chp-31

Contents

622

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A Textbook of Engineering Mechanics

C H A P T E R

Kinetics of Motion of Rotation

31

Contents 1. 2. 3. 4. 5. 6. 7.

8.

9. 10. 11. 12. 13.

31.1. INTRODUCTION

We have already discussed in chapter 21 the kinematics of motion of rotation and its ap plications. In this chapter, we had discussed the geometry of motion without taking into consideration the forces causing motion of rotation. But in this chapter, we shall also discuss the forces, which influence the motion of rotation. 31.2. TORQUE

It is the turning moment of a force on the body on which it acts. The torque is equal to the product of the force and the perpendicular distance from any point O to the line of action of the force. Mathematically, torque, T = F × l 622

14. 15. 16.

17. 18.

19.

20.

21.

Introduction. Torque. Work done by a Torque. Angular Momentum. Newton’s Laws of Motion of Rotation. Mass Moment of Inertia. Mass Moment of Inertia of a Uniform Thin Rod about the Middle Axis Perpendicular to the Length. Moment of Inertia of a Uniform Thin Rod about One of the Ends Perpendicular to the Length. Moment of Inertia of a Thin Circular Ring. Moment of Inertia of a Circular Lamina. Mass Moment of Inertia of a Solid Sphere. Units of Mass Moment of Inertia. Radius of Gyration. Kinetic Energy of Rotation. Torque and Angular Acceleration. Relation Between Kinetics of Linear Motion and Kinetics of Motion of Rotation. Flywheel. Motion of a Body Tied to a String and Passing Over a Pulley. Motion of Two Bodies Connected by a String and Passing Over a Pulley. Motion of a Body Rolling on a Rough Horizontal Plane without Slipping. Motio Mo tion n of of a Bo Body dy Ro Rolli lling ng Down a Rough Inclined Plane without Slipping.

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Chapter 31 : Kinetics of Motion of Rotation

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623

F = Forc Forcee acting acting on the the body body, and

where

l

Notes:1.

2.

= Perpendicular Perpendicular distanc distancee between between the the point point O and line of action of the force (known as arm or leverage). The units units of torque torque depend depend upon the units of force an d levera ge. If the forc e is in N and leverage in mm, then the unit of torque w ill be N-mm. Similar Similarly ly,, if the force force is in kN and leverage in m, then the unit of torque will be kN-m. The magnitude of the moment of a force is numerically equal to that of the torqu e, if the force and the arm is the same. The term torque is used for the moment of a force in the motion of rotation.

31.3. WORK DONE BY A TORQUE

Consider a body pivoted at O. Let a tangential force P be applied at a distance r from from the pivot as shown in Fig. 31.1. As a result of the force, let the body rotate through a small angle ( θ) in radians. We know that the length of the arc AB = r θ and work done by the force P in rotating the body from A to B = P ( AB) = P ( r θ) = Pr (θ) But (P × r ) is equal to the torque ( T ). ). Therefore work done by a torque is equal to the torque ( T ) multiplied by the angular displacement (θ) in radians.

Fig. 31.1. Work done by a torque.

31.4. ANGULAR MOMENTUM

It is the total motion possessed by a rotating body and is expressed mathematically as : Momentum Mome ntum = Mass moment moment of inertia inertia × Angular Angular velocity velocity = I ω 31.5. NEWTON’S LAWS OF MOTION OF ROTATION

Following are the three Newton’s Laws of Motion of Rotation : 1. Newton’s First Law of Motion of Rotation states, “ Every body continues in its state of rest or of uniform motion of rotation about an axis, unless it is acted upon by some external torque”. 2.

Newton’s Second Law of Motion of Rotation states, “ The rate of change of angular momentum of a body is directly di rectly proportional to the impressed torque, and takes place i n the same direction in which the torque acts”.

3.

Newton’s Third Law of Motion of Rotation states, “ To every torque, there is always an equal and opposite torque.”

