Mod 9

Module 9: ROTATIONAL KINEMATICS MOMENT OF INERTIA: We know that according to Newton’s first law of motion, “a body must continue in its state of rest or of uniform motion along a straight line, unless acted upon by an external force.” This inertness or inability of a body to change by itself its position of rest or of uniform motion is called inertia. We know by experience, the greater the mass of the body; the greater is its inertia or opposition to the desired change. The mass of the body thus is the measure of its inertia for linear motion. Exactly in the same manner, in case of rotational motion, also we find that, a body free to rotate about an axis opposes any change desired to be produced in its state of rest or of rotation, showing that it possesses inertia for this type of motion. It is the rotational inertia of the body, which is called moment of inertia. In case of linear motion, the inertia of a body depends on wholly on its mass. In case of rotational motion, the inertia depends not only on its mass of the body but also on the effective distance of its particles from the axis of rotation. KINETIC ENERGY OF ROTATION AND MOMENT OF INERTIA: The kinetic energy of system of particles of masses m1, m2…….mn is defined as

K.E =

1 1 1 1 2 2 2 2 m1v 1 + m 2v 2 + m 3 v 3 + - - - - - - - - - - + m n v n 2 2 2 2 n

=

1

∑ 2 mv 2 i =1

Substituting v = ωr [where ω is the angular speed of the rotating body], we have

 1 1 n 2  ∑ m i r i 2  ω 2 m ( r ω ) = ∑2 i i 2  i =1 i =1  n

K.E =

in which ω is the same for all particles. The quantity in the parentheses tells us how the mass of the rotating body is distributed about its axis of rotation. This quantity is known as moment of inertia. We may now write, n

I = ∑ mi ri 2 i =1

I = ∫ r 2 dm Thus,

K. E =

TORQUE: The ability of a force

1 2 Iω 2

 F to rotate a body depends not only on the magnitude of its tangential component

Ft but also on just how far from the axis of rotation at O it is applied. To include both these factors, we define a quantity called torque τ as 1

τ = [r ][F sin φ ] , ……………….. [1] Two equivalent ways of computing the torque are

τ = [r ][F sin φ ] = rFt ,......... ..[ 2 ] τ = [r sin φ ][F ] = r⊥ F,......... ...[ 3 ]

r⊥ is the perpendicular distance between the axis of rotation at O and an extended line running   through the vector F . This extended line is called the line of action of F and r⊥ is called the moment  arm of F . where

RADIUS OF GYRATRION: “The radial distance from a given axis at which the mass of a body could be concentrated without altering the rotational inertia of the body about that axis is called the radius of gyration”. Let k represent radius of gyration and it can be shown that

k= IM General Theorems on Moment of Inertia: There are two general theorems of great importance on moment of inertia which, in some cases, enables us to determine the moment of inertia of a body about an axis, if its moment of inertia about some other axis be known. (i) The Principle of theorem of Perpendicular Axes: According to this theorem, the moment of inertia of a plane lamina about an axis, perpendicular to the plane of the lamina, is equal to the sum of the moment of inertia of the lamina about two axes at right angles to each other, in its own plane and intersecting each other at the point where the perpendicular axis passes through it. Thus, if Ix and Iy be the moment of inertia of a plane lamina about the perpendicular axes OX and OY, which lie in the plane of the lamina and intersect each other at O, the moment of inertia about an axis passing through O and perpendicular to the plane of the lamina, is given by I = Ix+ Iy For, considering a particle of mass m at p, at distances x and y from OY and OX respectively, at distance r from O, we have I=

∑ mr 2 ,

So that,

I x = ∑ my 2

I x + Iy =

∑ mx 2 = ∑ m ( y 2 + x 2 ) = ∑ mr 2

and I y =

∑ my 2 + ∑ mx 2

I x + Iy = I (ii) The Principle of Theorem of Parallel Axes: We can often simplify the calculation of moments of inertia for various bodies by using the parallel-axis theorem, which relates the moment of inertia about an axis through the centre of mass of an object to the moment of inertia about a second, parallel axis. Let Icm be the moment of inertia about an axis through the centre of mass of an object of total mass M, and let I be that about a parallel axis a distance h away. The parallel-axis theorem states that

