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Home , Rigid body

RIGID BODIES - OF

The inability of a body to change by itself its of rest or uniform is called Inertia. The greater the of the body, the greater its inertia as greater is required to bring about a desired change in the body. Thus the mass of the body is taken as a measure of its inertia for translatory motion. Similarly a body, capable of about an axis, possesses inertia for rotational motion. The greater the or required to change the state of rotation of the body, the greater its rotational inertia. This rotational inertia of the body is called the M.I of the body about the axis of rotation. Thus M.I. is the rotational amalogue of mass in translatory motion. Definition of M.I of a particle of mass ‘m’ about a given axis of rotation is defined as the product of the

2 mass and the square of the ‘r’ of the particle from the axis of rotation. i.e., M.I I = Mr Consider a body of mass M capable of rotation about a fixed axis AOB. The body can be imagined to be made up of a no. of particles of mass m1, m2, m3,...... at A r1, r2, r3,...... from the axis of rotation. Then the M.I of the body about the axis of rotation = sum of the M.Is of all the particles about the axis of rotation r1 r i.e., I = m r2 + m r2 + m r2 +.....= Smr2 3 K m1 1 1 2 2 3 3 C m3 O r m2 2 2 Þ I = Smr -----(1)

Thus M.I of a body about an axis of rotation is defined as the sum of the B plroducts of the mass and square of the distance of all the particles constituting the body from the axis of rotation.

RADIES OF GYRATION Consider a body of mass M capable of rotation about an axis. Let the entire mass M of the body be imagined to be concentrated at a C. This point is called the Centre of mass of the body. Let C be at a distance K from the axis of rotation. Then M.I. of the body about the axis of rotation can be written as I = MK2 ---(2) where K is called the radius of gyration of the body w.r.t. the axis of rotation. Def : Radius of gyration of a body capable of rotation about an axis is defined as the distance of the point where the entire mass of the body is imagined to be concentrated from the axis of rotation. From (1) & (2) : MK2 = Smr2

mr2 or K2 = M

KINETIC OF ROTATION Consider a rigid body rotating with a constant angular w about a fixed axis AOB as shwon in the figure. As the body is made up of a number of particles of m1, m2, m3, ...... at distances r1, r2, r3, ...... from the axis of rotation, all these particles describe circular paths of radii r1, r2, r3, ...... with the same angular speed w. As the linear of the particles V = rw, V is different for different particles.

Let V1, V2, V3, ...... be the linear of the particles. Then the total K.E. of rotation of the body, 1 1 1 E = m v2 + m v2 + m v2 + ..... 2 1 1 2 2 2 2 3 3

1 1 1 = m r2 w2 + m r2 w2 + ..... 2 1 1 2 2 2

1 = w2 m r2 + m r2 +m r2 +.... 2 1 1 2 2 3 3

1 = w 2I , where I is the MI of the body about the axis rotation. 2

1 2 \ E = — Iw 2

THEOREM OF PARALLEL AXES Statement : The M.I of a body about any axis is equal to the sum of the M.I of the body about a parallel axis passing through its C.G. and the plroduct of the mass of the body and square of the distance b/w the two axes. This is called Steiner’s theorem. Proof : Consider a body of mass M whose C.G. is at G. Let the body rotate about an axis AOB and its M.I about about AOB be I. Let CGD be a parallel axis passing through G. Let the separation b/w the two axes be ‘r’. Consider a particle of mass ‘m’ at P at a distance x from CGD. Then M.I of this particle about AOB = m(r + x)2 M.I of the whole body about AOB is I = S m(r + x)2 = S mr2 + S mx2 + S 2mrx A C = Mr2 + I + 2 r S mx.....(1), where G r x O G P S m = M is the total mass of the body, S mx2 = I is the M.I of the body about CGD. G B D The weight of the particle of mass m at P = force acting on m = mg. The moment of this force about CGD = mgx. \ Sum of the moments of the weights of all the particles about CGD = Smgx. The algebraic sum of the moments of all the about an axis passing through the centre of of a body = 0. \ Smgx = 0 Q g ¹ 0, Smx = 0 \ 2r Smx = 0-----(2) Substituting (2) in (1),

2

I =IG + Mr Hence the theorem

Z THEOREM OF PERPENDICULAR AXES Statement : The M.I of a lamina about any axis ^r to its surface is equal to the sum r X of the moments of inertia about two ^ axes in the plane of the lamina, all the three 0 X axes passing through the same point on the lamina. r m Let OX & OY be two rectangular axes in the plane of the lamina and OZ, an axis Y p(x, z) through `O’ ^r to both OX & OY.. Consider a particle of mass m at a point r distant `r’ from O in the X - Y plane. Let its coordinates be x & y. Then r2 = x2 + y2. 2 M.I. of the particle about z-axis = mr2. 2 \ M.I of the lamina about z-axis, IZ = S mr . 2 2 Þ IZ = S m(x + y ) 2 S mx = M.I of the lamina about y-axis = Iy 2 S my = M.I of the lamina about x-axis = Ix

. Hence the theorem.

