J3010 Unit 3

HOISTING J3010/3/1

UNIT 3

HOISTING

OBJECTIVES

General Objective : To understand the concept of dynamics of rotation.

Specific Objectives : At the end of this unit you should be able to:

 apply basic principle on which all these machines are based.

 recognize the effect of combining a hoist drum of moment inertia
I with a hanging load of mass M and weight W = mg.

 sketch and recognize all force and torque that involve on these
machines.

 use suitable concepts to solve related problem.

 calculate all these topic questions correctly.

HOISTING J3010/3/2

INPUT

3.0 INTRODUCTION

We study the effect of combining a hoist drum of moment of inertia with a
hanging load of mass M and Weight W = Mg.

In this topic we are
concerned in dynamics of
rotation.

3.1 INERTIA COUPLE

Comparing the formulae P = Ma and T = Iα , it is seen that moment of inertia
I plays the same part in a change of angular motion as mass M does in
change of linear motion. By analogy with the idea of inertia force we may
regard the torque T as being balanced by inertia couple, Iα , which sense is
opposite to that of the angular acceleration α, (Fig. 3.1). The problem is then
in effect reduced to a static one.

The reality of the effect of an inertia couple will be appreciated by anyone
who has tried to accelerate a bicycle wheel rapidly by hand. Although the
weight may carried wholly by the bearings an effort is required to set the
wheel spinning. An inertia couple is, of course, reactive.

HOISTING J3010/3/3

Fig. 3.1

Example 3.1

A 30 kg flywheel, revolving at 5.24 rad/s has a radius of gyration of one
meter. Calculate the torque which must be applied to bring the flywheel to
rest in 10 seconds.

Solution 3.1

Moment of inertia of the flywheel, I  mk 2

 30x12

I  30kgm2

Deceleration, 1  0  t

0  5.24  ( )10

  0.524rad / s 2

Torque, T  I

 30x0.524

T  15.72 Nm

HOISTING J3010/3/4

3.2 ACCELERATED SHAFT
Consider a shaft (Fig. 3.2) carrying a rotor having a moment of inertia about
the shaft axis.

Fig. 3.2

If the bearing friction is equivalent to a couple Tf .
Then, in order to accelerate the shaft and rotor the driving torque T must
balance both the inertia couple I  and the friction couple Tf .

Thus,

T  I  T f

Example 3.2

A flywheel has a moment of inertia of 10 kg.m2 . Calculate the angular
acceleration of the wheel due to a torque of 8 Nm if the bearing friction is
equivalent to a couple of 3 Nm.

Solution 3.2:

Given: I = 10 kgm2 T = 8 Nm Tf = Iα = 3 Nm

T  I  T f

I  T  T f

I  8  3 Nm

I  5 Nm

5
 rad/s2
10

HOISTING J3010/3/5

3.3 SHAFT BEING BROUGHT TO REST

If the shaft is being brought to rest by a braking torque T the friction couple
Tf assist the braking action so that T and Tf together must balance the inertia
couple I  ;  is now a retardation its sense being opposite to that of the
motion (Fig.3.3).

Fig. 3.3

Thus,

T  T f  I

If there is no braking torque, the friction couple alone brings the shaft to rest.
Then,
T f  I

Note, in both cases, that
(a) the friction couple T f opposes the motion.
(b) the inertia couple I opposes the change of motion.

HOISTING J3010/3/6

Example 3.3

A flywheel together with its shaft has a total mass of 300 kg and its radius of
gyration is 900 mm. If the effect of bearing friction is equivalent to a couple
of 70 Nm, calculate the braking torque required to bring the flywheel to rest
from a speed of12 rev/s in 8 s.

Solution 3.3

Given: N = 12 rev/s = 12 x 2π = 75.4 rad/s


Thus, Retardation, α=
t

75.4
=
8

= 9.42 rad/s2

I of flywheel and shaft = Mk2

= 300 x 0.92

= 243 kg m2

Inertia Couple = Iα

= 243 x 9.42

= 2290 Nm

HOISTING J3010/3/7

Activity 3A

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE
NEXT INPUT…!

