The total moment of inertia is obtained by integrating over

SAQ 1 :Find the moment inertia of a thin spherical shell of mass M

SAQ 2 :I Using the above result, determine the moment of inertia of

Figure 6.6

)T from -R

: and radius

a solid sphe

Example 4Find the moment of inertia of a right circular cone of base radius R andheight h.Solution :The cone can be thought of as a collection of disks of variable radii (Rg. 6.6).The radius of such a disk at a height y from the base can be written as

hThe mass of the disk is given by

The momeilt of inertia of the cone is abtained by summing over the momentsof inertia of the disks


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6.2.3 Axis Transfer Theorems MomentollneruFixedAxls RotaUonThere are two theorems which are very useful when we know the momentinertia about some axis and wish to calculate the same about some other axis.(a) The parallel axis theoremThe more important of the two theorems is the parallel axis theorem. Thistheorem relates the moment of inertia of a rigid body about an axis passingthrough its centre of mass with the moment of inertia about another axiswhich is parallel to the former. Let I, be the moment of inertia about an axispassing through the centre of mass and I be the moment of inertia throughanother axis which is parallel to it. Let d be the distance between the abovetwo parallel axes. The theorem states tha t

A rigid body can be looked upon as consisting of a large number of slicesnormal t o the set of axes mentioned above. Thus we need to prove thetheorem only for a single slice of the body (Fig. 6.7) which is normal to a pair

:allelt hro

one of which passes through the centre of mas,point D on the body.

and

--Y

the

Figure 6 .7 : A rigid body illustrating the locations ofthe parallel axes and a typical mass element

Let us consider the plane of the body to be the-x-y plane. If we choose D asthe origin of the coordinate system, the moment of inertia of the body aboutthe axis through D is

ID = J dm(z2+ y2)Let the coordinate of the centre of mass C be (x,, y,). If we shift the origin ofthe coordinate system to the point C while keeping the axes parallel, thecoordinates of a mass element dm with respect to the new origin will be(x', y'), where


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RoQUondDynrnla In terms of these new quantities, the moment of inertia ID may be written as

The first term is the moment of inertia about the parallel axis through thecentre of ma s . The second term is Md 2 , where d is the distance between thetwo axes. By the definition of the centre of mass the last two terms areidentically zero. This completes the proof of the parallel axis theorem. Bysuccessive application of the theorem one can find moment of inertia about-many different axes.Example 6Find the moment of inertia of a sphere about a tangent.Solution :,Any tangent to a sphere is parallel to some axis through the centre of the/sphere. Thus we may directly apply the parallel axis theorem to get

Az* xample 6Calculate the moment of inertia of a solid cylinder about an axis which isno rm 4 to the axis of the cylinder and passes through the centre of mass.Solution :We may consider the cylinder as being made up of a large number of disksstacked one over the other. Consider one such disk of thickness dx at adistance of x from the origin (Fig. 6.8). The moment of inertia of the diskabout the y-axis (which is along a diameter of the disk) is d m R 2 / 4 where themass element dm = uR2&yp.Using the parallel axis theorem, the moment ofI Figure 6.8 inertia of this disk about y, axis is

The total moment of inertia is obtained by integrating over y frm - L / 2 to L /2

SAQ 3Using the result of the above example find the moment of inert ia of a pencillike rigid rod of length L about an axis through its centre and perpendicularto its length. Use parallel axis theorem to find its moment of inertia about anaxis which is perpendicular to it at one of the ends.


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(b ) The perpendicular axis theoremThis theorem is somewhat limited in its applicability and is useful for planarobjects only. It states that t& moment of inertia of the body about an axisperpendicular to its plane is given by the sum of the moments of inertia abouttwo mutually perpendicular axis in the plane of the object which intersect theperpendicular axis. If the body is in x-y plane, the theorem requires

The proof of this theorem is a straight forward application of the Pythagorastheorem as can be seen from Fig. 6.9 which shows a laminar object ofarbitrary shape. Taking the plane of the body to be the xy-plane, themoment of inertia about the z-axis is given by

The theore-


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RotatloallDynunl~ Case 1 : If the z-axis is awaxis of symmetry I,, and I, , vanish. This caneasily be seen by referri~lg o Fig. 6.10 wIlere it is shown that for each masselement dm at (2 , y, 2) there% a symmetrically placed mass element at( -2 , -y , a) so that

integral zero.Example 8Find the product oflies in the x-y plane

I,, = 1 rodm = 0Case 2 : If the plane containing the xy axes is a plane of symmetry, there is apair of mass elements dm, one at r'a.nd the other at - z . This also makes the

inertia! makin

I,, for a slender rod of lengtg an angle a with the x axis

,h L andas show

nass min Fig.