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I = ICM + Mh2 Particular cases of Moment of Inertia 1. Moment of Inertia of a Thin Uniform Rod (i)

About an axis through its center and perpendicular to its length:

Let AB be a thin uniform rod of length L and mass M, free to rotate about an axis CD through center O and perpendicular to its length. Then its mass per unit length =

C

M L

dx A

B

Let us consider a small element of length dx on it , at a distance x from O. The mass of the small element

dm =

O

x

D

M dx L M  dx L

2 Moment of inertia of the small segment dx about the axis = x2 dm = x 

Thus the moment of inertia of the whole body, L 2

L 2

L 2

M 2M 2M  x 3 2  I = 2 ∫ x dm = 2 ∫ x dx = x dx = ∫  3 L L L  0 0 0 (ii)

2

2

L

 2 ML2  =  12 0

About an axis passing through one end and perpendicular to its length:

The treatment is the same as above, except that since the axis CD here passes through one end B of the rod, the expression of the moment of inertia of the element of dx of the rod is now to be integrated between the limit, x = 0, at B and x = L, at A. C Thus dx

L

M M x3  ML2 I = ∫ x 2 dm = =   L L  3 0 3 0 L

A

B x

D

This problem can also be using the parallel-axis theorem:

I = Icm + Mh 2 =

1 1 ML2 + M[L / 2 ]2 = ML2 12 3

2. Moment of Inertia of a Thin Triangular Plate about one side Let ABC be a thin triangular plate of surface density ρ (mass per unit area), whose moment of inertia is to be determine about the side BC. If the altitude of the plate be AP = H,

3

Its area =

1 ×a×H 2

And its mass M =

[here BC = a]

1 aHρ 2

Now, let us consider that the triangular plate is made up by a number of thin strips, parallel to BC. Let us consider one such strip is DE of width dx at a distance x from BC. The area of the strip = DE. dx

A

The mass of the strip dm = DE. dx. ρ Now , in the similar triangles AQD and APB , we have

DQ AQ = BP AP H −x DQ = .BP H

B a

QE AQ = PC AP H−x QE = .PC H Now, DE = DQ+QE =

H−x H−x (BP + PC ) = .a H H

Mass of the strip dm =

H−x . a.dx.ρ H

Moment of inertia of DE about the side BC = x2 dm = x 2

H−x .a.dx.ρ H

Moment of inertia of the whole plate about BC, H

H

0

0

2 2 ∫ x dm = ∫ x

H−x aρ a dx ρ = H H

H

∫ (H − x) x

2

dx

0

aρ H aρ H 3 2 Hx dx − x dx = H ∫0 H ∫0 =

aρH 3 1 H2 = .a.H .ρ. 12 2 6

=

MH 2 6

3. Moment of Inertia of a Circular Disc

(i)

Q E

x

similarly. From similar triangles AQE and APC we have

I=

dx

D

about an axis through its center and perpendicular to its plane:

Let M be the mass of the disc and R be the radius. The axis of rotation is YY` .

4

P

C

The area of the disc =

πR 2

Mass per unit area of the disc =

M . πR 2

Let us consider a ring of the disc at a distance x from the axis. The radius of the ring is x and width is dx. The area of the ring = 2πx dx

Y

M 2M x dx The mass of the ring, dm = ( 2πx ) dx. = 2 R2 πR Moment of Inertia of the ring about the axis through O = x2 dm Moment of Inertia of the whole disc about the axis I=

(ii)

R

R

0

0

2 2 ∫ x dm = ∫ x

R

R

O

R 2M 2M 2M  x 4  MR 2 3 x dx = x dx = =   ∫ 2 R2 R2 0 R2  4 0

about its diameter:

Y’

Let AB and CD be two perpendicular diameters of a circular disc of radius R and mass M. Since the moment of inertia about AB is equal to the moment of inertia about CD. Let it be I.