3.1 A drum rotor has the moment of inertia 31.8 kg.m2 . Find the time taken for
the rotor to reach a speed of 3600 rev/min from rest if the driving force
torque is 55 Nm and the friction torque is 5 Nm.

3.2 The rotating table of the vertical boring machines has a mass of 690 kg and a
radius of gyration of 700 mm. Find the torque required to accelerate the table
to 60 rev/min in three complete revolutions from rest.

3.3 A light shaft carries a turbine rotor of mass 2 tonnes and a radius of gyration
of 600 mm. The rotor required a uniform torque of 1.2 kNm to accelerate it
from rest 6000 rev/min in 10 min. Find:

(i) the friction couple,
(ii) the time taken to come to rest when steam is shut off.

3.4 A winding drum of mass 200 tonne has a radius of gyration of 3 m. Find the
constant torque required to raise the speed from 40 to 80 rev/min in 60
seconds if the friction torque is 15 kNm. If the wheel is rotating freely at 80
rev/min and a brake is applied bringing it to rest in 120 rev. Find the brake
torque assuming uniform retardation.

HOISTING J3010/3/8

Feedback To Activity 3A

Have you tried the questions????? If “YES”, check your answers now

3.1 240s

3.2 354 Nm

3.3 (i) 446 Nm, (ii) 16.9 min

3.4 140.5 kNm, 68.8 kNm.

HOISTING J3010/3/9

INPUT

THE HOIST

Four cases will be considered: the load is rising or falling, being accelerated or
brought to rest. In every case two equations can be written down:
(a) the equation for the balance of couples at the hoist drum
(b) the equation for the balance of forces at the load.

In every case we recall that the friction couple at the bearing or rope will oppose the
rotation and the inertia couple will oppose the change of rotation.
For acceleration, a is upwards, hence the inertia force is downwards. If angular
acceleration,  is anticlockwise, the inertia couple is clockwise. If rotation of the
drum is anticlockwise the friction couple acts clockwise.

3.4 LOAD RAISED AND WITH ACCELERATING UPWARD

For rotation of the hoist drum the driving torque T must balance the friction
couple Tf, , the inertia couple I and the torque Pr due to the tension P in the
rope at the drum.

Fig. 3.4

HOISTING J3010/3/10

Thus,

Angular Motion:

T = I + Pr + Tf

For Linear motion of the load, the tension P in the rope at the load must
balance both the dead weight and the inertia force Ma.

Thus,

Linear Motion:

P = Mg + Ma

Example 3.4

A hoist drum has a moment of inertia of 85 kgm2 and is used to raised a lift of
mass 1 tonne with an upward acceleration of 1.5 m/s2. The drum diameter is
1 m.
Determine:
(a) the torque required at the drum
(b) the power required after accelerating for 3 seconds from rest.

Solution 3.4

(a) The torque required at the hoist drum is made up of three parts.

1. torque I required to accelerate the drum
2. torque Wr required to hold the dead weight of the lift.
3. torque Mar required to accelerate the lift.

M = 1000kg W = Mg I = 85 kgm2
= 1000 x 9.8
= 98000 N

a 1.5
=   3rad / s 2
r 0.5

HOISTING J3010/3/11

Thus,

Total torque = I + Wr + Mar

= (85 x 3) + (9800 x 0.5) + (1000 x 1.5 x 0.5)

= 5905 Nm

(b) After 3 seconds, the lift speed.

v = at

= 1.5 x 3

= 4.5 m/s (This is the speed of the drum circumference)

Therefore angular velocity of the drum,

v 4.5
   9rad / s
r 0.5

Power required = torque x angular velocity

=T 

= 5905 X 9

=53.15 Kw

This the power required at the instant after 3 seconds.

