General Dynamical Problems

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-\b{power} required to drive the car up a gradient of 1 in 25 at a speed of 25 m.p.h.

Let the friction and road resistance be $F$ lb.and the wind resistance be $Cv^2 lb.$ ( $v$ is in $ft./sec.$)

Then, Total resistance $X$ velocity = 550 X h.p

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

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

From equations (7) and (8), $C=\displaystyle\frac{134}{3430}=0.392$

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 $C$,

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

• The horse-\b{power} required is $28.4\;h.p.$

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.

The diagram shows diagrammatically the side elevation of the car. $R_1$ and $R_2$ are the normal reactions at the front and rear wheels respectively and $F$ is the friction at the rear wheels.

By Newton's second Law, the inertia force =$Ma=\displaystyle\frac{2500}{32.2}\times 2=15.5\;lb.$

This acts down the plane in the opposite direction to the acceleration and the problem can now be considered as being one of statical equilibrium.

Taking moments about the base of the rear wheel.

$R_1\times 8=15.5\times 2-40\times 2.5+\left ( \frac{2500}{20} \right )\times 2+2500\times 3$

$\therefore \;\;\;\;R_1=\frac{7960}{8}=995\;lb.$

Note that the weight has been resolved parallel and perpendicular to the plane. Its normal component being approximately 2500 lb.

Resolving normal to the plane

$R_2=2500-R_1=1505\;lb.$

Resolving parallel to the plane

$F=155+\frac{2500}{20}-40=240\;lb.$

The coefficient of friction =$\displaystyle\frac{F}{R_2}=\displaystyle\frac{240}{1505}=0.16$

• $R_1=995\;lb.$ and $R_2=1505\;lb.$
• The coefficient of friction is $0.16$

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 if:

• a) The cycle is moving in a straight path.
• b) It is going round a curve of 150 ft radius at 30 m.p.h. Neglect rotational inertia and obliquity when turning.

In the following diagram, the motor cycle is traveling from right to left and the inertia force $Ma$ is in the opposite direction to the acceleration. $F$ is the friction force at the rear wheel and acts in a direction to oppose motion.

• a) Cycle moving in straight line

Friction = $\mu\;R$ and hence $F=0.75R_2$

Resolving horizontally, $0.75R_2=Ma$ or $R_2=1.33Ma$

Taking Moments about the base of the front wheel: $R_2\times 4.75=W\times 2.75-2.5Ma$ (Note $\;4ft.9in.=4.75ft.$ etc.)

By substituting in the value of $R_2$, $6.32\;Ma=2.75\;Mg-2.5\;Ma$

From which the greatest deceleration $\displaystyle \hat{a}$, $\hat{a}=2.75\times \displaystyle\frac{32.2}{8.82}=10.0\;ft/sec.^2$

• b) Travelling around a curve

There is now a centrifugal force acting through the centre of gravity, $G$. This is balanced by a sideways frictional force at the front and rear wheels. This is distributed such that 2/4.75 is attributed to the front wheel and $\displaystyle\frac{2.75}{4.75}$ to the rear.

The Centrifugal Force =$M\displaystyle\frac{v^2}{r}=M\times \displaystyle\frac{44^2}{150}=12.9M$ at $G$

Hence the sideways frictional force at the rear wheel is:

$\left ( \frac{2.75}{4.75} \right )12.9M=7.47M$

But the total frictional force at the rear wheel is $0.75\;R_2$ Hence the component of frictional force in the direction of motion is given by:

Resolving Horizontally

Taking Moments about the front wheel. ( This is exactly the same as in case a) above)

Substituting into equation (9) for $F$ and $R$ from (10) and (11) and squaring

$M^2a^2=\left ( \frac{0.75}{4.75} \right )^2\left ( 2.75Mg-2.5Ma \right )^2-7.47^2M^2$

i.e. $a^2=195-11.1a+0.155a^2-55.7$

Or $0.845a^2+11.1a-139.3=0$

From which, $a=7.9\;ft./sec.^2$

• a) $\hat{a}=10.0\;ft/sec.^2$
• b) $a=7.9\;ft./sec.^2$

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.

• c) Find the horse-\b{power} being dissipated at the end of the whole period if the train weighs 500 tons.

• a) Retardation =$-\displaystyle\frac{dv}{dt}=kv$

Where $k$ is a constant to be determined from the given conditions.