31.6. MASS MOMENT OF INERTIA

In chapter 7 we have discussed the moment of inertia of plane figures such as T -section, -section, I -section, -section, L-section etc. But in this chapter, we shall discuss the moment of inertia o f solid bodies or mass moment of inertia. The mass moment of inertia of a solid body about a line is equal to the product of mass of the body and square of the distance from that line. Mathematically :













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where m1, m2, m3 ...... are the masses of the various elements of a body and r 1, r 2, r 3 ...... are distances of the various elements from the fixed line about which the moment of inertia is required to be found out. In the following pages, we shall discuss the mass moment of inertia of the bodies, which are important from the subject point of view. 31.7. MASS MOMENT OF INERTIA OF A UNIFORM THIN ROD ABOUT THE MIDDLE AXIS PERPENDICULAR TO THE LENGTH

Fig. 31.2. Uniform rod.

Consider a uniform thin rod AB of length 2 l with O as its mid-point as shown in Fig. 31.2. m = Mass per unit unit length length of the rod. rod.

Let

∴ Total mass of the rod, M = 2 m l

Now consider a small strip of length dx at at a distance x from from the mid-point O. We know that the mass moment of inertia of the strip about O = (m dx ) x 2 = mx 2 dx

...(i)

The mass moment of inertia of the whole rod may be found out by integrating the above equation for the whole length of the rod i.e. from – l to + l. Therefore +l

+l

⎡ x3 ⎤ ⎡ (+ l )3 (– l )3 ⎤ 2 – I = ∫ mx dx = m ⎢ ⎥ = m ⎢ ⎥ 3 3 3 ⎣ ⎦ ⎣ ⎦ –l –l =

2ml 3 3

=

Ml 2

3

...(Q M = = 2ml)

31.8. MASS MOMENT OF INERTIA OF A UNIFORM THIN ROD ABOUT ONE OF THE ENDS PERPENDICULAR TO THE LENGTH

Fig. 31.3. Uniform rod.

Consider a uniform thin rod AB of length 2l as shown in Fig. 31.3. Let

∴

unit length length of the rod. rod. m = Mass per unit Total mass of the rod M = 2 ml

Now consider small strip of length dx at a distance x from from one of the ends (say A) as shown in













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The mass moment of inertia of the whole rod may be found out by integrating the above equation for the whole length of the rod i.e. from 0 to 2l. Therefore 2l

⎡ x3 ⎤ ⎡ (2l )3 ⎤ = = m x d x m m – 0⎥ ⎢ ⎥ ⎢ I = ∫ ⎣ 3 ⎦0 ⎣ 3 ⎦ 0 2l

=

2

8 ml 3

=

3

4 Ml 2

...(Q M = = 2ml)

3

31.9. MASS MOMENT OF INERTIA OF A THIN CIRCULAR RING

Consider a thin circular ring of radius r and and O as centre as shown in Fig. 31.4. m = Mass per unit length length of the ring. ring.

Let

∴ Total mass of the ring M = 2 π r m

Now consider a small element of length dx as shown in Fig. 31.4. We know that mass moment of inertia of the strip about th e central axis = (m dx ) r 2 The mass moment of inertia of the whole ring about central axis (i.e. at right angles to the plane) I = (2 π r m) r 2 = M r 2 ZZ

...(Q M = = 2 π r m) Now as per theorem of perpendicular axis, the mass moment of inertia about X-X or or Y-Y axis axis I XX = IYY =

I ZZ

2

2

=

M r

2

Fig. 31.4. Circular ring.

= 0.5 Mr 2

31.10. MASS MOMENT OF INERTIA OF A CIRCULAR LAMINA

Consider a circular lamina of radius r with O as centre. m = Mass per unit area of the lamina.