which. 6.11.The mass element a t the position (z, ) is dm = pdr where dr is a lengthelement along the rod at a distance T from the origin. using'the polarI Figure 6.1 1 coorfinates x = r cos a, y = T sina , we have

I., = J P X Y ~ ~= lT/:2sina cos apdr

sin2a r R/2= P I [

-R/2- sin 2a L3 ML2- p = -2 12 24 sin 2a

Note that I,, = 0 if or = 0 or r/2, a result which could be obtained fromsymmetry. 0

6.3 ROTATION OF A RIGID BODY ABOUT- FIXED AXISIn the previous sections you have seen the relationship between the angularmomentum of a body and its moments of inertia. We will use these conceptsto study the rotational motion of a rigid body about an axis which is fwed inspace. Rotation of a rigid body about a fixed axis is very commonly used intechnology, for instance, in turbines, pumps and other rotating machineries.6.3.1 Equations of MotionConsider a rigid body of arbitrary shape shown in Fig. 6.12 which is rotatingabout the z-axis which is fixed in space. The x and the y axes are body fixedand these therefore rotate with the body. The components of the angularmomentum of the body about the origin are given by Eqns.(6.7). Since theangular velocity of the body has only the z component w, = w , we get

Lz = -I*,wL, = - IyzwLz = I,#


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Momentd nertla &so that the angular momentum vector may be written as FlxedAxis Rotation,f = -iIzzw - Iyzw + k lZzw (6.15)LThe torque acting on the body is obtained by differentiating the above with LJrespect to time. While differentiating, we must remember that the unit

vectors i and j rotate with the body and therefore are not constant. You mayshow, using the results obtained in Unit 5,

d ,- 8 =dt 4-j - - u iSubstituting these in Eqn.(6.15), we get Figure 6.12

d i? = - =dt a [-;I=, - Iyz + i ~ ~ ~ ]+iIYzw2- IZ,w2where a = h / d t is the angular acceleration. The components of the torquemay therefore be written as

= Iyzw2 - I%,&T, = - zZw2- YzaT, = I,,a (6.16)

The motion of the rigid body is completely described, if we supplement theequ;tion (6.16) with the three equations governing the motion of the centre ofmass

Fz = mazF, = ma ,Fz = 0

The rotation of the rigid body is described by the last of the equations (6.16).The remaining equations determine the forces of constraints which keep the zaxis fixed in space while the body is rotating.

-We have seen that for symmetric bodies I,, and I, , are zero. This givesT, = ry= 0, i.e. the torque only has t. component. This along with F, = 0tells us about the nature of reactions necessary t& maintain planar motion.The system of external forces ( loads plus reactions ) has a planar resultant.In any case, the rotational motion of the body is governed by the simpleequation T, = I, ,a. Note that this relation is very similar to the relationbetween the force and acceleration in case of linear motion. The role of inertia(mass) is deady played by the moment of inertia.Example 0A uniform sphere of mass M and radius R is spinning on an axis through itscentre at a rate of 5 rev/s. It comes to stop in 20 seconds. How large is thefrictional torque that brings the sphere to a stop.Solution :The angular deceleration a = 5 x 2x120 = n/2 rad/sa. The torque required is

T = I a2MR2r=-- 5 2= 0.2xMR2


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S A Q 5 :The wheel shown in Fig. 6.13 rotates freely on its axle and pulls a belt asindicated. The moment of inertia of the wheel is 18 kg-m2. The radius of thewheel is 0.2 m. There is appreciable tension only in the upper end of the beltwhile the tension in the lower end is negligible. When the wheel is spinning at2 rev/s, it is disconnected from its power source and coasts to rest in 15 s.What is the tension in the upper end of the belt ?