A

dx

x

C O

B

D

Now according to perpendicular axis theorem,

Moment of inertia of the disc about AB + M.I. of the disc about CD = Moment of inertia of the disc about an axis through O and perpendicular to its plane. I+I= 2I = ∴

(ii)

I=

MR 2 2

MR 2 2

MR 2 4

about a tangent to the disc in its own plane:

Let AB be the tangent to the circular disc of radius R and mass M, about which its moment of inertia is to be determined. Let CD be a diameter of the disc, parallel to tangent AB. The

MR 2 moment of inertia of the disc about this diameter is equal to . 4 So according to parallel axes theorem, Moment of inertia of the disc about AB = M.I. of the disc about CD + MR2

 MR 2  5MR 2  + MR 2 = I =   2  4 

A

C

O D

R

B

(iii) about a tangent to the disc and perpendicular to its plane: This tangent will obviously be parallel to the axis through the center of disc and perpendicular to its plane, the distance between the two axes being equal to the radius of the disc.

5

Hence, by the principle of parallel axes, we have M.I about the tangent = M.I about the perpendicular axis + MR2

MR 2 3MR 2 + MR 2 = 2 2

I=

4. Moment of Inertia of a solid sphere: (i) about its diameter: Figure represent a section of the sphere through its center O of mass M and radius R.

4 πR 3 3 M 3M = Mass per unit volume = 4 4πR 3 πR 3 3 The volume of the sphere =

Let us now consider a thin circular slice of the sphere at a distance x from the center and thickness dx. The radius of the circular disc =

(

R2 − x2

The surface area of the disc = π R 2 − x 2

(

)

)

The volume of the disc = π R 2 − x 2 dx Mass of the disc = volume of the disc

(

)

2 2 = π R − x dx .

×

mass per unit volume

3M 4πR 3

=

(

)

3M R 2 − x 2 dx 3 4R

Now moment of inertia of this disc about the diameter 2 × ( radius of the disc )

dI = mass of the disc

 R 2 − x 2  3M  2 2 R − x dx ×  3 2 4R

(

=

R

2

)

=

(

)

3M 2 2 2 R − x dx 8R 3

Moment of Inertia of the whole sphere about the diameter,

(

R

)

2 3M 2 I = 2∫ R − x 2 dx 3 0 8R

= 2.

3M 8R 3

R R R 4  2 2 R dx − 2 R x dx + x 4 dx  ∫ ∫ ∫  0  0 0

 3 3M  4 R 2 x  = R ( x ) − 2 R 0  3 4R 3    =

3M 4R 3

R2 + x 2

2

R R    x5    +      0  5 0  

 5 2R 5 R 5  + R −  3 5  

6

A

O

B

x

dx

=

2 MR 2 5

The moment of inertia of the sphere about one diameter is the same as any another diameter.

(ii) about a tangent: A tangent drawn to the sphere at any point, will obviously be parallel to one of its diameters and the distance between the axes is equal to R, the radius of the sphere. According to the parallel axes theorem we have, M.I of the sphere about a tangent = M.I of the sphere about a diameter + MR2

I=

2 7 MR 2 + MR 2 = MR 2 5 5

5. Moment of Inertia of a solid cylinder

(i)

about an axis passing through its center and perpendicular to its own axis of cylindrical symmetry:

Let us consider a solid cylinder of mass M, radius R and length L. Let YY1 be the axis passing through its center O and perpendicular to its own axis XX1 about which moment of inertia is to be determined. Let us consider the cylinder made up of a number of thin disc and one such disc is at a distance x from o and of thickness dx. Mass of the cylinder = M Volume of the cylinder = πR 2 L Mass per unit volume of the cylinder =

M πR 2 L

Y

Volume of the disc = πR 2 dx 2 Mass of the disc M ′ = πR dx .