Rearranging and integrating: $\displaystyle\int \frac{dv}{v}=-\displaystyle\int k\,dt+A$ or $ln v=-kt+A$

At $t=0$, $v=88ft./sec.$ , $A=\ln 88=4.47$

At $t = 40 sec.$, $v = 79.2 ft./sec.$ , $k=\displaystyle\frac{4.47-\ln 79.2}{40}=0.0026$

At $t = 120 sec.$, $\ln v=-0.0026\times 120+4.47=4.17$

From which, $v=65\;ft./sec.=44.3\;m.p.h.$

• b) From Newton's second Law:

The retarding Force = Mass of one tone $X$ the retardation

$=\left ( \frac{lb.wt.}{g} \right )\times kv=\left ( \frac{2240}{32.2}\right )0.0026v=0.181v\;lb./tone$

• c) The horse-power being dissipated at the end of the whole period.

Horse-power =(Force) $\times \displaystyle\frac{v}{550} =500\times 0.181\times \displaystyle\frac{65^2}{550}=695\;h.p.$

• a) The velocity is $v=65\;ft./sec.=44.3\;m.p.h.$
• b) The retardation force is $0.181v\;lb./tone$
• c) The horse-\b{power} is $695\;h.p.$

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.

• 1) The value of $b$

At a uniform speed on the level, Horse-power= Resistance $X$ Speed / 550 i.e. $50=(60+b\times 80^2)80\times \displaystyle\frac{88}{60\times 550}$

$\therefore \;\;\;\;\;b=\frac{234-60}{6400}=0.0272$

• 2) To find distance car will travel

It is advisable to work in ft. and seconds units by writing $a+_bV^2 =a+b_1v^2$ where $v$ is in ft./sec. and $b_1=(60/88)^2b=0.0126.$

A constant torque on the road wheels implies that there is a constant propulsive force which is: $a+0.0272\times 80^2$ (i.e. its value at 80 m.p.h.)

On the upgrade of the equation of motion is:

$a+0.0272\times 80^2-(a+0.0126v^2)-\frac{3000}{10}=\left ( \frac{3000}{32.2} \right )v.\frac{dv}{ds}$

Rearranging and integrating:

$s=-\frac{3000}{32.2}\int_{\frac{80\times 88}{60}}^{\frac{40\times 88}{60}}\frac{v.dv}{126+0.0126v^2}=\frac{3000}{32.2\times2\times 0.0126}\left [ \ln(126+0.0126v^2) \right ]_{58.7}^{117.4}$

$=3710\ln\frac{126+174}{126+43.5}=3710\;\ln1.77=2120\;ft.$

• 1) The value of the constant $b$ is $0.0272$
• 2) $2120\;ft.$

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.

If $W$ is the total weight of the chain, let $x$ in. be the length overhanging the table at any instant. The force accelerating the chain is $\displaystyle\frac{Wx}{24}$ and applying Newton'\b{s second Law}:
$\displaystyle\frac{Wx}{24}=\displaystyle\frac{W}{g}\times a$

$\therefore \;\;\;\;\;a=16.1x\;\;\;\;(g=32.2\times 12\;in./sec^2)$

Writing $a = v\;\displaystyle\frac{dv}{dx}$ And integrating gives

$v^2=16.6(x^2-36)$ ($v=0$ when $x=6$)

This can be re-written as $\displaystyle\frac{dx}{dt}=4.02\sqrt{(x^2-36)}$

Rearranging:

$t=\int_{6}^{24}\frac{dx}{4.02\sqrt{x^2-36}}$

$=\frac{1}{4.02}\left [ \cosh^{-1}\left ( \frac{x}{6} \right ) \right ]_6^{24}=\frac{2.06}{4.02}=0.513\;sec.$

• The time is $0.513\;sec.$

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: $\omega ^2=\frac{2ga}{k^2+a^2}(\cos\theta-\cos\alpha)$ 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 force exerted by the pivot on the bar at the instant it passes through the lowest position.

For rotation about a fixed axis, Kinetic Energy =$\displaystyle\frac{1}{2}I\omega^2$

Using the parallel axis theorem, $I=I_G=\left ( \displaystyle\frac{W}{g} \right )a^2=\displaystyle\frac{W}{g}(k^2+a^2)$

Loss of Potential Energy =$Wa(\cos\theta-\cos\alpha)$

Equating the loss of P.E. to the gain in K.E.,

$\frac{1}{2}\left ( \frac{W}{g} \right )(k^2+a^2)\omega^2=Wa(\cos\theta-\cos\alpha)$

Differentiating, $2\omega\;\displaystyle\frac{d\omega}{dt}=-\displaystyle\frac{2ga}{k^2+a^2}\sin\theta\;\displaystyle\frac{d\theta}{dt}$

And noting that $\displaystyle\frac{d\theta}{dt}=-\omega$

The angular acceleration, $\displaystyle\frac{d\omega}{dt}=-\displaystyle\frac{ga}{k^2+a^2}\sin\theta.$

In the lowest position the force on the pivot is the centrifugal force plus the weight.