Let

∴

Total mass of the lamina

M = = π r 2 m

Now consider an elementary ring of thickness dx at at a radius x as as shown in Fig. 31.5. We know that mass of this thin ring x dx dx = m 2 π x

We know that the mass moment of inertia of a thin ring (of mass m.2π x .dx ) about central axis = (m 2 π x dx ) x 2 = m2 π x 3 dx The mass moment of inertia of the whole section ab out the central axis, may be found out by integrating the above equation for the whole radius of the circle, i.e. from 0 to r . Therefore













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A Textbook of Engineering Mechanics r

∫x

= m 2π

3

dx

0 r

⎡ x 4 ⎤ mπr 4 π = m 2 ⎢ ⎥ = 2 ⎣ 4 ⎦0 2

Mr

= 0.5 Mr 2

...(Q M = = πr 2m) 2 Now as per theorem of perpendicular axis, the mass moment of inertia about X-X or or Y-Y axis axis =

I XX = IYY =

I ZZ 2

2

=

0.5 M r 2

= 0.25 M r 2

Note: The above formula holds good for the mass, moment of inertia of a solid cylinder also. In this case, M is is taken as the mass of the solid cylinder. 31.11. MASS MOMENT OF INERTIA OF A SOLID SPHERE

Consider a solid sphere of radius r with with O as centre. m = Mass per unit volume volume of the sphere sphere

Let

∴

Total mass of the sphere M =

m 4 π r 3

3 Now consider an elementary plate PQ of thickness dx and and at a distance x from from O as shown in Fig. 31.6. We know that the radius of this plate y = r 2 – x 2

∴

Mass of this plate

= mπ y2 dx = = mπ (r 2 – x 2) dx

and moment of inertia of this plate about X - X axis axis

Mass × = Mass = mπ (r =

mπ

2

2

(Radius) 2 2 (r – x 2 ) dx ×

2

2 2

4

4

( r – x ) dx

mπ

2

2

– x )

Fig. 31.6. Sphere.

2

2 2

dx (r + x – 2 r x ) dx 2 The mass moment of inertia of the whole sphere may now be found out by integrating the above equation from – r to to + r . Therefore

=

+ r

I =

∫

mπ

– r

mπ

2

dx (r 4 + x 4 – 2r 2 x 2 ) dx

+ r

∫(

4

4

2

2 2

x )d













Contents

Chapter 31 : Kinetics of Motion of Rotation =

mπ ⎡

x5

4

– ⎢ r x + 2 ⎣ 5

2r 2 x 3 ⎤ 3

+ r

⎥ ⎦ – r

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⎛ m 4 πr ⎞ 2 Mr 2 2 M ... = ⎜ ⎟⎟ Mr 0.4 = = = ⎜ 3 15 5 ⎝ ⎠ Note: Since the sphere is symmetrical, therefore the mass moment moment of inertia of a sphere about any axis is the same. 3

8 m πr 5

31.12. UNITS OF MASS MOMENT OF INERTIA

We have already discussed that the mass moment of inertia of a body is nu merically equal to its mass and the square of distance between the centre of gravity of the mass and the point about which the mass moment of inertia is required to be found out. Therefore units of mass moment of inertia depend upon the mass of the body and distance. If mass is in kg and the distance in metres, then the units of mass moment of inertia will be kg-m 2. Similarly, it may be kg-mm 2 etc. 31.13. RADIUS OF GYRATION

If the entire mass of a given body be assumed to be concentrated at a certain point, at a distance k from the given axis, such that Mk 2 = I

∴

k =

...(where I is is the mass moment of inertia of the body)

I

m The distance k is is called radius of gyration. Thus the radius of gyration of a body may be defined as the distance from the axis of reference where the whole mass (or area) of a body is assumed to be concentrated. Y-Y Y are, The suffixes such as X-X or or Y are, usually attached to k , which indicate the axis about which the radius of gyration is evaluated. Thus k XX will indicate the radius of gyration about X-X axis. axis. In the following lines, we shall discuss the radius of gyration of some important sections : 1. Radius of gyration of a thin circular ring