THEOREM OF PERPENDICULAR AXES IN THREE DIMENSIONS Let OX, OY & OZ be three rectangular axes with the origin at O in the given body. Consider a particle of mass m at a point P. Let OP = r. From P draw PQ ^r to the XY plane. Draw QB parallel to OX & QA parallel to OY. Then OA = x is the x-- coordinate and OB = y is the y-coordinate of P. Join OQ, OQ is ^r to OZ.. Z N Draw PN parallel to OQ. Then PN is also ^r to OZ. ON = PQ = z is the Z-- coordinate of P. and r2 = x2 + y2 + z2 ----- (1) p(x, y, z) 2 M.I of the particle at P about the Z-axis = m × NP r z 2 2 x A = m × NP = m × OQ = 0 X 2 2 m(OA + OB ) y y = m(x2 + y2) B x Q 2 2 \ M.I of the body about the Z-axis, IZ = S m(x + y ) Y 2 2 Þ IZ = S mx + S my -----(2) 2 2 ||| ly IY = S mx + S mz -----(3) 2 2 and IZ = S my + S mz -----(4) 2 2 2 2 2 2 IX + IY + IZ = S my + S mz + S mx + S mz + S mx + S my = 2S mx2 + 2S my2 + 2S mz2 = 2S m(x2 + y2 + z2) 2 Þ IX + Iy + IZ = 2S mr {from (1)

Þ IX + Iy + IZ = 2 I 1 or I = dIx + Iy + Iz i 2 Thus the M.I of a three dimensional body about any axis passing through a point in the body is equal to half the sum of the M.I of the body about three mutually ^r axes passing through the same point.

1a) M.I of a thin rod about an axis perpendicular to its length and passing through its C.G. M Consider a thin uniform rod XY of length l and mass M, then its mass per unit length m = . Let l BGB be an axis ^r to its length and passing through its C.G.`G’. Con- sider a small element of thickness dx at a distance x from the axis of A rotation. Then mass of the element = mdx. M.I of this element about L AGB = mdx × x2 G 2 = mx dx x \ M.I of the rod about AGB is B

3 l l 2 2 x3 O m Ll3 l 3 O ml3 ml × l2 Ml 2 I = mx2dx = m = + = = = z 3 P 3 M 8 8 P 12 12 12 l QP NM QP - l 2 - 2

If K is the rodius of gyration of the rod about the axis of rotation, then I = MK2

2 2 Ml l l \ MK = Þ K = = 12 12 2 3 b) M.I of the rod about an axis at one end of the rod and perpendicular to the rod Let AXB be an axis at one end and ^r to the length of the rod.

l l mx3 O ml 3 ml ´ l2 Then I = mx2dx = P = = z 3 P 3 3 0 Q0 A

Ml 2 \ I = . dx 3 We know I = MK2. X Y X Ml 2 l \ MK2 = Þ K = 3 3 l

B 2. M.I of a rectangular lamina (a) About an axis passing through its centre and parallel to breadth Consider a rectangular lamina ABCD of mass M, length l and breadth b. Let m be its mass per unit area M Y dx i.e., m = , or M = m(l × b). A B l ´ b YY/ is an axis ||el to its breadth. Consider a small strip G of thickness dx at a distance x from the axis. b X' X Its mass = m × (b × dx) / M.I of the strip about YY = mbdx × x2 = mbx2 dx D / C \ M.I of the rectangular lamina about YY is X l l ' l Y 2 mb I = mbx 2 dx = x3 2 Y z l l 3 - - 2 2

mb F l3 l3 I mb 2l3 m(l ´ b)´ l2 Ml 2 = + = ´ = = 3 HG 8 8 KJ 3 8 12 12

Ml2 I = Y 12

4 b) Axis Passing through its Centre and parallel to length : / XX is an axis passing through its centre and ||el to the length. Consider a small strip of thickness dx at a distance x from the axis. Its mass = m × (l × dx). / dx M.I of the strip about XX = m l d x × x2 1 x x = m l x2 dx X / G \ M.I of the rectangular lamina about XX is b b l 2 2 IX = z m l x dx b - 2

ml ml 2b3 m(l ´ b)´ b2 Þ I = x3 = ´ = x 3 3 8 12

b ml 2 ml 2b3 m(l ´ b) ´ b2 Mb2 3 = ´ = = Þ IX = x b 3 - 3 8 12 12 2

mb 2 Þ I = X 12 c) Axis Perpendicular to the plane of the lamina and passing through G

According to the theorem of ^r axes, M.I of the lamina about an axis ^r to the plane of the lamina and passing through G = M.I about two mutually ^r axes through G in the plane of the lamina. / Let I be the M.I of the lamina about an axis ZGZ passing through its CG and ^r to its plane.

