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Chapter 8 - Conservation of Energy

The Law of Conservation of Energy

The law of conservation of energy states that within a closed system the total amount of energy is always conserved.

Another way of this is saying this is that energy can be neither created of destroyed, it can only be converted from one form to another.

There are, however, many different forms of energy.

For mechanics problems it is useful to think about a more restricted law, which considers mechanical energy to be conserved.

Conservation of mechanical energy can be assumed whenever the forces which apply to a system are entirely conservative.

Gravity is a conservative force

$\Delta U_{G}=\int_{y_{1}}^{y_{2}}mg\,dy=mg(y_{2}-y_{1})$

The change in potential energy depends only the change in height.

Gravitational Potential Energy

We've seen already that potential energy due to gravity should have the form $mg(y_{2}-y_{1})$. We can measure potential energy relative any position that makes sense to us, the most sensible place to measure from will depend on the problem. We can then say that the gravitational potential energy of an object is given by the height $h$ of that object above that reference point is $mgh$. An object below the reference point will have negative gravitational potential energy.

Problem 8.91

Nail Drop Demo

If I drop the weight from twice the height what is the ratio of the velocity measured at the photogate to the velocity measured in the first drop?

If I drop the weight from twice the height what is the ratio of the distance the nail goes in to the distance it went in on the first drop?

Nail Drop Explained

During the drop of the weight only conservative forces act. So whatever potential energy we lose, $mg\Delta h$, gets converted in to kinetic energy, $\frac{1}{2}mv^{2}$.

If we drop the weight from twice the height we see that the ratio of the kinetic energy should be $\frac{K_{2h}}{K_{h}}=2$ and as the kinetic energy is proportional to the square of the velocity the ratio $\frac{v_{2h}^{2}}{v_{h}^2}=2$ and $\frac{v_{2h}}{v_{h}}=\sqrt{2}$

When the weight hits the nail the force is used to push the nail in to the block. We should note that this a non-conservative force, the force is used to deform the block in an irreversible fashion! However the work-energy theorem still applies, so the amount of work done is equal to the kinetic energy lost.

As $W=Fd$ the distance the nail goes in should be twice as much when the kinetic energy is twice as much, $\frac{d_{2h}}{d_{h}}=2$.

Total Energy Conservation

Non conservative forces remove mechanical energy from the system, but it is not destroyed, it is simply converted to a different form of energy (frequently, but not always, heat).

The total energy conservation law can also be useful, for example when a frictional force $\vec{F}_{fr}$ is acting and an object travels a distance $d$ while it goes from a height $h_{1}$ to $h_{2}$, changing it's velocity from $v_{1}$ to $v_{2}$, conservation of total energy tells us

$\frac{1}{2}mv_{1}^2+mgh_{1}=\frac{1}{2}mv_{2}^2+mgh_{2}+F_{fr}d$

Snow Bike

Snow Bike Video

If the cyclist starts from rest then what is his velocity at the bottom of the hill?

How far will he travel down the road before coming to a stop?

Gravitational Potential Energy over Long Distances

As we saw previously the force due to gravity on an object of mass $m$ due to an object of mass $M$ at distance $r$ is $-\frac{GmM}{r^2}\hat{r}$

If we raise an object from the surface of a planet to a distance $h$ above it the change on potential energy is

$\Delta U = \int_{R_{E}}^{R_{E}+h}\frac{GmM}{r^2}\,dr=-\frac{GmM}{R_{E}+h}+\frac{GmM}{R_{E}}$

We should consider where a suitable zero of potential energy is

Choosing $r=\infty$ as our reference position is attractive as then the potential energy change in bringing an object from infinity to a position $r$

$\Delta U = \int_{\infty}^{R}\frac{GmM}{r^2}\,dr=-\frac{GmM}{r}$

is the same as the potential energy, so generally we say the gravitational potential energy at point $r$ from the center of a mass $M$.

$U = -\frac{GmM}{r}$

Elastic Potential Energy

Another example of a conservative force is the force provided by a spring

$F_{s}=-kx$

where $x$ is the distance of the end of the spring from the equilibrium end position of the spring. The change in potential energy associated with compressing or extending a spring is

$\Delta U = -\int_{x_{i}}^{x_{f}}F_{s}\,dx=\int_{x_{i}}^{x_{f}}kx\,dx=\frac{1}{2}k(x_{f}^2-x_{i}^2)$

Escape velocity using force

$\frac{dv}{dt}=\frac{-GM_{E}}{(R_{E}+x)^{2}}$

$\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=\frac{-GM_{E}}{(R_{E}+x)^{2}}$