$=\left ( \frac{W}{g} \right )a\omega^2+W$

Using equation (12) and noting that $\theta+0$ =$W\left [ \displaystyle\frac{2a^2}{k^2+a^2} (1-\cos\alpha)+1 \right ]$

As $k^2=\displaystyle\frac{a^2}{3}$, and $a+60^0$

$=12\left [ \frac{2\times 15^2}{15^2/3+15^2}(1-0.5)+1 \right ]=21\;lb.$

• The angular acceleration is $\displaystyle\frac{d\omega}{dt}=-\displaystyle\frac{ga}{k^2+a^2}\sin\theta$
• The force is $21\;lb.$

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.

Initially, since the the angular deceleration is uniform, $\alpha_1$ is given by:-

$\alpha_1=\frac{200-150}{14}\times \frac{2\pi}{60}=0.374\;rds./sec^2$

The frictional torque is $I\alpha_1=0.374\;I$

from "The theory of Machines" Mechanisms paragraph 6 equation (3). It can be seen that the frictional torque is also approximately $Wr\mu$

Equating, $Wr\mu=100\times \displaystyle\frac{3}{2}\times \mu=0.374I$

After adding the cast-iron ring: $\alpha_2=\displaystyle\frac{200-150}{20}\times \displaystyle\frac{2\pi}{60}=0.262\;rad/sec^2$

The weight of the cast-iron ring is: $\pi(9^2-7^2)\times 3\times 0.26=78.3\;lb.$

The moment of Inertia of the ring is ( See paragraph 3):-

$\frac{78.3}{32.2\times 12}\times \frac{18^2+14^2}{8}=13.2\;lb.in.sec^2$

The frictional torque for the flywheel and ring is now:

$(I+13.2)\alpha_2=(100+13.2)1.5\mu$

Substituting in the value of $\alpha_2$

Subtracting equation (13) from equation (14), $\mu=\displaystyle\frac{13.2}{618}=0.0213$

From equation (13), $I=402\times 0.0213=8.55$

Also $I=\displaystyle\frac{W}{g}\times k^2=\displaystyle\frac{100}{32.2\times 12}\times k^2=8.55$

Re-arranging, $k=\sqrt\displaystyle\frac{8.55\times 32.2\times 12}{100}$

i.e. The radius of gyration of the flywheel is $5.75 in.$

• The coefficient of friction is $\mu=0.0213$
• The radius of gyration of the flywheel is $5.75 in.$

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.

When the steady speed of 75 rds. has been reached, the applied torque must equal the friction torque. i.e. The applied torque $=10\times 75 =750\;lb.ft.$

The equation of motion then

$750-10\omega=I\alpha= Mk^2\alpha=\frac{180}{32.2}\times \left ( \frac{16}{12} \right )^2\frac{d\omega}{dt}\approx 10\frac{d\omega}{dt}$

Rearranging and integrating: $t=\displaystyle\int_{0}^{\omega}\displaystyle\frac{d\omega}{75-\omega}= \ln\displaystyle\frac{75}{75-\omega}$

This can be written as: $(75-\omega)e^t=75$

From which: $\omega=75(1-e^{-t})$

When $t = 1 sec.$, $\omega=75(1-0.368)=47.4\;rad./sec.$

Under the second set of conditions and assuming that the friction torque remains the same, the equation of motion now becomes:-

$90(75-\omega)-10\omega=10.\;\frac{d\omega}{dt}$

A steady speed is reached when the acceleration is zero. i.e. $\displaystyle\frac{d\omega}{dt}=0$

Hence, $\omega=\displaystyle\frac{90\times 75}{100}=67.5\;rad./sec.$

As above this can be rearranged and integrated to give an expression for $t$.

$t=\displaystyle\frac{1}{10}\int_{0}^{\omega}\displaystyle\frac{d\omega}{67.5-\omega}$ Or $10t=\ln\displaystyle\frac{67.5}{67.5-\omega}$

This can be rearranged and solved for $\omega$

$\omega=67.5(1-e^{-10t})$

When $t=0.1sec.$, $\omega=67.5(1-0.368)=42.7\;rad./sec.$

• The torque is $750\;lb.ft.$
• The speed at any time is $\omega=75(1-e^{-t})$, The speed after 1 sec. is $47.4\;rad./sec.$
• The final steady speed is $67.5\;rad./sec.$, The speed after 0.1 sec. is $42.7\;rad./sec.$

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 ?

