# General Dynamical Problems

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

**Contents**

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

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 acting on a mass of will produce an acceleration of
- If other units are used ( or ) then it is better to use the form and substitute values in consistent units throughout.

- If the unit of force is to be , and the acceleration is measured in , then the unit of mass is ; sometimes called a

### 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

- The unit of Force is

### 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= 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.- is called the
**Radius of Gyration**and - 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,

This is known as the

**Parallel Axis Theorem**. is the radius of gyration about a parallel axis through the centre of gravity.The product 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: where is the perpendicular distance from the axis onto the line of action of . About an axis of rotation , and the Moment of Momentum reduces toThe equation of

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

**Angular Motion**is: Moment of Forces = Rate of Change of Angular Momentum (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 for and 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.

### Acceleration As A Function Of Distance.

- Since Then, Since (This corresponds to the equation of energy) (Between limits) i.e. The Work done = The gain in Kinetic EnergyNote. A further integration of equation (5) can be carried out by writing . This allows the time interval to be determined. ( See Example 6)

### Acceleration As A Function Of Velocity.

- Writing and re-arranging, (See Example 4) or Which gives (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.

If is the force (or resolved part) in the direction and it moves its point of application a distance , then,

Work done= Similarly, the work done by a Couple turning through an angle

Work done= Similarly, the work done by a Couple turning through an angle

## 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 and its centre of gravity has a linear velocity , then its total

If the body is rotating about a fixed axis and since and
the above equation is:
**Kinetic Energy (K.E.)**is given by:## 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)
i.e., for one body the

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.
**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**).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,
Down the gradient, (Zero velocity)
From equations (1) and (2),
Up Gradient,
Substituting in the value for ,

Solution

- The
**horse-****power**required is

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