6.3.2 Rotational Kinetic Energy and PowerYou have seen that for a rigid body rotating about a fixed axis, every masselement dm traces out a circle of radius r about the axis, where T is th'edistance of the mass element from the axis. The velocity of the mass elementis therefore tangential to the circle and has a magnitude wr. The kineticenergy of the mass element is therefore given by

~ h bum of the kinetic energy of all the mass elements constituting the bodyis therefore the kinetic energy of the rigid body and is given by

where I is the moment of inertia of the body about the axis. If you comparethe above expression for the kinetic energy with the familiar expression mv2/2for the kinetic energy of translation, you will observe the similarity betweenthe moment of inertia I and the mass m. You should note that 1 o ~ / 2s notany new type of kinetic energy. This simply is the part of the kinetic energyof the system C rn;v; which may be specifically assigned to rotat ional motion.As in the case of angular momentum, it is possible to associate a part of thekinetic energy with the motion of particles about a parallel axis through thecentre of mass ( which is body-fixed ) and a second part with the rotation ofthe centre of mass itself about the axis. Consider Fig. 6.14 in which a sectionof the rigid body rotating about the fixed axis AA is shown. Let CC be anaxis passing through the centre of mass and parallel to the axis AA. Thedistance between the two axes is denoted by r, . An element dm which is at adistance T from AA is at a distance r' from the axis CC, the relationshipbetween T and T I is given by elementary trigonometry to be


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The kinetic energy may therefore be written as Moment of nertlajkFlredArls Rotation

- 1 1- -w2 J r r 2 dm+ -r:w2 M - cJ r' cos0dm2 2The last term in the above equation is J x'dm where x' is the x-coordinate ofd m with respect to the centre of mass as the origin. By the definition of thecentre of mass this is identically zero. Thus

If the kinetic energy is varying with time, we take the time derivative ofEqn.(6.17) t o obtain d K-t = Iw awhere a is the angular acceleration. Using the relationship between the torqueand the angular acceleration this can be written as

If the change in th e kinetic energy is due t o a force F , the rate of change ofkinetic energy is equal to the rate of doing work and we have

which implies that the rotational power P is given by the product of thetorque and the angular velocity. You may observe the si~pilarity f thisrelationship with the expression P = - v' for the case of translational motion.Using w = d e l d t , we can write

which gives dW = r d 8 . From this it follows that if a constant torque isapplied during an angular displacement 0 , the work done will be

Example 10A wheel of radius T having a moment of inertia I about an axis through itscentre and perpendicular to its plane-has an angular velocity w . At time t = 0a tangential force F is applied to it s rim decelerating it (Fig. 6.15 ). Find theangular distance moved by the wheel and the time taken before it is brought

, to rest. PSolution :The change in the angular momentum of the body from Iw to zero is due tothe torque F r acting on the body for a time t.Thus

Frt = Iwwhich gives t = I w l F r . The angular distance moved is found from the workenergy relationship Figure 6.151Fr0 = - I W ~2 ,which gives 8 = b 2 / 2 F r 35


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Rotational Dyna mics 6AQ 6 :A wheel with moment of inertia I = 10 kg-m2 is spinning at 2 rev./s on its' axis. How large is the frictional torque if the wheel makes 40 revolutions

before it comes to a stop ?

6.3.3 Impulse Momentum PrincipleThe work-energy theorem is useful in solving problems when kinematicalquantities like the position and velocity appear in the problems. When wehave information on quantities like momentum or angular momentum, it is farmore practicable to talk in terms of these quantities.You. have seen that when external forces act on a system of particles, thelinear momentum of the centre of mass changes. The impulse momentumprinciple states that the impulse of the external forces acting on the system isequal to the change in the momentum of the system during the time intervalin which such forces act.

where Fl and F2refer to the initial and final momenta of the centre of mass.In a very similar way we may obtain a relationship between the angularimpulse and the angular momentum by directly integrating the equationT = d z l d t ,

4d 2 i d t = ~2 - & (6.18)The left hand side of the above equation is the angular impulse in ~ a l o g ywith the definition of linear impulse j.Equation (6.18) states that the changein the angular impulse is equal to the change in the angular momentum: ,Example 11A solid cylinder is rotating about its own axis with an angular speed w,. Thecylinder is suddenly caught along the edge at BB (Fig. 6.16). Find theangular velocity of the cylinder immediately afterwards and the impulseexerted by the disk on the axis BB.Solution :If the edge BB is brought to rest, the cylinder would rotate about this edgewith an angular speed w. The velocity of the centre of mass becomes wR.Since the disk was previously rotating about an axis through the centre ofmass, the velocity of the centre of mass was zero. By the impulse momentumprinciple, the change in the momentum of the centre.of mass is equal to theimpulse J acting on the system during the time in which the edge was

Figure 6.16 arrested.J = m(w R - 0) = mvw R

The angular impulse is J R which must result in the change of angularmomentum of the disk. mR2J R = -(w, -W)2Combining these we get w = w0/3 and J = w , R / 3 .