M M = dx 2 πR L L

X

O

The moment of inertia of this disc about Y Y ′ is dI.

dx

According to Parallel axes theorem, Y’

1 M ′R 2 + M ′x 2 4 1 M M = πR 2 dx . 2 R 2 + πR 2 dx . 2 .x 2 4 πR L πR L

dI =

=

1 MR 2 M dx + x 2 dx 4 L L

The moment of inertia of the whole cylinder,

I = 2∫ [

X’

x

1 MR 2 M dx + x 2 dx ] 4 L L

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L

L

1M 2 2 M  x3 =2 R [ x ] 0 + 2  4 L L 3 =

2   0

 R 2 L2  MR 2 ML2  + = M  +  4 12 4 12  

TABLE1: COMPARISON OF LINEAR AND ANGULAR MOTION WITH CONSTANT ACCELERATION

LINEAR MOTION

ROTATION ABOUT A FIXED AXIS

a = cons tan t

α = cons tan t

v = v 0 + at

ω = ω 0 + αt

x = x0 + v0t +

1 2 at 2

θ = θ0 + ω0 t +

v 2 = v 20 + 2a[ x − x 0 ] x − x0 =

1 2 αt 2

ω 2 = ω 20 + 2α[ θ − θ 0 ]

1 [ v + v 0 ]t 2

θ − θ0 =

1 [ ω + ω 0 ]t 2

TABLE 2: ANALOGS IN ROTATIONAL AND LINEAR MOTION LINEAR MOTION Displacement

ROTATION ABOUT A FIXED AXIS X

Angular displacement

θ

velocity

v=

dx dt

Angular velocity

ω=

dθ dt

Acceleration

a=

dv dt

Angular acceleration

α=

dω dt

Mass [translational inertia]

M

Force

F = Ma

Work

W = ∫ Fdx

Rotational inertia Torque Work

Kinetic energy

1 Mv 2 2

Kinetic energy

Power

P = Fv

Power

Linear momentum

Mv

Angular momentum

8

I τ = Iα

W = ∫ τdθ 1 2 Iω 2 P = τω Iω

Table 3: Moments of inertia for various common bodies. The 'M' in each case is the total mass of the object.

slender rod:

axis through center

axis through end

rectangular plane:

axis through center

axis along edge

sphere

thinwalled hollow

solid

cylinder

hollow

solid

thinwalled hollow

axis through tangent

9

I=

3MR 2 2

Solved Examples: Example 1: A hydrogen chloride molecule consists of a hydrogen atom whose mass mH is 1.01 u and a chlorine atom whose mass mCl is 35.0 u. The centres of the two atoms are a distance d = 1.27 x 10-10 m = 127 pm apart. What is the rotational inertia of the molecule about an axis perpendicular to the line joining the two atoms and passing through the centre of mass of the molecule? Solution: Let x be the distance from the centre of mass of the molecule to the chlorine atom. Then, we can write

0= which yields

x=

mH d, m Cl + mH

− m Cl x + mH [d − x ] m Cl + mH

……………. [1]

The rotational inertia about an axis through the centre of mass is

I = ∑ miri2 = mH [d − x ]2 + m Cl x 2 , ……………..[2] i

Substituting the value of x from Eq. [1] , we get

I = d2

mHm Cl [1.01 u][35.0 u] = [127 pm]2 = 15.800 u. pm 2 mH + m Cl 35.0 u + 1.01 u

Example 2: With modern technology, it is possible to construct a fly wheel that stores enough energy to run an automobile. The energy is stored as rotational kinetic energy when the fly wheel is initially made to spin by a machine. The stored energy is then gradually transferred to the automobile by a gear system as the automobile is being driven. Suppose that such a wheel is a solid cylinder whose mass M is 75 kg and whose radius R is 25 cm. If the wheel is spun at 85,000 rev/min, how much rotational kinetic energy can it store? Solution: The rotational inertia of the cylinder wheel is

I=

1 1 MR 2 = [ ][75 kg][0.25 m]2 = 2.34 kg. m 2 2 2

The angular velocity of the wheel is

ω = [85000 rev / min][2π rad / rev ][1 min/ 60 s ] = 8900 rad / s The kinetic energy of rotation is then