Let $n$ r.p.m. be the speed of the flywheel immediately after closing a rivet. Then:

The loss of kinetic energy of the flywheel = The energy used to close the rivet - The energy supplied by the motor.

i.e. $\displaystyle\frac{1}{2}(\omega{_{1}}^{2}-\omega{_{2}}^{2})=8000-5\times 550$

Hence, $\displaystyle\frac{1}{2}\left ( \displaystyle\frac{1500}{32.2} \right )\left ( \displaystyle\frac{\pi^2}{30^2} \right )\left ( 240^2-n^2 \right )=5250$

$\therefore \;\;\;\;\;\;240^2-n^2=20,600$

Giving $n=\sqrt{57,600-20,600}=192.5\; r.p.m.$

i.e. A reduction in speed of $47.5 r.p.m.$

The maximum rate of closing rivets = The output of the motor per minute / The energy required to close a rivet

$\frac{5\times 33000}{8000}=20.6\;\;\;\;\;\;\text{say}\;\;20$

• The reduction in speed is $47.5 r.p.m.$
• The maximum rate of closing rivets is $20.$

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.

Let $W$ be the weight of the spring and $L$ its axial length.

Consider the kinetic energy of an element the spring at an axial length $l$ from the fixed end and of axial length $\delta l$

If $v$ is the velocity of the vibrating mass at any instant, then the velocity of the element is $\displaystyle\frac{vl}{L}$

The kinetic energy of the element is given by: K.E.=$\displaystyle\frac{1}{2}\left ( \displaystyle\frac{W\delta l}{gL} \right )\left ( \displaystyle\frac{vl}{L} \right )^2$

The total K.E. of the spring =$v^2\int_{0}^{L}\left ( \displaystyle\frac{Wl^2}{2gL^3} \right )dl=\displaystyle\frac{1}{2}\left ( \displaystyle\frac{1}{3}\displaystyle\frac{W}{g} \right )v^2$

This is the same as that of a weight of $\displaystyle\frac{1}{W}$ moving with the vibrating mass.

• a) Let $\omega$ be the angular velocity of the disc at any instant, then the angular velocity of the element will be $\omega l/L$ and if the coil radius is $R$ , its linear velocity will be $R\omega l/L$

K.E. of element =$\displaystyle\frac{1}{2}\left ( \displaystyle\frac{W\delta l}{gL} \right )\left ( \displaystyle\frac{R\omega l}{L} \right )^2$

The total K.E. of the spring =$\omega^2\displaystyle\int_{0}^{L}\left ( \displaystyle\frac{WR^2l^2}{2gL^3} \right )dl=\displaystyle\frac{1}{2}\left (\displaystyle \frac{1}{3}\times \displaystyle\frac{WR^2}{g} \right )\omega^2$

i.e. Add on to the inertia of the disc one-thirde of the moment of inertia of the spring about its axis.

• b)This is similar to the above with the exception that the radius of gyration of a circular rod about its axis is $\displaystyle\frac{d}{\sqrt{8}}$ and hence:

The K.E. of the rod =$\displaystyle\frac{1}{2}\left ( \displaystyle\frac{1}{3}\times \displaystyle\frac{Wd^2}{8g} \right )\omega^2$

i.e. Add on to the inertia of the disc one-third of the moment of inertia of the rod about its axis.

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.

From the answers to Example (11) it can be seen that the weight moved : $=8+\frac{1}{3}(1.25)=8.417\;lb.$

Let $x$ be the distance moved from the fully open position.

Using Newton'\b{s second law} and neglecting gravity, the equation of motion can be written. $\left ( \frac{8.417}{32.2\times 12} \right )\ddot{x}=(1.825-x)\times 60$

i.e. $\ddot{x}+2750x=5020$

The complete solution to this differential equation is : $x=A\sin\sqrt{2750.t}+B\cos\sqrt{2750.t}+\frac{5020}{2750}$ $=A\sin52.5\;t+B\cos52.5\;t+1.825$

When $t=0$, $x=0$ and $B=-1.825$

When $t=0$ $\dot{x}=0$ and $A=0$

Substituting in the above values for $A$ and $B$:

When $x=0.75$, $\cos52.5t=1-0.411=0.589$

From which, $t=\displaystyle\frac{0.943}{52.5}=0.018\;sec.$

Differentiating equation (15), $\dot{x}=52.5\times 1.825\sin52.5\;t$ $\therefore \;\;\;\;\;\dot{x}=52.5\times 1.825\times 0.808=77.5\;in./sec.$

• The time taken to close the valve is $0.018\;sec.$
• The velocity at the moment of impact is $77.5\;in./sec.$

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 :

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

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

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.