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When no external torque acts on the system d i l d t = 0, and the angular Moment o f Inertla&Fixed Axis Potatlonmomentum is constant. This is the principleof conservation of momentum. Avery important point regarding the applicability of Eqn. (6.18) is that theangular impulse is equal to the change in the angular momentum of any body- not just of rigid bodies. If we are interested in using a specific expression for-the angular momentum, for instance , he expression L = Iw which isapplicable for a rigid body, it is only necessary that the body behaves rigidlyat the start and the end of the angular impulse. An example of this is that ofa figure skater who increases her angular speed by folding in arms. Theimpulse - momentum principle can still be used even though during theprocess of folding in the arms the skater may not be considered as rigid.Example 12A figure skater rotating freely with an angular speed w draws in her arms. In

-doing so her moment of inertia about the axis of rotation decreases by a factorof 2. Find the new angular speed and account for the change in the kineticenergy.Solution :Before the skater draws in her arms, we may treat her as a rigid body so thatthe initial angular momentum is Iw,. After she has drawn her arms in, shecan once again be considered as a rigid body. If we neglect the small frictionaltorque at the skates and the torque due to air drag, the angular momentum isconserved. If w is her new angular speed, we have

so that w = 2w0, i.e. the angular speed has doubled.Note tha t the initial kinetic energy is I 4 1 2 while the final kinetic energybecomes IW;. The kinetic energy has increased. Since no work is done by theexternal forces and torques, this must be due to the internal work done by hermuscles in pulling in the arms. The skater has lost internal energy in theprocess.

.

SAQ 7 :A rotor of radius R1 with a moment of inertia Il rotating with a constantangular speed wl is brought into contact with a second rotor of radius R2 andmoment of inertia I2which is initially at rest (Fig. 6.17). Find the commonangular speeds of the rotors and determine the time required to at tain thisspeed, if the frictional force between the rotors is F. Figure 6.17

1 Example 13Consider an Atwood's machine with a heavy pulley. The frictionles pulley(Fig 6.18) haa a mass m and a radius R . Calculate the accelerations of themasses ml and mz.Solution :The free body diagrams of the masses ml, m2 and the pulley P re shown inFig. (6.18). Notice that unlike the case of mapless pulley, the tensions in the


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RobUonalDynlarla string on two sides of the pulley are now different. If ml > rnp the equationsof motion can be written as follows :

m l g - T I = m laT2 - m2g = m2a7

1 Figure 6.18 : Atwood's Machine and the free body diagrams of the masses

For the pulley, we will write down the torque equation. The torque r on thepulley is T IR- T2R .If I is the moment of inertia of the pulley about ahorizontal axis (we will assume the pulley to be like a disk with I = ~ ' / 2 ) ,

where a is the angular acceleration of the pulley, which is a rigid bodyrotating about a fixed axis. Since the string does not slip we must have

Solving these we get

SAQ 8 :In the above problem, calculate the tensions on either side. Which side hasmore tension and why?

SAQ 9 :Comment on the following problem which was first mentioned by LewisCanol. A long light cord passes over a pulley. At one end of the cord a bunchof bananas is tied, and at the other there is a monkey whose mass is equal to


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tha t of the bananas. If the system starts a t rest, with the bananas higher than ManentolInertl8 &Fbrd AxbRatahathe monkey, will be monkey be able to reach the banana by climbing up thecord?