K=

1 2 Iω = [1/ 2 ][2.34 kg. m 2 ][8900 rad / s ]2 = 9.3x10 7 J = 26 kW . h. 2

This amount of energy, used with the expected efficiency, would run a small car about 200 mi. Example 3: A uniform disk of radius R = 20 cm and mass M = 2.5 kg is mounted on a fixed horizontal axle. A block whose mass m is 1.2 kg hangs from a mass-less cord that is wrapped around the rim of the disk. Find [a] the acceleration of the falling block [assuming that it does fall], [b] the tension in the cord, [c] the angular acceleration of the disk and [d] the tangential acceleration of a point on the rim of the wheel [disk]. The cord does not slip and there is no friction at the axle. Solution: [a] Let us assume that the block accelerates downward, so the magnitude mg of its weight must exceed the tension T in the cord. Therefore, from Newton’s second law, we get T − mg = ma , ……………. [1] where a is the acceleration of the falling block.

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The torque acting on the disk is TR. The rotational inertia I of the disk is

1 MR 2 . [Two other forces also 2

act on the disk, its weight Mg and the normal force N exerted on the disk by its support. Since both these forces act at the axis of the disk, they exert no torque on the disk.] Applying Newton’s second law in angular form [ τ = Iα ] to the disk, we get

TR =

1 MR 2 α , ……………………… [2] 2

Combining [1] & [2] leads to

[2 ][1.2 kg] 2m = [ 9 .8 m / s 2 ] = 4. 8 m / s 2 M + 2m 2.5 kg + [2 ][1.2 kg] [b] Using the relation a = Rα , we can write from [2] 2T = Ma a=g

The tension T in the cord is then

T=

1 1 Ma = [2.5 kg][ 4.8 m / s 2 ] = 6.0 N 2 2

[c] The angular acceleration of the disk is

a 4. 8 m / s 2 α= = = 24 rad / s 2 R 0.20 M [d] The tangential acceleration of a point on the rim is

a = Rα = [24 rad / s 2 ][0.20 m] = 4.8 m / s 2 Example 4: Assuming that the disk of Example 3 starts from rest, compute the work done by the applied torque on the disk in 2.0 s. Compute also the increase in rotational kinetic energy of the disk. Solution: Since the applied torque is constant, the resulting angular acceleration is constant. The total angular displacement in constant angular acceleration is

θ = ω0 t +

1 2 1 αt = 0[ t ] + [24 rad / s 2 ][2.0 s ]2 = 48 rad 2 2

For constant torque, the work done in a finite angular displacement is

W = τ[ θ 2 − θ 1 ]

in which

τ = TR = [6.0 N][0.20 m] = 1.2 N. m,

and Therefore,

θ 2 − θ 1 = θ = 48 rad W = [1.2 N. m][ 48 rad] = 57.6 J

This work must result in an increase in rotational kinetic energy of the disk. Starting from rest the disk acquires an angular speed ω . The rotational energy is

To obtain

ω we use

1 2 1 1 Iω = [ MR 2 ]ω 2 2 2 2 ω = ω 0 + αt = 0 + [24 rad / s 2 ][2.0 s ] = 48 rad / s

Then

1 2 Iω = [0.5 ][2.5 kg][0.20 m]2 [ 48.0rad / s ]2 = 115.2 J 2 Hence the increase in kinetic energy of the disk is equal to the work done by the resultant force on the disk.