Mb2 Ml2 i.e., I = I + I = + X Y 12 12

M(l2 +b2 ) Þ I = 12

(d) Axis along one end (AD)

l mb mbl l2 I = m b x2 dx = l 3 = b g z 3 3 0 Z Y Ml2 I = 3 A B / G X X b 2 M.I about AB : I = z m l x dx D C 0 / / Y Z

5 mb mlb = b3 = b2 3 3

Mb 2 Þ I = 3 e) Axis passing through the mid point of AD or BC and perpendicular to the plane of the lamina ABCD is the lamina. Let 0 be the mid point of AD; let MN be an axis through 0 ^r to the plane of / the lamina. Let M.I of the lamina about MN be I, YGY is a ||el axis through the centre of the lamina and / to its plane. Let the M.I of the lamina about Y0Y be IG. M(l 2 + b2 ) Then I = G 12 By ||el axes theorem,

l 2 I = I + MF I G HG 2KJ M 2 Y M(l 2 + b2 ) F l I A Þ I = + MG J B 12 H 2 K 0 G Ml2 Mb2 Ml 2 X = + + 12 12 4 D C 22 l Ml Mb I = + / 3 12 N Y ||| ly M.I about an axis through the mid point of AB or CD and ^r to the plane of the lamina

2 2 b l I = M + N 3 12

3. M.I of a circular ring about an axis a) through its centre and perpendicular to its plane / Consider a thin circular ring of mass M and radius R. Let Y0Y be the axis through its centre and ^r to its plane. Consider a small ele- / Y ment of mass m. The M.I of this element about Y0Y = mR2. / \ M.I of the ring about Y0Y , I = SmR2. R 2 O m Þ I = MR Q S m = M and / R is the distance of each of the elements from 0 Y b) M.I of the ring about an axis along its diameter / / Let X0X and Y0Y be two axes along two diameters of the ring. Then IX = IY / M.I of the ring is the same about all diameters. Let I and I be the M.I of the ring about X0X and / X Y Y0Y respectively. Then IX = IY . Let the M.I of the ring about an axis through 0 and ^r to its plane be I. Then I = MR2 where M is

6 the mass and R is the radius of the ring. r By the theorem of ^ axes, I = IX + IY Þ I = 2I = 2I . X Y Y I MR2 Þ I = I = = . x y 2 2 / O X X MR2 \ M.I of the ring about its diameter = 2 / Y

4) M.I of a circulr lamina (disc) about an axis passing through its centre and perpendicular to its plane. / Let Y0Y be an axis through its centre and ^r to its plane. Consider a circular lamina of mass M and radius R. Y Area of the lamina = pR 2 . R O M Mass per unit area of the lamina, m = . pR2 Fig.(a) / The disc can be considered to be made up of a number of annular rings. Y Consider one such ring of thickness dx and radius x. Area of the ring = 2px dx Mass of the ring = 2px dx m = 2pmx dx. / M.I of the ring about Y0Y = 2 p m x dx × x2 = 2 p m x3 dx

R / M.I of the lamina about Y0Y I = z 2pmx3 dx is 0

R dx x4 O pmR4 Þ I = 2pm P = I O 4 2 QP0

2 2 2 B (pR m)R MR = Þ I = ——— 2 2 Fig.(b) b) M.I about its diameter / / Let X0X and Y0Y be two ^r axes along two ^r diameters of the disc. / / Let IX and IY be the M.Is of the disc about X0X and Y0Y respectively MR2 Then I = I . M.I of the disc about an axis passing through its centre and ^r to its plane, I = X Y 2 r By the theorem of ^ axes, I = IX + IY = 2IX = 2IY

MR2 Þ = 2I = 2I . 2 X Y 7 MR2 Y \ I = I = X Y 4 / 0 X X c) M.I. about a tangent Let AB be an axis along a tangent to the disc. Let the M.I of the disc about AB / / be I. Let YGY be an axis parallel to AB and along the diameter (through the Y MR2 C.G ‘G’). Let the M.I of the disc about YGY / be I . Then I = , where M is the mass and R G G 4 is the radius of the disc. By the theorem of parallel axes, 2 I = IG + MR MR2 A Y = + MR2 4