6.4 OSCILLATIONS OF RIGID BODIESYou are already familiar with oscillating bodies such as a simple pendulum. Inanalysing the motion of a simple pendulum, we had assumed the pendulumbob to be a particle. When the pendulum bob is displaced from itsequilibrium position through small angles, restoring forces will tend to bring itback t o the equilibrium. If the amplitude is small enough, the restoringacceleration is proportional to the displacement from the mean position andthe bob executes simple harmonic oscillations. In the following we willconsider the oscillatory motion of extended bodies. The principle involved isidentical to that in the case of a simple pendulum. In this case the extendedbody executes oscillations about a fixed axis.6.4.1 Compound (Physical) PendulumA rigid body of mass m and moment of inertia I which can freely rotate in avertical plane about a horizontal axis through the point of suspension A (Fig.6.19) is known as a compound or a physical pendulum. A familiar example ofsuch a pendulum is a grandfather's clock. In the following we will obtain theequation of motion of such a pendulum. Note that the reaction forces passthrough the point of suspension. The only force on the rigid body which doesnot pass through the point A is the gravitational force mg. Since the reactionforces are not known at this stage, we take torque of forces about the point A X%.o that the unknown forces do not contribute to the torque. The net torque istherefore due to the gravitational force and is Figure 6.19 :A compound PendulumT = -mga sin 8where a is the distance between A and the centre-of mass C, and 9 is theangular displacement from the vertical. The moment of inertia about the axisthrough A is I, = I , + ma2, where I , is the moment of inertia about a parallelaxis through C. Using Eqn. (6.16), the angular acceleration a is given by

The equation of motion therefore is

For s m d amplitudes of oscillation we may replace sin 8 by the radian measureof the angle B itseh. The equation that we obtain is that for a simpleharmonic motion

which has a solution39


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Rotational Dynamics mga'sin(fi4)where A, and 4 re constants to be obtained from initial conditions. Theperiod of oscillation of the compound pendulum is given by

whnma we have used I, = I, +ma2 = m(k2 + a2 ), where kc is the radius ofgyration of the body about the horizontal axis through C. The period ofoscillation therefore depends on the distance of the point of suspension fromthe centre of mass and has its minimum value if the body is supported at a -distance kc from its centre of mass.It is interesting to note that one has another set of parallel axes about whichthe compound pendulum has the same time period as has been calcillatedabove. If b is taken as the distance of such an axis from the centre of mass, inorder that a compound pendulum has the same time period as that about anaxis at a.distance a, we have from Eqn.(6.19)

which gives b = k : / a . An interesting property of this second set of axes is asfollows. Suppose the pendulum is suspended from the point A. Consider apoint B which is located at a distance b from the cenl~e f mass C lying onthe line joining AC a.s shown in Fig. 6.20. If an impulsive blow is struck onthe body such that the impulse fl i e s in the plane of motion, the body exertsno impulsive reaction at A.

Figure 6.20 : The centre of percussionIf there is no impulsive reaction at A, the change in the angular momentum ofthe body about the point C is given by the product of the impulse with thedistance b of the point of application of the impulse from the point C. Thisresults inan angular velocity bJSince the support a t A is fixed, the velocity of the centre of mass is given by.vc = awHowever, from the relationship between the linear impulse and the centre ofmass velocity

J = m v ,Combining these we get b = I Jmn . Writing I , in terms of the radius ofgyration we get ab = kz . The point B is known as the centm of percussion forthe support point A. Clearly, the point A is the centre of percussion if thesupport point becomes B.


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Momentof Inertia &SAQ lO : FlxedAXIS RotatlonA thin homogeneous circular disk of radius R and mass M is suspended as acompound pendulum in a vertical plane from a point located at a distance afrom the centre. Find the distance a which gives the maximum frequency ofoscillation and determine this frequency. 1

\

SAQ 11:A uniform bar of mass m is suspended from a thin axle that passes through ahole near the top end A of the bar (Fig. 6.21). How far from A should a blowbe applied at right angles to the bar in order to start the bar rotating withoutbreaking the axle. Figure 6.21

6.4.2 Torsional PendulumA homogeneous disk of radius R is supported by a thin wire which is rigidlyattached to the disk and the support (Fig. 6.22 ). If the disk is twisted fromits equilibrium position through an angle 0, the wire will exert a restoringtorque on the disk which is proportional to the angle of twist 0. If the disk istwisted initially through an angle 8, and then released, it will execute simpleharmonic motion. This can be seen as follows. If 6 is the instantaneous angle Figure 6.22 : A torsionalof twist, the restoring force is -KO, where K is the torsional spring constant. pendulumWe may use equation (6.16) to obtain the equation of motion as this is a caseof fixed axis rotation. We get

where I s the moment of inetria of the disk about the wire. Eqn (6.20) is theequation of motion fi r a simple harmonic motion. Since at t = 0,8 = 00, thesolution is given by -

8 = 0, cos f i tThe disk therefore performs oscillations with a time period ( 1 / 2 r ) m .Torsional oscillations have many practical applications in engineeringproblems.