$\int^{v}_{v_{0}}v\,dv=\int^{x}_{0}\frac{-GM_{E}}{(R_{E}+x)^{2}}\,dx$

$\frac{1}{2}(v^{2}-v_{0}^2)=\frac{GM_{E}}{R_{E}+x}-\frac{GM_{E}}{R_{E}}$

$v^{2}=v_{0}^2+2(\frac{GM_{E}}{R_{E}+x}-\frac{GM_{E}}{R_{E}})$

as $x\to\infty$ the object will escape if

$0\geq v_{0}^{2}-\frac{2GM_{E}}{R_{E}}$

For the object to escape

$v_{0}\geq \sqrt{\frac{2GM_{E}}{R_{E}}}$

Escape velocity using potential energy

At the Earth's surface the total mechanical energy of a rocket is

$\frac{1}{2}mv^{2}-\frac{GmM_{E}}{R_{E}}$

The condition for escape is that at infinity the velocity equals zero (just avoids being negative). As at infinity the gravitational potential energy is equal to zero the condition for the escape velocity is

$\frac{1}{2}mv_{escape}^{2}-\frac{GmM_{E}}{R_{E}}=0$

and

$v_{escape}=\sqrt{\frac{2GM_{E}}{R_{E}}}$

Space Elevator

The escape velocity is very large and launching a rocket is not a very sophisticated solution. (And it doesn't always work..apparently quite frequently).

A Space Elevator is a possible alternative, and is being increasingly seriously considered.

A space elevator would not reduce the amount of work needed to be done to get something in to space. The appeal of a space elevator is the ability to control the rate at which work is done.

Power

Power is defined as the rate at which work is done.

Average power is given by

$\bar{P}=\frac{W}{t}$

Instantaneous power is given by

$P=\frac{dW}{dt}$

Recall that $dW=\vec{F}\cdot d\vec{l}$, which means that

$P=\frac{dW}{dt}=\vec{F}\cdot \frac{d\vec{l}}{dt}=\vec{F}\cdot\vec{v}$

Units of Power are J/s or W (Watts).

Power to drive up a slope

How much power is required from the engine to drive a car up a plane (with friction) at a constant velocity?

We can assume the the applied force from the engine is directed up the plane. It needs to be equal and opposite to the forces down the plane.

$F_{A}=mg \sin\theta + \mu mg \cos \theta$

$P=(mg \sin\theta + \mu mg \cos \theta)v$

Bicycles

The amount of power required to do a certain amount of work in a certain amount of time is not adjustable. However a bicycle allows you to adjust the manner in which you produce the power. Our legs are not very good at moving at very high speed, because it is difficult for us to move them more than a certain number of times per minute and the stride length is obviously limited.

Suppose the net force opposing your motion is $F_{opp}$. To move the wheel of a bike at a velocity $v_{w}$ you need to produce an amount of power $F_{opp}v_{W}$, this will be the same as the amount of power on the pedals $F_{P}v_{P}$.

$F_{P}v_{P}=F_{opp}v_{W}$

But because the pedals have a smaller radius than the wheel we can move them at a slower velocity with a larger force. The use of adjustable freewheels and chainwheels allow the rider to adjust the force and the velocity of their feet to suit their biomechanics.

$F_{P}=F_{opp}\frac{v_{W}}{v_{P}}=F_{opp}\frac{r_{W}}{r_{FW}}\frac{r_{CW}}{r_{P}}$

Of course as you reduce the force using gears you also reduce the velocity of the wheel by the same factor. If you don't want to do this, you need to increase either the force with which you push or the speed at which you turn the pedals.

Cadence

There is an ideal pedal speed, or cadence, to pedal at. It is usually higher than most people use!

Cadence Video

Homework Tips

Problem 8.31

An inclined plane with friction. The total mechanical energy is not conserved, but we can quantify the work done by friction which is lost to the motion.

The equation

$\frac{1}{2}mv_{1}^2+mgh_{1}=\frac{1}{2}mv_{2}^2+mgh_{2}+F_{fr}d$

can be applied to both parts of the motion

Problem 8.90

This problem combines ideas about conservation of energy and centripetal acceleration.

Part A. Determine the minimum release height.

The minimum required velocity is when the centripetal acceleration at the top of the loop is equal to the gravitational force. (Examine the vertical circle or looping plane diagram to see why.) Then use conservation of energy to relate the mechanical energy at the top of the loop to the original potential energy, mgh. Don't forget that the potential energy is not zero at the top of the loop!

Part B,C,D. The condition at any point in the circle is that the net force in the radial direction must be equal to the centripetal force. Conservation of energy will tell you what the velocity is, and hence what the centripetal force is. You'll also need to take into account the direction of other forces on the object. Also you'll need to you results from part A to remove g from the problem.

Problem 8.50

For A refer to the equation for gravitational potential energy at large distances.

For B get the velocity from the equation for the velocity of a circular orbit.

For C the work done should be equal to the total change in energy from orbit to another.

Problem 8.105

Part A

Use conservation of energy, kinetic energy at bottom is the converted to potential energy at top.

Part B

Find the amount of energy released in one second. (You had to calculate the energy of one rock in Part A.)

phy131studiof17/lectures/chapter8.txt · Last modified: 2017/08/04 13:41 (external edit)
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