PHY 121/123 LECTURE NOTES - Week 5

We are ready for another new concept, or really pair of concepts, energy and its conservation. As with forces, we may start with gravity, and define a quantity which does not change, or equivalently, is exactly conserved, when gravity is the only cause of acceleration. Near the surface of the earth, where the acceleration of gravity is effectively constant, consider the expression

$$E/m \equiv v^2/2 + gy ,$$

where y is the height of the object one is watching. [The unit of energy E, kgm²/s², is called the joule (symbol J), after a British scientist and engineer.] First, suppose there are no other forces, and the motion is just in the vertical direction. Then we already know this quantity is constant. That is just a rule derived in the text and used in some CAPA problems, $\Delta (v^2) = 2 a\Delta y$, where $\Delta y=h$ is the displacement in the direction of the acceleration, namely, the change in height.

Now let's make this a little more complicated, by supposing that the object is moving in the horizontal direction. We know the horizontal component of the velocity v_x does not change, because there is no horizontal acceleration. Therefore, for any projectile motion in both horizontal and vertical directions, we again conclude that the expression for E/m is constant, because now v² must be treated as the square of the magnitude of the vector velocity, which means v²=v_x²+v_y², so $\Delta (v^2)=\Delta(v_y^2)$, exactly what we studied just above. We have found that the change in v² depends only on the height, regardless of the change in horizontal position.

This result is true even if we introduce other forces, as long as those forces never push or pull the object along the direction in which it is moving. For example, suppose a box is sliding down a very smooth ramp, inclined at an angle $\theta$ to the horizontal direction. Then the gravity force vector, which points straight down, has a component $mg{\rm sin}\theta$ which pushes the box along the ramp surface. The component of gravity perpendicular or normal to the ramp does not push the box, because it is exactly balanced by the normal force of the ramp on the box. Consequently, the acceleration of the box down the ramp is $a=g{\rm sin}\theta$. How far does the box go for a change in height h? The travel distance $\ell$ is the hypotenuse of a right triangle with angle $\theta$ to the horizontal, and the height h is the side of the triangle opposite to the angle $\theta$, giving $h={\ell\rm sin}\theta$, and therefore $a\ell = gh$, so once again we get $\Delta (v^2) = 2gh$, and therefore v²/2 + gy is constant.

An even more complicated example is suggested by a CAPA problem, where a skier is coming down a hill which has several ups and downs, but friction is negligible. Once again, the result is the same, v²/2 + gy = constant. We can go beyond the book in a way which is quite modern, related both to Einstein's view of gravity and to the study which comes in the second semester of sytems with electric forces. If we call gy=V_grav the gravitational potential, then we are saying that for any object only pushed or pulled by gravity, the kinetic energy mv²/2 plus the potential energy mV_grav make a constant.

The idea of a gravitational potential works on a larger scale too. If we consider a satellite in orbit around a planet, we already know the acceleration of gravity has a magnitude g(r)=g_S R_S²/r², where the subscript S refers to the value at the surface of the planet, and of course the acceleration always is directed towards the center of the planet. If we imagine for simplicity a satellite moving radially in or out, we have dv²/2dt = vdv/dt= va= -vg(r). Now, let us check that the gravitational potential may be written V_g(r)=-gR_S²/r. By standard rules of taking derivatives $dV_g(r(t)/dt= dV_g(r)/dr \times dr/dt =(gR_S^2/r^2)v=+g(r)v$.Therefore, again we find $v^2/2-gR_S^2/r = {\rm constant}$, just as is true near the surface of the planet where g(r) is approximately constant.

We can use this result to compute an interesting quantity, the minimum speed a rocket must have to leave the earth and never come back, known as the escape velocity. If such a rocket slows more and more, approaching zero speed as it goes farther and farther away, then both v² and V_g(r) approach zero. That means the sum of the two, which is constant, must be zero everywhere. At the surface this means v²/2-g_sR_S²/R_S=0, or v²=2g_SR_S. Going back to an earlier discussion of the velocity required for a satellite to float around the earth just above the equator, we see that the escape velocity is $\sqrt{2}$times the float velocity, or about 24,000 mi/hr.

