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Engineering › Theory Of Machines ›

General Dynamical Problems

A series of Dynamical Problems that come within the study of machines.

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Contents

Introduction
Force, Mass And Momentum
Moment Of Inertia, Angular Motion
Motion Under Variable Acceleration.
Kinetic Energy
Impact, Impulse.
Page Comments

Introduction

In the study of Machines there are a number of problems that do not fit neatly into a specific topic, but which at the same time are essential to the understanding of machines. In addition, some notations have multiple applications and are useful in deriving solutions to common spectrum of problems associated with machines. Such problems have been grouped together in this section.

Force, Mass And Momentum

The force

required to give an acceleration

to the centre of gravity of a mass

is obtained from Newton 's Second Law:

(1)

is the weight of the body (i.e. the force exerted by gravity upon it) then:

$W=Mg\;\;\;$ or $\;\;\;M=\displaystyle\frac{W}{g}$

A force is any influence that causes a free body to undergo a change in speed, a change in direction, or a change in shape.

Centre of gravity is the point in or near a body at which the gravitational potential energy of the body is equal to that of a single particle of the same mass located at that point and through which the resultant of the gravitational forces on the component particles of the body acts.

From this it follows that:

Imperial Units

If the unit of force is to be , and the acceleration is measured in , then the unit of mass is ; sometimes called a slug.
i.e. A force of $1\;lb.wt.$ acting on a mass of $1\;slug$ will produce an acceleration of $1\;ft/sec.^2$
If other units are used ( or ) then it is better to use the form $F=\displaystyle\frac{W}{g}\;a$ and substitute values in consistent units throughout.

Mks Units

The unit of Force is Newton and the unit of mass Kilogram or Kilo
Thus a force of 1 Newton acting on a mass of 1 Kilo will produce an acceleration of $1\;metre/sec.^2$

Conversion Of Units

A complete list of the corresponding units in the Imperial and MKS systems is given in "Engineering General". Here it is simply mentioned that:

A force of 1 lb = 4.45 Newtons.

The product Mv is called the Momentum of the body and the general form of Newton's second Law (which has to be used when the mass is varying) is:

The rate of change of Momentum=

$F=\left ( \frac{d}{dt} \right )M\,v}$

(2)

Note that

, and

are all vector quantities and must all be measured in the same direction.

Moment Of Inertia, Angular Motion

The Moment of Inertia

of a body about a given axis is the sum of the products of mass and distance squared for all the particles of the body, i.e.

$I=\int r^2\;dm=Mk^2$

is called the Radius of Gyration and $M=\displaystyle\frac{W}{g}$
The units of are slugs. : i.e. or or in any consistent units.
The following Table gives and for a number of common shapes.

It must be stressed that all the shapes shown above are of uniform thickness and composition, and that the Moments of Inertia are about the axis shown.
Where it is necessary to know the Moment of Inertia about an axis of distance from the Centre of Gravity,

$k^2=k{_{g}}^{2}+h^2$

This is known as the Parallel Axis Theorem.

is the radius of gyration about a parallel axis through the centre of gravity.

The product $I_g\omega$ is called the Angular Momentum about an axis through the centre of gravity. It is also known as the spin couple.

The total Moment of Momentum about any other axis is given by:

$I_g\omega+Mv_gh$

where

is the perpendicular distance from the axis onto the line of action of

About an axis of rotation

, $v_g=h\omega$ and the Moment of Momentum reduces to

$\left ( I_g+Mh^2 \right )\omega=I_o\omega$

(3)

The equation of Angular Motion is: Moment of Forces = Rate of Change of Angular Momentum

$\left ( \frac{d}{dt} \right )\left ( I\;\omega \right )=I\alpha$

(4)

(For a constant

)
This can be applied about any fixed axis of rotation, or about an axis through the centre of gravity.

Motion Under Variable Acceleration.

There are many problems in which the resultant force(or torque) acting on a body is not constant and consequently the acceleration produced will vary. General methods of solution are given below and are applicable to linear and angular motion by interchanging $\theta$ for

and $\omega$ for

Acceleration is the rate of change of velocity as a function of time. It is vector and it is the second derivative of position with respect to time or, alternately, the first derivative of the velocity with respect to time.

It is assumed that expressions for acceleration have been obtained by substituting into equations (1) or (4).

Acceleration As A Function Of Distance.

