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This section covers a series of Dynamical Problems which come within the study of machines

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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. Such problems have been grouped together in this section.

Force, Mass and Momentum

The force F required to give an acceleration a to the centre of gravity of a mass M is obtained from Newton's Second Law;-

$\boldsymbol{F=Ma}$

(1)

If W is the weight of the body (i.e. the force exerted by gravity upon it} then:-

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

(2)

From this it follows that:-

Imperial Units

If the unit of force is to be lb. wt and the acceleration is measured in $\boldsymbol{ft./sec^2}$ and the unit of mass is $\boldsymbol{lb.wt./32.2}$ which 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 $\boldsymbol{1 ft/sec.^2}$

If other units are used (tons or ${ in./sec.^2}$ ) then it is better to use the form $F=(W/g)\;a$ and substitute values in consistent units throughout.

MKS units

The unit of Force is the 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 just 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:-

$\boldsymbol{F=\text{The rate of change of Momentum}\;=\left ( \frac{d}{dt} \right )M\,v}$

(3)

Not that F, v, and a are all vector quantities and must all be measure in the same direction.

Moment of Inertia, Angular motion

The Moment of Inertia I 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.

$\boldsymbol{I=\int r^2.dm=Mk^2}$

(4)

k is called the Radius of Gyration and $\boldsymbol{M=W/g}$

The units of I are slugs.ft^2: i.e. $lb.ft.^2/g\;\;\;or\;\;\;lb.ft.sec^2\;\;\;or\;\;\;Wk^2/g$ in any consistent units.

The following Table gives $k^2"$ and I 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 is is necessary to know the Moment of Inertia about an axis of distance h from the Centre of Gravity,. Then:-

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

(5)

This is known as The Parallel Axis Theorem. $k_g$ 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:-

$I_g\omega+Mv_gh$

(6)

where h is the perpendicular distance from the axis onto the line of action of $v_g$

About an axis of rotation O, $v_g=h\omega$ and the Moment of Momentum reduces to

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

(7)

The equation of Angular Motion is:-

Moment of Forces = Rate of Change of Angular Momentum

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

(8)

Which 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 boidy 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 s and $\omega$ for v.

It is assumed that expressions for acceleration have been obtained by substituting into equations (1) or (8)

(a) Acceleration as a function of Distance.

$\text{since}\;\;\;\;\;\;\;a=\frac{dv}{dt}= \left ( \frac{dv}{ds} \right )\left ( \frac{ds}{dt} \right )=v.\frac{dv}{ds}$

(9)

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

(10)

$\text{since}\;\;\;\;a=\frac{F}{M}\;\;\text{This corresponds to the equation of energy}$

(11)

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

(12)

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

Note. A further integration of equation (10) can be carried out by writing $v=ds/dt$ . This allows the time interval to be determined. ( See Example 6)

(b) Acceleration as a function of time.

Writing

$a=\frac{dv}{dt}\;\;\;\;\;\text{and re-arranging}$

(13)

$v=\int a.dt+B$

(14)

Which corresponds to the momentum equation

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

(15)

i.e. The change of Momentum = Impulse.

A further integration of equation (14) will give the distance traversed.

Acceleration as a function of velocity.

Writing

$a=\frac{dv}{dt}$

(16)

and re-arranging

$t=\int \frac{dv}{a}+C_1\;\;\;\;\;\;\;(\text{See Example 4})$

(17)

$a=v.\frac{dv}{ds}$

(18)

Which gives

$s=\int v.\frac{dv}{a}+C_2\;\;\;\;\;(\text{See Example 5})$

(19)

Work and Power

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

$\text{Work done}=F.x$

(20)

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

$=\tau. \theta$

(21)

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.

Kinetic Energy

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

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

(22)

If the body is rotating about a fixed axis O and since $v=h\omega\;\;\;\;and\;\;\;\;k{_{0}}^{2}=k{_{g}}^{2}+h^2$

$\text{The above equation is}\;\;K.E.=\frac{1}{2}I_o\omega^2$

(23)

Impact, Impulse.

When two bodied collide, each exerts an equal force on the other and for the same period of time. By the integration of equation (3)

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

(24)

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 |{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.

Worked Examples.

The following worked examples have been "hidden". They can be viewed by clicking on the button. Remember that 60 m.p.h. is equivalent to 88 ft./sec.

Example 1

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. (U.L.)

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Example 2

A motor car weighing 2500 lb. has axles 8 ft apart. When standing on a level road, the centre of gravity of the car is 2 ft. above ground level and 3 ft. in front of the rear axle .

Calculate the normal reaction on each wheel when the car is moving down a gradient of 1 in 20 against a wind resistance of 40 lb. acting parallel to the road and 2 ft. 6 in from it, with the engine switched off and the rear wheel brakes applied so as to give a deceleration of $2\;ft./sec^2$ . What must be the coefficient of friction between the rear wheels and the road if the rear wheels are not to skid under these conditions. Neglect the rotational inertia of the wheels and engine. (U.L.)