Still with pure gravity, back on the surface of the earth, take another look at a CAPA problem of a ball on the end of a string going in a vertical circle. From Newton's second law, we know that at the top the tension in the string is T= mv²/r-mg, while at the bottom it is T=mv²/r+mg. In words, at the top gravity is helping to keep the ball going in a circle, so less tension is needed than without gravity. At the bottom, gravity is pulling the ball away from the center, so more tension is needed. However, there is still another ingredient, namely, the speed v should be greater at the bottom than the top because gravitational potential energy is converted to kinetic energy. A spectacular case is when at the top T=0, so gravity is exactly the right force to hold the ball in its circular path. This means v_top²=gr. At the bottom, we have $v_{bot}^2=v_{top}^2+2g\Delta y= v_{top}^2+2g\times 2r=5gr$. Therefore, at the bottom, T=mv²/r+mg=6mg! This means that if a 60kg man were going around in a loop-the-loop vertical circular ride, all the time sitting on a scale, and the scale read zero at the top, then it would read 360 kg at the bottom!

You can try this out yourself by whirling a ball on a string, first horizontally, in which case your hand just moves in a little circle as the ball moves in a big circle. Now if you do it vertically, with the ball nearly stalling (i.e., the string almost going slack) at the top, you will feel your hand jerked up and down as the ball goes from bottom to top and back again. The reason is that the tension is so much greater when the ball is at the bottom than when it is at the top.

Now it's time to start introducing other forces. You might say we already have done that, because we considered sliding on an inclined plane with a normal force but no friction present, and also going in a circle held in by a string. In both these cases, there is no force except gravity acting along the direction of motion, which is why we could do our energy calculations entirely in terms of transfer between kinetic energy and gravitational potential energy. Suppose now we add friction to the list. Then energy transfer, or work, can be done to convert mechanical (kinetic plus potential) energy into something else, namely heat. If total energy is to be conserved, then we should be able to check that what was lost from one form of energy has reappeared as another form. Later in the term, we'll discuss this more quantitatively, but each of us can do a simple qualitative check. If you rub your hands together, the surfaces warm up, and the explanation is that friction has converted kinetic energy associated with the moving hands into heat energy. A quantitative test of this would include checking that if twice as much mechanical energy is lost, then twice as much heat is produced - which as mentioned will come later.

Let's look at work and rate of work in more detail. Suppose a man is pushing a 100 kg box along a surface with a coefficient of friction $\mu=1/2$. To keep steady speed (no acceleration) he must just balance friction, which means a force of 500 N. If the man were pushing the box at 10 m/s, a feasible speed for a champion runner not pushing something, then this would mean supplying a power $P=Fv_{\parallel}=500\times 10=5000$ W. [Here the symbol W, for the unit of power, is called a watt, after another British scientist and inventor. As a unit of energy transfer per unit time, 1 W = 1 J/s.] No human could do this for more than a few seconds without collapsing. In daily life, our average power expenditure is roughly 100 W, the power associated with a standard light bulb. In any case, the man is doing work on the box, but the box is not gaining kinetic energy. Instead, it is doing work on the box-floor interface, heating it by friction.

We may also come back to the inclined plane problem, including friction. Say the coffecient of friction is $\mu=1/2$, and the angle is 45^o as in the practice exam. Then the dependence of kinetic energy mv²/2 on height is given by $mgy(1-\mu/{\rm tan}\theta)+mv^2/2 =$ constant, so only half the gravitational potential energy goes into kinetic energy. The rest goes into heat.

Finally, we now have just enough tools to explain the `physics magic' of the demonstration in which a tennis ball and a basketball are dropped simultaneously from a height of about 1 m, and the tennis ball, which was following the basketball down, bounces up to the ceiling. Ignoring small corrections from the diameters of the balls and from inelasticity (loss of mechanical energy to heat), the basketball drops down to the floor, developing a speed such that v_bounce²= 2gh. By conservation of kinetic energy, the ball bounces up with the same speed, but now in the up direction. The tennis ball is still coming down, also with speed v_bounce. Because the basketball is much more massive, the tennis ball bounces from it almost as the basketball bounced off the floor. Therefore we want to describe this collision from the viewpoint of someone travelling up with the basketball, who sees the tennis ball coming down with 2v_bounce. After hitting, the tennis ball will be going up with 2v_bounce. For someone sitting and watching the basketball still rising with approximately v_bounce, the tennis ball has velocity up of 2v_bounce+v_bounce=3v_bounce. Therefore the tennis ball will keep going up until all its speed is gone, meaning (3v_bounce)²=gh_final=9gh_initial=9 m or about 30 feet, enough to hit even a very high ceiling. Consequently, all the corrections left out don't really change anything.