HOISTING J3010/3/12

3.5 LOAD FALLING AND ACCELERATING DOWNWARDS (NO
DRIVING TORQUE ACTING)

The load is allowed to fall freely, resisted only by friction and inertia force
and couples. The rotation of the hoist drum, the accelerating torque Pr due to
rope tension must balance both the friction couple Tf and the inertia couple
I.

Fig. 3.5

Thus,

Angular Motion:

Pr = Tf + I

For linear motion of the load the accelerating force due to the weight must
balance the upward tension P in the rope and the inertia force Ma.

Thus,

Linear Motion:

P = Mg - Ma

HOISTING J3010/3/13

Example 3.5

A hoist drum has a mass of 360 kg and a radius of gyration of 600 mm. The
drum diameter is 750 mm. A mass of 1 tonne hangs from a light cable
wrapped round the drum and is allowed to fall freely. If friction couple at the
bearings is 2.7 k Nm. Calculate the runaway speed of the load after falling
for 2 seconds from rest.

Solution 3.5

Given:

Md = 360 kg k = 600 mm = 0.6 m

Dd = 750 mm = 0.75 m Rd = 0.375 m Tf = 2700 Nm

M = 1 tonne = 1000 kg ω0 = 0 rad/s t=2s

I = Mk2 = 360(0.6)2 = 129.6 kgm2

Linear Motion: P = Mg - Ma = M(g - a) a = rα

P = 1000(9.81 - rα) = 9810 - 375α

Angular Motion:

Pr = Tf + I = 2700 + 129.6

2700  129.6
P=
0.375

 = 3.62 rad/s2

Then, ω1 = ω0 + t = 0 + 3.62 (2)

= 7.24 rad/s

v = r ω1
= (0.375) 3.62
= 2.71 m/s

HOISTING J3010/3/14

3.6 LOAD FALLING AND BEING BROUGHT TO REST

We now consider the braking of the hoist drum as the load falls. The
accelerations are therefore reversed as compared with the previous case. For
rotation of the drum the braking torque T is assisted by the friction couple to
balance the accelerating torque Pr due to the rope tension and the inertia
couple I.

Fig. 3.6

Thus,

Angular Motion:

T + Tf = Pr + I

Linear Motion:

P = Mg + Ma

Example 3.6

The maximum allowable pull in a hoist cable is 200 kN. Calculate maximum
load in tones which can be brought to rest with a retardation of 5 m/s2 . The
hoist drum has a moment of inertia of 840 kgm2 and a diameter of 2.4 m.
What is the corresponding braking torque on the drum?

HOISTING J3010/3/15

Solution 3.6

Given: Pmax = 200 x 103 Nm a = - 5 m/s2

I = 840 kgm2 Dd = 2.4 m

rd = 1.2 m

a = rα

a 5
α= = = 4.16 rad/s2
r 1 .2

Linear Motion:
P = Mg + Ma

200 x 103 = M (g + a)

= M (9.81 + 5)

200 x10 3
M = kg
14.81

M = 13.5 tonne

Angular Motion:

T + Tf = Pr + I

T = Pr + I - Tf = 200 x 103 (1.2) + 840 (4.16) – Tf

There is no braking torque, Tf = I

T = 200 x 103 (1.2) + 3494.4 – 3494.4

T = 240 k Nm

HOISTING J3010/3/16

3.7 LOAD RISING: COMING TO REST UNDER FRICTION ONLY

Since there is no braking torque applied, the drum is retarded by the torque Pr
due to the rope tension and the friction couple Tf . These two couples must
balance the inertia couple of the drum.

Fig. 3.7

Thus,

Angular Motion:

I = Tf + Pr

Linear Motion:

P + Ma = Mg

HOISTING J3010/3/17

Example 3.7

In an experiment, a hoist drum has a diameter is 500 mm. It is used to raised
load 50 kg and coming to rest under friction. The upward acceleration is 3.0
m/s2 . The friction couple is 0.35 Nm. Find the moment of inertia of the
drum.