In traveling a distance $\delta s$ the mass of water picked up is $m\deltas$, and the momentum given to it is $(m\delta s )(\delta s/\delta t)$. Hence the rate of change of momentum is given by:

$=\left ( \frac{m\delta s}{\delta t} \right )\left ( \frac{\delta s}{\delta t} \right )$

Let $\delta s}/{\delta t} =V$ and let the initial value of $V$ be $u$.

Then the rate of change of momentum= $mV^2$

This produces a retarding force on the total mass $(M_1+M_2+ms)$ and using Newton'\b{s second law}: $-(M_1+M_2+ms)\left ( V\displaystyle\frac{dV}{ds} \right )=mV^2$

$\frac{dV}{V}=-\frac{m\;ds}{M_1+M_2+ms}$ Integrating,

$\int_{u}^{v}\frac{dV}{V}=-\int_{0}^{S}\frac{m\;ds}{M_1+M_2+ms}$ i.e. $\ln\displaystyle\frac{v}{u}=\ln \displaystyle\frac{M_1+M_2}{M_1+M_2+mS}$
Or

If $P$ is the initial diminution of the draw-bar pull, the impulse on the locomotive, then:

$P\;\delta t=M_1\; \delta u+mu\;\delta s$ Or

The Impulse on the train: $P\;\delta t = -M_2\;\delta u$ Or

Substituting from equation (18) into equation (17), $P=-\displaystyle\frac{M_1}{M_2}P+mu^2$

From which, $P=\displaystyle\frac{M_2mu^2}{M_1+M_2}$

Writing $v=\displaystyle\frac{d s}{d t}$ from equation (16)

$t=\frac{1}{u}\;\int_{0}^{S}\frac{M_1+M_2+ms}{M_1+M_2}\times ds=\frac{1}{u}\left [ S+\frac{mS^2}{2(M_1+M_2)} \right ]$

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.

Let $V$ ft./sec. be the velocity of the hammer just before impact with the wedge. Then:

The kinetic energy of the hammer = The loss of potential energy by the hammer

i.e.$\displaystyle\frac{WV^2}{2g}=Wh$ Or $V=\sqrt{2\times 32.2\times 3(1-\cos60^0)}=9.82\;ft./sec.$

Now let $v$ ft./sec. be the common velocity of the wedge and hammer just after impact ( The hammer does not rebound).

Equating the momentum before and after impact ( The conservation of Momentum) $(12+2)v=12V$

• a) To find the value of $R$

The work done by driving the wedge into the block against the resistance must equal the loss of kinetic energy .

i.e. $\displaystyle\frac{1}{2}R\times \displaystyle\frac{1}{4\times 12}=\displaystyle\frac{1}{2\times }\displaystyle\frac{12+2}{ 32.2}\times 8.42^2$

Hence $R=1475\;lb.$ $\therefore \;\;\;\;v=8.42\;ft./sec.$

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

If $x$ ft. is the distance that the wedge has moved, then the resistance is $1475\times 48x$ and the equation of motion is:

$1475\times (48)x=-\frac{12+2}{32.2}\ddot{x}$

i.e. $\ddot{x}=-163,000x$

To solve this differential equation multiply both sides by $\2\dot{x}$ . Now:

$2\dot{x}\;\ddot{x}=- 2\dot{x}\times 163000\;x^2$

Integrating with respect to $t$:

$\dot{x}^2=-163000\;x^2+Constant$

When $x=0$, then $\dot{x}=8.42\;ft/sec.$ from which The Constant is $70.9$

$\therefore \;\;\;\;\;\;\dot{x}=\sqrt{70.9-163000\;x^2}$

From which:

$t=\int_{0}^{\frac{1}{48}}\frac{dx}{\sqrt{70.9-163,000x^2}}$ $=\left ( \frac{1}{\sqrt{163,000}} \right )\left [ \sin^{-1}x\sqrt{\frac{163,000}{70.9}} \right ]_{0}^{\frac{1}{48}}=\left ( \frac{1}{404} \right )\left ( \frac{\pi }{2} \right )=0.0039\;sec.$

• a) $R=1475\;lb.$
• b) $t=0.0039\;sec.$