2 5MR Þ I = ——— R 4 G

5) M.I of a thin sphirical shell about diameter / / Consider a thin spherical shell of a mass M and radius R. Let Y0Y be an axis B Y along a diameter of the shell. Area of the shell = 4pR2. M \ Mass per unit area, m = 4p R2 The shell may be considered to be made up of a number of rings. / Let PQSR be one such ring of thickness dx and radius y at a distance x from Y0Y . Area of the ring = 2py dx y From the figure, sinq = Þ y = R sinq. R x cosq = Þ x = R cosq ; d x = – R sinq dq. R PR = dx = R dq and R2 = x2 + y2 Þ y2 = R2 – x2. \ Area of the ring = 2 p (R sinq)( R dq) Y P = 2p R (R sinq dq) dq R = 2 p R dx (in magnitude) \ Mass of the ring = 2 p R dx m R = 2p m R dx q Y / / 0 I X X M. of the ring about X0X (diameter) X = Mass × (radius)2 = 2p m R dx y2 = 2p m R dx (R2 – x2). S / M.I of the shell about X0X is Q Y / R 2 2 I = z 2p m R (R - x ) dx -R

8 R R 3 2 = z 2p m R dx — z 2p m Rx dx -R -R

2p m R = 2p m R3 2 R - 2 R3 3

4p m R4 = 4p m R4 — 3

2 x 4 p m R4 4p R2m ´ 2 R2 = = 3 3

2 2 Þ I = — MR 3

b) M.I of a thin spherical shell about a tangent Let M.I of the shell about a tangent be I and about a ||el axis along its diameter (i.e., through its centre of gravity be IG.) Then by the theorem of parallel axes, 2 I = IG + MR , where M is the mass of shell and R is its radius.

2 2 2 = MR + MR 3

5 2 Þ I = — MR 3 a) M.I. of a solid sphere about its diameter Consider a solid sphere of radius R and mass M. Let r be the density of the material of the sphere. 4 Volume of the sphere = pR3 3 4 Mass of the sphere = pR3r 3 The solid sphere may be considered to be made up of a number of concentric spherical shells of different radii. Consider one such shell of radius x and thickness dx. Surface area of the shell = 4 p x2. Volume of the shell = 4 p x2 dx Mass of the shell, = vol. × density = 4 p x2 dx r

/ 2 M.I of the shell about Y0Y = Mass × (radius)2 3 2 = 4 p x2 dx r x2 3

9 8 Y = p r x4 dx 3

R / 8 4 M.I of the sphere about Y0Y , I = z p r x dx x 0 3 0 8p r 5 F 4 3 I 2 2 dx = R - 0 = G p R rJ ´ R 3 ´ 5 e j H 3 K 5

2 2 R I = — MR Þ / 5 Y b) M.I about a tangent

By parallel axes theorem, A Y 2 M.I about the tangent, I = IG + MR 2 = MR2 + MR2 5 R G

7 2 I = — MR 5

M.I. of a solid cylinder about an axis passing through its centre and ^r to its B / length Y / Consider a solid cylinder of mass M, radius R and length l. Let YGY be an axis passing through its centre of gravity and ^r to its length. Let m be the mass per unit length of the cylinder M i.e., m = . l Y The cylinder may be considered to be made up of a number of thin discs of rod R. / R X 0 X Consider one such disc of thickness G x / dx at a distance x from YGY . Mass of M the disc = mdx = dx l l / / Y M M.I of the disc about an axis / M0M along its own diameter

Mass ´ (r ´ d)2 M R2 MR2 = = dx = dx 4 l 4 4l

/ / M.I of the disc about YGY = M.I of the disc about M0M + Mass × x2 (by ||el axes theorem) 10 MR2 M = dx + dx x2 4l l

MR2 M = dx + x2 dx 4l l

M LR2 O M dx + x2 dxP = l NM 4 QP

/ M.I of the cylinder about YGY is given by

l 2 M LR2 O dx + x2 dx I z M P = l l M 4 P - N Q 2

l l 2 3 2 4 M M R + X = l 3 l 4 l l 2 2

MR4 M L l3 l3 O + M + P = 4 3 l NM 8 8 QP

MR2 M l3 = + 4 12 l

MR2 M l 3 = + 4 12

LR2 l2 O Þ I = M M + P NM 4 12 QP b) M.I about the diameter of one of its faces

2 / F l I I = M.I about YGY + Mass × G J H 2K

2 2 2 LR l O M l A Y Þ I = M M + P + NM 4 12 QP 4

LR2 l2 + 3 l 2 O = M + M P G NM 4 12 QP O

LR2 l2 O I = M M + P l M 4 3 P 2 N Q / B Y

11 c) M.I about its own axis M.I of the disc about an axis pasisng through its centre and ^r to its plane =

2 / Mass of the disc ´ rad M.I of the disc about the axis XOX of the cylinder = 2

MR2 = 2 / M.I of the cylinder about XOX is

mR2 R2 I = Þ I = bSmg å 2 2

2 MR Þ = I = 2

12