We know that mass moment of inertia of a thin circular ring about the central axis, I = Mr 2 ZZ

Mk 2 = Mr 2

or

k = r ∴ 2. Radius of gyration of a circular lamina

We know that mass moment of inertia of a circular lamina about the central axis, 2

I = ZZ

∴

Mr

k =

2 r

2

2

2

M k =

or

2 k =

or

2

r

2

Mr

2 = 0.5 r 2

3. Radius of gyration of a solid sphere

We know that the mass moment of inertia of a solid sphere about any axis, I = 0.4 Mr 2

or

Mk 2 = 0.4 Mr 2















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31.14. KINETIC ENERGY OF ROTATION

We have already discussed in Art. 30.15 that energy is the capacity to do some work. Thoug h the energy exists in many forms, yet the kinetic energy is important from the subject point of view. The kinetic energy of rotation, possesed by a body for doing work by virture of its motion of rotation. Now consider a rotating body, which is brought to rest by a uniform angular retardation due to some torque. I = Mass moment moment of inertia inertia of the body, body, and

Let

velocity y of the body. body. ω = Angular velocit The *kinetic energy of rotation, 2

E =

I ω

2

r.p.m. Example 31.1. A circular wheel of mass 50 kg and radius 200 mm is rotating at 300 r.p.m. Find its kinetic energy. M ) = 50 kg; Radius of the wheel ( r ) = 200 mm = 0.2 m Solution. Given: Mass of the wheel ( M and angular velocity ( ω ) = 300 r.p.m.

We know that mass moment of inertia of the circular wheel, I = 0.5 Mr 2 = 0.5 × 50 (0.2)2 = 1 kg-m2

and angular velocity of the wheel, r.p.m. = 5 r.p.s. r.p.s. = 10 π rad/s ω = 300 r.p.m.

∴

Kinetic energy of the rotating wheel, E =

I ω2

2

=

1 (1 (10 π) 2 2

= 493.5 N-m = 493·5 J

31.15. TORQUE AND ANGULAR ACCELERATION

Consider a body rotating about its axis. Let

M = Mas Masss of the the body body

velocity ty of the body ω = Angular veloci acceleration ion of the body, body, and α = Angular accelerat T = Torque acting acting on the the body

Ans.













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Similarly, moment of the force for particles of mass m2 = m2r 22α

∴

Total moment for all the particles ( i.e. torque) T = Σ m1r 12α = α Σ m1r 12

But Σ m1r 12 is the* mass moment of inertia (if r is taken as the radius of gyration). Therefore T = I α 31.16. RELATION BETWEEN KINETICS OF LINEAR MOTION AND KINETICS OF MOTION OF ROTATION

Following are the relations between the kinetics of linear motion and kinetics of motion of rotation : S.No.

Linear motion

Motion of relation

1.

Mass ( M )

Moment of inertia ( I )

2.

Force (P)

Torque (T )

3.

Force equation ( P = M α)

Torque equation ( T = = I α)

4.

Linear motion ( v)

Angular motion ( ω )

5.

Linear momentum ( Mv)

Angular momentum ( I ω ω)

6.

Linear kinetic energy

Rotational kinetic energy

⎛ Mv2 ⎞ ⎜⎜ E = ⎟ 2 ⎟⎠ ⎝

⎛ I ω 2 ⎞ ⎜⎜ E = ⎟ 2 ⎟⎠ ⎝

7.

Distance traversed ( s)

Angular displacement ( θ)

8.