Since $a=\displaystyle\frac{dv}{dt}= \left ( \displaystyle\frac{dv}{ds} \right )\left ( \displaystyle\frac{ds}{dt} \right )=v\;\displaystyle\frac{dv}{ds}$

Then,

$\int a.ds=\frac{dv}{dt}= \int v.dv+A=\frac{1}{2}v^2+A$

(5)

Since $a=\displaystyle\frac{F}{M}$ (This corresponds to the equation of energy)

$\displaystyle\int F.ds=\displaystyle\frac{1}{2}M(v_2^2-v_1^2)$ (Between limits)

i.e. The Work done = The gain in Kinetic Energy

Note. A further integration of equation (5) can be carried out by writing $v=\displaystyle\frac{ds}{dt}$ . This allows the time interval to be determined. ( See Example 6)

Acceleration As A Function Of Time

Writing $a=\displaystyle\frac{dv}{dt}$ and re-arranging

$v=\int a\;dt+B$

(6)

Which corresponds to the momentum equation:

$M(v_2-v_1)=\int F.dt$

i.e. The change of Momentum = Impulse.

A further integration of equation (6) will give the distance traversed.

Acceleration As A Function Of Velocity.

Writing $a=\displaystyle\frac{dv}{dt}$ and re-arranging, $t=\displaystyle\int \displaystyle\frac{dv}{a}+C_1$ (See Example 4)

or $a=v\;\displaystyle\frac{dv}{ds}$

Which gives $s=\displaystyle\int v\;\displaystyle\frac{dv}{a}+C_2$ (See Example 5)

Power is the rate of work. In the f.slug.sec. system 1 horse-power (h.p.) is 550 ft.lb./sec. 0r 33,000 ft.lb./min.

is the force (or resolved part) in the direction

and it moves its point of application a distance

, then,
Work done= $F\;x$

Similarly, the work done by a Couple $\tau$ turning through an angle $\theta =\tau\; \theta$

Kinetic Energy

The kinetic energy of an object is the energy which it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.

The energy possessed by a body is a measure of its capacity to do work. If a body has an angular velocity $\omega$ and its centre of gravity has a linear velocity

, then its total Kinetic Energy (K.E.) is given by:

$\frac{1}{2}Mv^2+\frac{1}{2}I_g\omega^2=\frac{1}{2}M\left ( v^2+k_g^2\omega^2 \right )$

If the body is rotating about a fixed axis

and since $v=h\omega$ and $k{_{0}}^{2}=k{_{g}}^{2}+h^2$

the above equation is: $K.E.=\displaystyle\frac{1}{2}I_o\omega^2$

Impact, Impulse.

When two bodies collide, each exerts an equal force on the other and for the same period of time. By the integration of equation (2)

$\int F.dt=M(v_2-v_1)$

i.e., for one body the impulse force is equal to the change of momentum. It follows that in a closed system of bodies the total Momentum remains constant (i.e.The Conservation of Momentum).

Similarly the moment of the impulse force about a fixed axis ( or axis through the centre of gravity) is equal to the change of angular momentum about that axis.

Example:

[imperial]

Example - Example 1

Problem

The speed of a car on the level is 40 m.p.h., the engine indicating 25 h.p. and the weight of the car being 2,500 lb. The car will just run down a gradient of 1 in 25 when the clutch is in but ignition cut off.

Assuming the engine and transmission friction and the road resistance to be independent of speed and the wind resistance to be proportional to the square of the speed, determine the horse-power required to drive the car up a gradient of 1 in 25 at a speed of 25 m.p.h.

Workings

Let the friction and road resistance be

lb.and the wind resistance be

(

is in

)

Then, Total resistance

velocity = 550 X h.p

On the level,

$\left [ F+C\left ( 40\times \displaystyle\frac{88}{60} \right )^2 \right]40\times\displaystyle\frac{88}{60}= 550 \times 25$

$\therefore \;\;\;\;\;F+3430C=234$

(1)

Down the gradient, $F-\displaystyle\frac{2500}{25}=0$ (Zero velocity)

$\therefore \;\;\;\;\;F=100\;lb.$

(2)

From equations (1) and (2), $C=\displaystyle\frac{134}{3430}=0.392$

Up Gradient, $h.p.=\left [ F+C\left ( 35\times\displaystyle\frac{88}{60} \right )^2+\displaystyle\frac{2500}{25} \right ]\times \displaystyle\frac{35\times 88}{60\times 550}$

Substituting in the value for

$h.p.=(100+104+100)\times 0.0934=28.4\;h.p.$

Solution

The horse-power required is $28.4\;h.p.$

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