Solution 3.7
a 3
a = rα α= α= α = 12 rad/s2
r 0.25

For the linear motion, P + Ma = Mg

P = Mg – Ma

P = M (g - a)

P = 50 (9.81 – 3)

P = 340.5 N

For the angular motion, I = Tf + Pr

I = 0.35 + 340.5 (0.25)

I = 0.35 + 85.12

I = 85.47

85.47
I =
12

I = 7.12 kgm2

NOTE:

Students are required to grasp firmly the following rules:

1. the friction couple opposes the rotation
2. the inertia couple opposes the change of rotation
3. the inertia force opposes the change of linear motion.

It may remarked also that in every case the direction of the rope tension P
and the load weight W is unaltered, although their effect may be to accelerate
or to retard the load.

HOISTING J3010/3/18

Activity 3B

NEXT INPUT…!

3.5 A load of mass 8 tonne is to be raised with a uniform acceleration of 1.1 m/s2
by means of a light cable passing over a hoist drum of 2 m diameter. The
drum has a mass of 1 tonne and a radius of gyration of 750 mm. Find the
torque required at the drum if friction is neglected. What is the power exerted
after 4 seconds from rest.

3.6 A mine cage of mass 4 tonne is to be raised with an acceleration of 1.5 m/s2
using a hoist drum of 1.5 m diameter. The drum’s mass is 750 kg and its
radius of gyration is 600 mm. The effect of bearing friction is equivalent to a
couple of 3 kNm at the hoist drum. What is the power required when the load
has reached a velocity of 6 m/s? What is the power required at a uniform
velocity of 6 m/s?.

3.7 A hoist has a winding drum 0.9 m effective diameter and a radius of gyration
of 0.35 m, the mass of the drum being 100 kg. A load of 320 kg is to be
raised 36 m, the mass of the lifting rope being 1 kg/m. If the acceleration is
1.8 m/s2 until a constant velocity of 6 m/s is reached, find the power
necessary just at the end of the acceleration.

3.8 A winding drum raises a cage of mass 500 kg through a height of 120 m. The
winding drum has a mass of 250 kg and an effective radius of 0.5 m and a
radius of gyration of 0.36 m. The mass of the rope is 3 kg/m. The cage has at
first an acceleration of 1.5 m/s2 until a velocity of 9 m/s is reached after hich
the velocity is constant until the cage nears the top, when the final retardation
is 6 m/s2 . Find :
(i) the time taken for the cage to reach the top
(ii) the torque which must be applied to the drum at starting
(iii) the power at the end of the acceleration period.

HOISTING J3010/3/19

Feedback To Activity 3B


3.5 87.8 kN m, 386.4 kW

3.6 299.5 kW, 259 kW

3.7 25.45 kW

3.8 (i) 17.08 s (ii) 4957 Nm (iii) 82.2 kW

HOISTING J3010/3/20

INPUT

3.8 LOAD BALANCING SYSTEM

We shall now consider some simple cases of the motion of two masses
connected by a light inextensible string. We note that a string connecting two
masses in motion is in a state of tension and that the string exerts forces on
the masses equal to the tensions at its ends.

If the string is light (that is, if its weight is neglected) the tension is the same
throughout its length. On the other hand, if the string is heavy the tension will
in general vary from point to point, depending upon the weight per unit
length. If the string is extensible the tension will vary with the extension.

Also, if the string passes round a pulley the tension is only the same on the
two sides if the pulley is smooth and the string is light. Otherwise the tension
in the string where it leaves the pulley depends upon the coefficient of
friction and the length of string in contact. In such an ideal case the tension
throughout the string will be constant.

Example 3.8

A load of mass 230kg is lifted by means of a rope which is wound several
times round a drum and which then supports a balance mass of 140 kg. As
the load rises the balance mass falls. The drum has a diameter of 1.2 m and a
radius of gyration of 530 mm and its mass is 70 kg. The frictional resistance
to the movement of the load is 110 N, and that to the movement of the
balance mass 90 N. The frictional torque on the drum shaft is 80 Nm.

Find the torque required on the drum, and also the power required at an
instant when the load has an upward velocity of 2.5 m/s and an upward
acceleration of 1.2 m/s2 .