Work done (W) = Ps

Work done (W) = T × × θ

31.17. FLYWHEEL

It is a circular heavy wheel, generally, fitted to a rotating engine shaft to control variation in its speed during each cycle. Strictly speaking, it serves as a reservoir to store and restore energy by virtue of its inertia. Or in other words, it shares its energy during the period, when the supply of energy is more than the requirement; and releases it during the period when the supply













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A Textbook of Engineering Mechanics N 2 = Minim Minimum um speed of flywheel flywheel during during a cycle,

Maximum um angular velocity velocity of flywheel flywheel during a cycle, cycle, and ω 1 = Maxim Minimum um angular velocity velocity of flywheel during during a cycle. ω 2 = Minim We know that average speed of the flywheel, N =

1

( N1 + N 2 ) ω = =

2 and average kinetic energy of the flywheel, E =

∴

1 2

(ω1 + ω2 ) = 0.5 (ω1 + ω2 )

I ω 2

2 Fluctuation of energy, 2 2 I ω1 I ω 2 I – = (ω12 – ω 22 ) E = Maxim Maximum um K.E. – Minimu Minimum mK K.E. .E. = 2 2 2 I (ω1 + ω2 ) (ω1 – ω2 ) = I ω = ( ω 1 – ω 2) ...[Q ω = = 0.5 (ω 1 + ω 2)] ω ( 2

=

I × 2πN ⎛ 2π N1

60

⎜ 60 ⎝

–

2πN 2 ⎞ 4 π2 × I N ( N1 – N 2 ) ⎟ = 60 ⎠ 3600

π2

× I N ( N1 – N 2 ) . 900 lywheel heel of an engi ne has a mass m ass of 6.5 ton tonnes nes and radi us of gyra gyra-Example 31.2. A f lyw tion 1.8 metres. If the maximum and minimum speeds of the flywheel are 120 r.p.m. and 118 r.p.m. respectively, find the fluctuation of energy. =

Solution. Given: Mass of flywheel ( M ) = 6.5 t = 6500 kg; Radius of gyration ( k ) = 1.8 m: Maximum speed of flywheel ( N 1) = 120 r.p.m. and minimum speed of flywheel ( N 2) = 118 118 r.p.m. We know that average speed of the flywheel, N =

1 2

( N1 + N 2 ) =

1 2

(120 + 118) = 119 r.p.m.

and mass moment of inertia, = M k 2 = 6500 (1.8)2 = 21 060 kg-m2 I =

∴

Fluctuation of energy,













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We know that maximum speed of the wheel N 1 = 360 + ( 0.02 × 360) = 367.2 r.p r.p.m. .m.

and minimum speed of the wheel,

∴

N 2 = 360 – (0.02 × 360) = 352.8 r.p r.p.m. .m. Fluctuation of energy ( E ), ),

30 000 =

= ∴

I =

π2 900

π2 900

× I N ( N1 – N 2 ) × I × 360 (367.2 – 352.8) = 56.85 I

30 000 000 56.85

= 527.7 527.7 kg-m kg-m2

Ans.

(ii) Mass of the wheel We know that mass of the wheel, I 527.7 = 651. 651.5 5 kg M = 2 = Ans. k (0.9) 2 Example 31.4. A flywheel flyw heel of o f mass ma ss 8 tonnes starts from rest, and gets g ets up a speed of 180 1 80 r.p.m. in 3 minutes. Find the average torque exerted on it, if the radius of gyration of the flywheel is 60 cm.

Solution. Given: Mass of the flywheel ( M ) = 8 t = 8000 kg; Initial angular speed ( ω 0) = 0 (because it starts from rest); Final angular speed ( ω ) = 180 r.p.m. =

180 × 2 π

= 6π rad/s.; Time 60 (t ) = 3 min = 180 s and radius of the gyration of the flywheel ( k ) = 60 cm = 0.6 m. Let

Constantt angular accelerati acceleration on of the flywheel α = Constan

We know that the mass moment of inertia of the flywheel, I = M k 2 = 8000 × (0.6)2 = 2880 kg-m 2

and final angular velocity of the flywheel ( ω ), ), 6 π = ω 0 + αt = = 0 + α × 180

α =

or

∴

6π

= 0.105 0.105 rad/s rad/s 2

18 0 Average torque exerted by the flywheel.