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6.5 SUMMARYIn this unit we discussed the physical principles which govern the rotationalmotion of rigid bodies about a fixed axis. The equations of motion weredeveloped by treating the body as a collection of particles and applying thegeneral principles of dynamics as formulated in the earlier units. Thecondition for a rigid body, that the distances between particles remain fixed,was used to simplify the general equations. While the translational motion ofthe centre of mass is given by a straight forward generalization of theNewton's second law, this equation does not tell us anything about theorientation or the angular motion of the body. In order to establish theorientation we have to solve an equation of motion involving the torque andthe angular acceleration. In the case of a rigid body rotating about an axisfixed in space, the torque is proportional to the angular acceleration, therelationship between the two has the same form as that between the force andthe linear acceleration. The rotational inertia however, depends both on themass and its distribution about the axis of rotation. An alternative, but inpractice more useful form of the equation of motion was derived which relatesthe angular impulse with the change in the angular momentum of a body.This and the equation connecting the corresponding linear quantities wereused in solving a variety of dynamical problems on the rotation of rigidbodies. When no external torque acts on the system, the angular momentumof the rigid body remains conserved. The rotational kinetic energy of the rigidbody was calculated and it was found that a work-energy theorem is valid inthis case also. This along with the principle of conservation of angularmomentum was used to solve several interesting dynamical problems.

Figure 6.23

6.6 ANSWERS TO SAQs1. A spherical shell can be considered as consisting of rings as shown in

Fig. 6.23 . The moment of inertia of such a ring of radius R sin 0 isd l -= ( R in B)2drn. The mass of the ring is dm = M sin 8 d 0 / 2 . Onintegrating over the rings (i.e. over 0 from 0 to a ) we get I = 2 ~ ~ ~ 1

/2 . We consider the solid sphere to be made up of a large number ofspherical shells of radius R and thickness d r . The moment of inertia of ashell element is ( 2 / 3 ) r 2 d m ,where d m , the mass of the shell is

Integrating over r from 0 o R , the result for a solid sphere follows.3 . Since the object is pencil like L > R . Thus the moment of inertiaabout an axis through its centre and perpendicular to it is M L 2 / 1 2 .

Transferring to a parallel axis located at the end of the rod the momentd nertia is M L 2 / 1 2 + ~ ( ~ 1 2 ) 'M L 2 / 3 .

4. The moment of inertia of a rectangular bar about an axis through itscentre of mass and perpendicular to its face has been shown to be( M / 1 2 ) ( L 2+ b2) . For a square object L = b = a so that the moment ofinertia becomes ~ a ~ / 6 .sing the perpendicular axis theorem, thenioment of inertia about a diagonal is ~ a ~ / 1 2 .


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5 . The torque acting is T R = l a . The angular acceleration is2 ( 2 r ) / 1 5= 4 / 1 5n . Substituting the values the tension T works out tobe 75.36 N.

6 . The work done is re. The initial kinetic energy of the wheel is(1 /2 )Zw2= 80n2 . Equating this to the work doneT = 80n2/40(2?r)= 3.14 N-m.

7. If F is the frictional force between the two rotors, the torque of the forceis used to reduce the angular speed of the first rotor from w l to a valuew . Simultaneously, the.angular speed of the second rotor is increasedfrom-zero to w . Thus F R i = I l a l and FR2 = Z2a2so that, we get,using kinematical equations w =w, +at,

Solving these,

8. The tensions are easily calculated by solving the equations of theexample 13 to be

The tension on the heavier side is more so that the net torque turns thepulley in a direction which will bring the heavier mass downwards.

9. Since equal masses are on the either side of the pulley, the gravitationaltorque acting on the system is zero. The angular momentum is thereforeconserved. If the monkey starts climbing up with a speed v with respectto the fixed pulley, in order that the angular momentum remains zero,the banana also moves up with the same speed. Thus the monkeycannot reach the banana. This assumes the pulley to be massless. Canyou think of what would happen if the mass of the pulley is considered ?

10. Time period of the compound pendulum is given by 2r , / (k2 + a2 ) l ga .For the disk kc = R / f i the minimum time period is obtained fora = R /& The frequency is ( 1 / 2 r ) { i z .

11. The radius of gyration of the rod about the centre of mass isk, =~/e.istance of A from C is LIZ. In order that a blow at Bcauses no reaction at the axle (which is thin and hence may breakotherwise), the distance CB is given by b = k z / a = L / 6 . The distanceAB therefore is 2L / 3 .

axis of rotation

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