11

Example 5: Consider a solid cylinder of M and radius R rolling down an inclined plane without slipping. Find the speed of its centre of mass when the cylinder reaches the bottom. Solution: The cylinder is initially at rest. In rolling down the incline the cylinder loses potential energy of an amount Mgh, where h is the height of the incline. It gains kinetic energy equal to

1 1 Icm ω 2 + Mv 2 2 2 where v is the linear speed of the centre of mass and ω is the angular speed about the centre of mass at the bottom. We have then the relation

1 1 Icm ω 2 + Mv 2 2 2 1 v = MR 2 and ω = . 2 R Mgh =

I cm

in which Hence

1 1 v 1 1 1 [ MR 2 ][ ]2 + Mv 2 = [ + ]Mv 2 , 2 2 R 2 4 2 4 v 2 = gh , 3 4 or v = gh 3 The speed of the center of mass would have been v = 2gh if the cylinder had slid down a frictionless Mgh =

incline. Example 6: A sphere and a cylinder, having the same mass and radius, start from rest and roll down the same incline. Which body gets to the bottom first? Solution:

2 MR 2 . Using the dynamical method, we get for the sphere 5 Mg sin θ − f = Ma, translation of cm , ………. [a] 2 fR = Icm α = [ MR 2 ][ a / R ], rotation about cm , …… [b] 5 where f is the force of static friction acting along the incline at the point of contact.

For a sphere

Icm equals

From [b] we can write

f= Putting this value of

2 Ma 5

f in [a], we get

5 g sin θ 7 2 In Example 5, we have already proved for the cylinder, a = g sin θ . 3 a=

Hence the acceleration of the centre of mass of the sphere is at all times greater than the acceleration of the centre of mass of the cylinder. Since, both bodies start from rest at the same instant, the sphere will reach the bottom first.

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Problems 1.

A circular disc of mass m and radius r is set rolling on a table. If its total energy E is given by E =

ω is its angular velocity, show that

3 mr 2 ω 2 4

2.

A circular disc of 49 kg weights and of radius 50 cm is rotating. Calculate the kinetic energy possesses when executing 120 rotations per minute.

3.

Show that the kinetic energy of a thin rod of length l and mass per unit length is m rotating about an axis through the middle point and perpendicular to its length with angular velocity

ω is 1 mω 2 l 3 24

4.

Assume the earth to be sphere of uniform density. [a] What is its rotational kinetic energy? [b] Suppose this energy could be harnessed for our use. For how long could the earth supply 1.0 kW of power to each of the 4.2 x 109 persons on earth? Take the radius of the earth to be 6.4 x 10 3 km and the mass of the earth to be 6.0 x 1024kg.

5.

[a] Show that a solid cylinder of mass M and radius R is equivalent to a thin hoop of mass M and radius R/√2, for rotation about a central axis. [b] The radial distance from a given axis at which the mass of a body could be concentrated without altering the rotational inertia of the body about that axis is called the radius of gyration. Let k represent radius of gyration and show that k

= IM.

6.

A body of radius R and mass m is rolling horizontally without slipping with speed v. It then rolls up a hill to a maximum height h. If h = 3 v 2 / 4g , [a] what is the body’s rotational inertia? [b] What might the body be?

7.

A cylinder of length L and radius R has a weight W. Two cords are wrapped around the cylinder, one near each end, and the chord ends are attached to hooks on the ceiling. The cylinder is held horizontally with the two cords exactly vertical and is then released. Find [a] the linear acceleration of the cylinder as it falls and [b] the tension in each cord as they unwind.

8.

A tape of negligible mass is wrapped around a cylinder of mass M, radius R. The tape is pulled vertically upward at a speed that just prevents the centre of mass from falling as the cylinder unwinds the tape. [a] What is the tension in the tape? [b] How much work has been done on the cylinder once it has reached an angular velocity ω ? [c] What is the length of tape unwound in this time?

9.

Find the moment of inertia about the axis of rotation for the object pictured below. The masses are m1 = m2 = 1.0 kg and m3 = 4.0 kg. If the axis of rotation is moved from centre to end, what is the new moment of inertia?

Axis of rotation

10. Four particles of mass m are connected by massless rods to form a rectangle of sides 2a and 2b as shown. The system rotates about an axis in the plane of the figure through the centre. Find the moment of inertia about this axis. If the axis of rotation is moved from centre to end, what is the new moment of inertia?

2b

Diameter 35 mm

11. The problem consists of a circular plate with an offset circular hole cut into it. Find the moment of inertia of the disc.

13

O

31.25 mm

X

Diameter 125 mm

2a