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and mass moment of inertia,

I = Mr 2 = 395 × (1)2 = 395 kg-m2

We know that frictional couple (or torque T ) 100 = I α = 395 α

∴

α =

100

395 and final angular velocity of the flywheel,

= 0.253 0.253 rad/s rad/s2

0 = ω 0 – α t = = 10 π – 0.253 t

∴

t =

10π 0.253

= 124. 124.2 2s

...(Minus sign due to retardation)

Ans.

EXERCISE 31.1 1.

A wheel has a string of length 4 m wrapped round its shaft. The string is pulled with a constant force of 150 N. It is observed that when the string leaves the axle, the wheel is rotating at 3 revolutions in a second. Find the moment of inertia of the wheel. [Ans. 3.38 kg-m2]

2.

The flywheel of a steam engine of mass 1000 kg has radius of gyration as 1 m. If the maximum and minimum speed of the flywheel is 80 r.p.m. and 78 r.p.m. respectively, find the fluctuation of energy. [ Ans. 1732.7 N-m]

3.

A flywheel of an engine has a mass of 1250 kg and radius of gyration 600 mm. Find the angular acceleration of the wheel, when it is subjected to a torque of 12 500 N-m. [Ans. 27.78 rad/s2]

4.

A constant torque of 2 kN-m is exerted on a crankshaft to start the engine. Th e flywheel has a mass of 1800 kg and rad ius of gyration 1 m. If there is a resisting torque of 1 kN-m, find the speed of the engine after 1 minute. [ Ans. 320.9 r.p.m.] [Hint. Effective torque = 2 – 1 = 1 kN-m]

5.

A flywheel of mass 400 kg and radius of gyration 1 m losses its speed from 300 r.p.m. to 240 r.p.m. in 120 seconds. Determine the retarding torque acting on it.[ Ans. 20.8 N-m]

6.

A retarding torque of 600 N-m is applied to a flywheel rotating at 240 r.p.m. Find the moment of inertia of the flywheel, if it comes to rest in 100 seconds.[ Ans. 2390 kg-m2]

31.18. MOTION OF A BODY TIED TO A STRING AND PASSING OVER A















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First of all, consider the motion of the body, which is coming down. We know that the forces acting on it are m.g. (downwards) and P (upwards). As the body is moving downwards, therefore resultant force acting on it = mg – P

...(i)

Since the body is moving downwards, with an acceleration ( a), therefore force acting on it. = ma

...(ii)

Equating equations ( i) and (ii), mg – P = ma

..(iii)

Now consider motion of the pulley, which is rotating about its ax is due to downward motion of the body tied to the string. We know that linear acceleration of the body is equal to the angular acceleration of the pulley.

∴ and torque,

a = ra

...(iv)

string × Radius of the pulley T = Tension in the string = P × r

...(v)

We also know that torque on the pulley, T = I α

...(vi)

Equating equations ( v) and (vi), P × r = I α P r 2 = I α r

= I a

∴

P =

I a

...(Multiplying both sides by r ) ...(Q a = r α) ...(vii)













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diameter, whose axis Example 31.6. A homogeneous solid cylinder of mass 100 kg and 1 m diameter, is horizontal, rotates about its axis, in frictionless bearings under the action of a falling block of mass 10 kg, which is carried by a thin rope wrapped around the cylinder . What will be the angular velocity of the cylinder two seconds after the motion? Neglect the weight of the rope.