HOISTING J3010/3/21

Solution 3.8

1.2 m/s2

T1 T2

230 kg 140 kg

Fig. 3.8.

Let T1 and T2 be the tensions in the rope (Fig. 3.8).

Then, T1  m1 g  m1a  R f (frictional resistance)

T1  230 x9.81  230 x1.2  110

and T2  m2 g  m2 a  mb (balance mass)

T2  140 x9.81  140 x1.2  110

 T1  T2  1527 N

Torque to accelerate loads = 1527 x 0.6
= 916.2 Nm
1 .2
Torque to accelerate drum = 70 x 0.532 x + 80
0 .6
= 119.3 Nm

Total torque, = 916.2 + 119.3
= 1035.5 Nm

2.5
Power, = 1035.5 x W
0. 6
= 4.32 kW

HOISTING J3010/3/22

Activity 3C

NEXT INPUT…!

3.9 Two particles of masses m1 and m2 kg are connected by a light inextensible
string passing over a small smooth fixed pulley. Find the resulting motion of
the system and the tension in the string.

3.10 Two particles of masses 6 and 10 kg are connected by a light string passing
over a smooth pulley.
(i) their common acceleration
(ii) the tension in the string
(iii) the force on the pulley.

3.11 A lift of mass 900 kg is connected to a rope which passes over a drum of 1 m
diameter and a balance mass of 450 kg is attached to the other end of the
rope. The moment of inertia of the drum is 100 kgm2 and it is driven by a
motor through a reduction gear (25 to 1) of 90 percent efficiency. Neglecte
the inertia of the gears, calculate the motor torque for a lift acceleration of 3
m/s2. If the maximum output of the motor is 15 kW, what will be the
maximum uniform speed of the lift?

HOISTING J3010/3/23

Feedback To Activity 3C


2m1 m2
3.9 T g
m1  m2

1
3.10 (i) 2.45 ms-2 (ii) 7 gN (iii) 15 g N
2

3.11 215 Nm, 3.06 m/s

HOISTING J3010/3/24

SELF-ASSESSMENT 3

You are approaching success. Try all the questions in this self-assessment section
and check your answers with those given in the Feedback on Self-Assessment 3
given on the next page. If you face any problems, discuss it with your lecturer.
Good luck.

1. The rotor of an electric motor of mass 200 kg has a radius of gyration of 150
mm. Calculate the torque required to accelerate it from rest to 1500 rev/min
in 6 seconds. Friction resistance may be neglected.

2. The flywheel of an engine consists essentially of a thin cast-iron ring of mean
diameter 2 m. The cross-section of the ring is 50 mm by 50 mm. Calculate
the moment of inertia of the flywheel and find the change in speed of the
flywheel if a constant torque of 110 Nm acts on it for 5 seconds. Density of
cast iron = 7850 kg/m3 .

3. The flywheel of an engine has a mass of 80 kg and a radius of gyration of
220 mm. If the engine rotating parts have a moment of inertia of 4.2 kg m2 .
Find the torque necessary to accelerate the engine and flywheel from rest to
1500 rev/min in 20 seconds. Assume a constant friction torque of 6 Nm.

4. A loaded mine skip has a mass of 13.5 Mg. When at rest at the bottom of the
pit, it is supported by a length of 1.5 km of rope of mass 10 kg/m. The
moment of inertia of the rotating parts of the winding gear is 500 Mg m2 and
the radius of the winding drum is 2.4 m. During winding, the skip has a
constant acceleration of 0.9 m/s2 at the beginning and the same retardation at
the end, with a constant velocity of 15 m/s between these periods. Find the
maximum power required from the driving motor.

HOISTING J3010/3/25

Feedback to Self-Assessment 3


1. 118 N m

2. 123.5 kgm2, 42.4 rev/min

3. 69.4 Nm

4. 5.36 MW.

If all your answers are
correct,
CONGRATULATIONS!!!!….

J3010 Unit 3

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