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Example 3

A motor-cycle has wheels 4ft.9in. apart. The centre of gravity of the cycle and rider is 2ft.^in. above ground level and 2 ft. in front of the rear axle. The coefficient of friction between the tyres and the road is 0.75

If the rear wheel only is braked, find the greatest deceleration that can be obtained (a) if the cycle is moving in a straight path, (b) If it is going round a curve of 150 ft radius at 30 m.p.h. Neglect rotational inertia and obliquity when turning. (U.L.)

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Example 4

The velocity of a train traveling at 60 m.p.h. decreases by 10% in the first 40 seconds after the application of the brakes.

(a) Calculate the velocity at the end of a further 80 seconds, assuming that during the whole period of 120 seconds, the retardation is proportional to the velocity.

(b) Derive an expression for the retardation force in lb.per ton weight of train.

(U.L.)

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Example 5

A car weighing 3000 lb. has when running on the level, a resistance to motion of $a+bV^2\;lb.$ where a is 60 lb., b is a constant and V is the speed in miles per hour: it is also found that the car maintains a speed of 80 m.p.h. with an effective horse-power of 50 when running on the level.

Find:-

(1)The value of the constant b.

(2)How far the car will run up a slope of 1 in 10 before the speed drops to 40 m.p.h. assuming that it starts at 80 m.p.h. and that a constant torque is maintained on the road wheels throughout.

(U.L.)

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Example 6

A frictionless flexible chain, of total length 24 in. hangs over the edge of a table by an amount 6 in. and is held there in that position. Determine the time taken for the chain to just slide off the table, if released. (U.L.)

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Example 7

A thin bar of length 2a turns freely about a horizontal hinge at the upper end. It is released from rest when making an angle $\alpha$ with the downwards vertical. By equating the loss of potential energy to the kinetic energy, show that the angular velocity $\omega$ when making an angle $\theta$ with the downwards vertical is given by:-

$\omega ^2=\frac{2ga}{k^2+a^2}(\cos\theta-\cos\alpha)$

(79)

where k is the radius of gyration about a parallel axis through its centre of gravity. Hence find an expression for the angular acceleration when in this position.

If a - 15 in., W = 12 lb., and $\alpha=60^0$ find the forces exerted by the pivot on the bar at the instant it passes through the lowest position.

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Example 8

A flywheel weighing 100 lb. is mounted on a 3 in. diameter shaft supported in two bearings, one on either side of the wheel. Under the action of the friction of the bearings,the speed of the flywheel falls from 200 r.p.m. to 150 r.p.m. in 14 sec. with uniform deceleration.

A plain cast-iron ring 18 in. outside diameter, 14 in. inside diameter and 3 in. thick is now bolted on to the side of the flywheel, concentrically with it.

The cast-iron weighs 0.26 lb./cu.in. The effect of friction is now to reduce the speed uniformly from 200 r.p.m. to 150 r.p.m. in 20 sec.

Assuming that the coefficient of friction is constant, find its value. Find also the radius of gyration of the flywheel (U.L.)

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Example 9

A flywheel of weight 180 lb. and radius of gyration 16 in.,is accelerated from rest to a steady speed of 75 rad./sec. by the application of a constant torque, the friction torque being $10\omega%$ lb.ft. at any instant.

Find the applied torque and obtain an expression for the speed at any time t sec. from the start. Calculate the speed after 1 sec.

If the wheel had been acted upon by a torque of magnitude $90(75-\omega)\;lb.ft.$ find the final steady speed and obtain an expression for the speed at time t sec. Calculate this speed after 0.1 sec.

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Example 10

A riveting machine is driven by a 5 h.p.motor. The moment of inertia of the rotating parts of the machine is equivalent to $1500 \;lb.ft.^2$ at the shaft on which the flywheel is mounted. At the commencement of an operation the flywheel is making 240 r.p.m. If closing a rivet occupies 1 sec. and corresponds to an expenditure of 8000 ft.lb. of energy, find the reduction of speed of the flywheel. What is the maximum rate at which rivets can be closed. (U.L.)

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Example 11

In an experiment to determine the modulus of rigidity by a mass vibrating at the bottom of a vertical helical spring, it is usual to add one-third of the mass of the spring to the vibrating mass in calculating the results.

Justify this rule.

State and justify the corresponding rule for the method of finding:- (a)The modulus of elasticity by the torsional oscillations of a disc at the end of a similar spring.

(b) The modulus of rigidity by the torsional oscillation of a disc at the end of a circular rod. (U.L.)

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Example 12

A valve is opened mechanically, released by a trip gear and closed by a close-coiled helical spring. The maximum valve opening is 0.75 in., the weight of the valve and attached parts is 8 lb., the weight of the spring is 1.25 lb. The stiffness of the spring is 60 lb./in. of compression and the spring is compressed by 1.825 in. when the valve is fully open

Determine the time taken to close the valve and the velocity at the moment of impact. Neglect friction and assume that there is no dash-pot.