Solution. Given: Mass of cylinder ( M ) = 100 kg; Diameter of the cylinder ( D) = 1 m or radius ( r ) = 0.5 m; Mass of the block ( m) = 10 kg and time ( t ) = 2 s. We know that linear acceleration of the solid cylinder, a =

2m g 2m + M a

=

2 × 10 10 × 9. 9 .8 (2 × 10) + 100

2 = 1.63 1.63 m/s m/s

1.63

= 3.26 3.26 rad/ rad/ss 2 0.5 ∴ Angular velocity of the cylinder 2 seconds after the motion.

and angular acceleration,

α =

r

=

= 0 + (3.26 × 2) = 6.52 rad/s ω = ω 0 + αt =

Ans.

Example 31.7. A body of mass 6 kg is suspended by a light rope wound round a solid disc of 60 kg and diameter 50 cm, the other end of the rope being fixed to the periphery of the pulley. Find (i) acceleration of the descending mass, (ii) pull in the rope, and (iii) velocity after the mass has descended 15 m. Take g as 9.8 m/s 2.

Solution. Given: Mass of the body (m) = 6 kg; Mass of the solid disc ( M ) = 60 kg; *Diameter of the disc ( D) = 50 cm and distance ( s) = 15 m. (i) Acceleration of o f the descending descendi ng mass













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and kinetic energy of the wheel, I ω2

I (6π) 2

= 177.7 I N-m 2 2 Now equating the work done and kinetic energy, E =

=

...(ii)

177.7 I = 529. 529.2 2 529.2 Ans. = 2.98 .98 kg-m g-m 2 I = ∴ 177.7 Example 31.9. A solid cylindrical pulley of mass 800 kg, having 0.8 m , radius of gyration and 2 m diameter, is rotated by an electric motor, motor, which exerts a uniform unif orm torque of 60 kN-m. A body of mass 3 t is to lifted by a wire wrapped round the pulley. Find (i) acceleration of the body; and (ii) tension in the rope.

Solution. Given: Mass of pulley pulley ( M ) = 800 kg; Radius of gyration k d ( ) = 0.8 m; Diameter of pulley ( ) = 2 m or radius ( r ) = 1 m; Torque (T ) = 60 kNm = 60000 N-m and mass of the body to be lifted ( m) = 3 t = 3000 kg. (i) Acceleration of the body Let

a = Acceler Acceleration ation of the body, body,

acceleration on of the pulley, pulley, α = Angular accelerati P = Tens ension ion in rope. rope.

First of all, consider the motion of the hanging body, which is going upwards due to the torque. We know that forces acting on it are m g = 3000 × 9.8 = 29 400 N (downwards) and P (upwards). As the body is going upwards, therefore the resultant force

Fig. 31.9.













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(ii) Tension of the rope Substituting the value of a in equation ( iii), 3000 0 × 8.71 8.71 = 26 13 130 0 P – 29 400 = 300 P = 26 130 + 29 400 = 55 530 N = 55.53 55.53 kN

Ans.

31.19. MOTION OF TWO BODIES CONNECTED BY A STRING AND PASSING OVER A PULLEY

Consider two bodies connected by an inextensible light string and passing over a simple pulley as shown in Fig. 31.10. Let bodies, m1 and m2 = Masses of the two bodies, M = Mass of the pulley pulley,, I = Moment of inertia inertia of of the pulley, pulley, r = Radius of the pulley pulley,, k = Radius of gyration gyration of the the pulley, pulley, a = Acceler Acceleration ation of the the two bodies, bodies,

α = Angular accelera acceleration tion of the pulley pulley, and Fig. 31.10. Motion 31.10. Motion of two bodies P1 and P2 = Pull Pullss in the strings strings.. First of all, consider the motion of body 1 of mass m1, which is coming dow n. We We know that the forces acting on it are m1g (downwards) and P1 (upwards). As the body is moving downwards, therefore, resultant force P

...(i)













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Equating equations ( viii) and (ix ), ), (P1 – P2) r = I α (P1 – P2) r 2 = I αr = Ia (P1 – P2) =

∴

...(Multiplying both sides by r ) ...(∴ a = r α)

Ia r 2

Adding equations ( iii) and (vi) m1g – P1 + P2 – m2g = m1a + m2a g (m1 – m2) – ( P1 – P2) = a (m1 + m2)

Substituting the value of ( P1 – P2) in the above equation, I .a g (m1 – m2 ) – 2 = a (m + m ) 1 2 r ⎡ I ⎤ g (m1 – m2) = a ⎢ 2 + ( m1 + m2 ) ⎥ ⎣ r ⎦

∴

Nottes: 1. No

a =

g (m1 – m2 )

⎡ I ⎤ ⎢⎣ r 2 + ( m1 + m2 ) ⎥⎦

The values of pulls (or tensions) in the two strings ( P1 and P2) may now be found out from the equations ( iii) and (vi) i.e. P1 = m1g – m1a = m1 (g – a)

and

P

(

)















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Pulls on either side of the rope

We know that pull in the rope with 15 kg mass, P1 = m1 ( g – a) = 15 (9.8 – 3.82) = 89.7 N

Ans.

and pull in the rope with 5 kg mass, P2 = m2 ( a + g) = 5 (3.82 + 9.8) = 68.1 N

Ans.

Example 31.11. Two bodies A and B of masses 800 kg and 600 kg are attached at the ends of a flexible rope. The rope passes over a pulley of 800 mm diameter. The pulley has a mass of 100 kg with a radius of gyration as 400 mm about its axis of rotation. Find the torque, which must be applied to the pulley to raise the 800 kg body with an acceleration of 1 m/s2. Neglect friction of the spindle.

Solution. Given: Mass of the body A ( m1) = 800 kg; Mass of the body B ( m2) = 600 kg; Diameter of pulley = 800 mm = 0.8 m or radius ( ) = 0.4 m; Mass of the pulley ( M ) = 100 kg; Radius













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Chapter 31 : Kinetics of Motion of Rotation Let

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P = Tension in the rope a = Linear accelera acceleration tion of the train, and acceleration tion of the the drum. α = Angular accelera

First of all, consider the motion of the cage, which is being pulled by the rope due to the torque on the drum.













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Example 31.13. Two bodies A and B of masses 30 kg and 10 kg are tied to the two ends of a light string passing over a composite pulley of radius of gyration as 70 mm and mass 4 kg as shown in Fig. 31.12.













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Now consider the mass 10 kg, which is going upwards. We know that the forces acting on it are m2 g = 10 × 9.8 = 98 newtons (downwards) and P2 newtons (upwards). As the body is moving upwards, therefore, resultant force = P2 – 98 body

...(iv)

Since the mass is moving upwards with an acceleration ( a2), therefore force acting on the = 10 a2 Equating equations ( iv) and (v), P2 – 98 = 10 a2

...(v) ...(vi)

Now consider the motion of the pulley, which is rotating about its axis due to downward motion of the 30 kg mass tied to the string. We know that the linear acceleration of the body is equal equ al to the angular acceleration of the pulley.

∴

a1 = r 1α = 0.05 α













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Example 31.14. Two bodies A and B, of mass 150 kg and 75 kg respectively are supported by a string of negligible mass and pass over a composite pulley. The bodies rest on two smooth inclined planes as shown in Fig. 31.13.















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Since the mass is moving upwards with an acceleration ( a2), therefore force acting on the body 75

...( )













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A Textbook of Engineering Mechanics EXERCISE 31.2

1.

A body of mass 20 kg is suspended by a light string wound round a pulley of mass 50 kg.













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Chapter 31 : Kinetics of Motion of Rotation Let

m = Mass of of the hanging body A, P = Tension in in the string, string, M

Mass of of the rollin rolling g body B

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Now equating the couple (responsible for rolling) and torque on the body, × r = I α F × 2

...(Multiplying both sides by )













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31.21. MOTION OF A BODY ROLLING DOWN A ROUGH INCLINED PLANE WITHOUT SLIPPING







































chp-31

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