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Example 13

A Steam Locomotive of mass $M_1$ , including its tender, starts to pick up water from a water trough when drawing a train of mass $M_2$ at a steady speed u. During the period of watering it may be assumed that:-

The mass of water m picked up per unit distance is constant.

The tractive effort of the engine and the resistance of the engine and train are unaffected.

Obtain an expression for the speed of the train after it has picked up water for a distance S and show that the initial instantaneous diminution of draw-bar pull is :-

$\frac{M_2\;m\,u^2}{M_1+M_2}$

(139)

Find also an expression for the time taken by the train to travel a given distance (U.L.)

Note. Steam Locomotives are not that common these days. There was no method of condensing and re-using the water and as a result they needed to take on fresh water at regular intervals. This produced problems for long distance locomotives and the solution was to have narrow, shallow, and long water troughs placed between the rails. The engine lowered a scoop into the trough and the speed of the locomotive swept water into its water tanks.

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Example 14

A haulage rope winds onto a drum at a radius of 1.75 ft., the free end being attached to a truck. The truck weighs 1120 lb. and is initially at rest. The drum is equivalent to 2500 lb with a radius of gyration of 1.5 ft. and a rim speed of 2.5 ft./sec. before the rope tightens. By considering the change of momentum of the truck w and the angular momentum of the drum, find the speed of the truck when the motion becomes steady. Find also the energy lost to the system.

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Example 15

The diagram shows a hammer weighing 12 lb. and pivoted at A. It falls against a wedge weighing 2 lb. which is driven forward 0.25 in. by the impact into a heavy rigid block.

The resistance to the wedge varies from uniformly with the distance through which it moves varying from zero to R lb.

Neglecting the small amount by which the hammer rises after passing through A and assuming that the hammer does not rebound find:-

(a) The value of R

(b) The time interval during which the wedge is moving forward.

(U.L.)

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Example 16

In the diagram the square threaded screw A is attached rigidly to the circular base plate B which rests on a plane horizontal surface C; Two weighted arms are attached to the nut N which can turn freely on the screw.

The screw and base weigh 50 lb. and have a radius of gyration of 4.5 in.; the under-surface of B has a mean contact radius of 6 in. The nut N with attached arms has a weight of 36 lb. and a radius of gyration of 8 in. The screw has a mean radius diameter of 2.5 in. and a pitch of 2.25 in. The coefficient of friction between the base plate B and the surface C os 0.35 and between the nut and the screw may be neglected.

If the nut and arms are given a rotational speed of 60 r.p.m. when 9 in. from the base find:- (a) The angular velocity of the nut and arms just before reaching the base.

(b) The angular velocity of the nut and arms just after impact with the base, assuming that the nut does not rebound. (U.L.)

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Example 17

A railway carriage is standing in a siding, with a door open at an angle $\theta$ to the train. Another carriage is shunted into it causing it to start moving with a velocity v in the direction which causes the door to shut.

The mass of the door is m, its centre of gravity is a from the hinge and the radius of gyration round a vertical line through the centre of gravity is k.

Obtain expressions for:-

(a)The angular velocity with which the door starts to move.

(b) The impulsive reaction at the hinge.

Taking the door as a uniform rectangle 2 ft. 6 in. wide and weighing 150 lb., v as 3 m.p.h. and $|theta=60^0$ calculate the angular velocity and the total kinetic energy of the door immediately after impact. (U.L.)

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Example 18

The tail-board of a lorry is 5 ft. long and 2.5 ft. high. It is hinged along the bottom edge to the floor of the lorry. Chains are attached to the top corners of the board and to the sides of the lorry so that when the board is in a horizontal position the chains are parallel and inclined at $45^0$ to the horizontal. A tension spring is inserted in each chain so as to reduce the shock and these are adjusted to prevent the board from dropping below the horizontal. Each spring exerts a force of 300 lb./in of extension

Find the greatest force in each spring and the resultant force at the hinges when the board falls freely from the vertical position. Assume that the tail board is a uniform body weighing 60 lb. (U.L.)

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Example 19

A hammer B suspended from pin C and an anvil A suspended from pin D are just touching each other at E when both hang freely.

B weighs 14 lb., its centre of gravity is 10 in. below C and its radius of gyration about C is 10.75 in. A weighs 48 lb., its centre of gravity is 7 in. below D and its radius of gyration about D is 7.5 in.

The hammer B is rotated $20^0$ to the position shown dotted and released. Assume that at the points of contact move horizontally at the instant of impact and that their relative linear velocity of recoil is 0.8 times their relative linear velocity of impact.

Find the angular velocities of the hammer and of the anvil immediately after impact. (U.L.)

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Example 20

A solid uniform cylinder rolls, without slipping down a plane AO which is inclined at $45^0$ to the horizontal. It then rolls up a plane inclined at $\theta$ to the horizontal. If the centre of gravity of the cylinder descends a vertical distance of 2 ft, whist the cylinder is rolling down AO claculate:-

(a) The minimum value of $\theta$ which will cause the cylinder to be brought to rest when it strikes OB

(b) The vertical distance through which the cylinder will rise as it rolls up OB is $theta=45^0$ (U.L.)