Energy is a conserved quantity in nature. This just means that it may be traded between different places and systems in various forms, but if you add it all up, the sum is constant in time. We'll see in third semester that over very brief time intervals governed by Heisenberg's uncertainty, that the rule of conservation of energy is constantly being broken in nature, and that breaking the rule leads to some of the most interesting phenomena. In fact, breaking of the rule leads to the very color of our sky and of sunsets. It should be noted that even when energy conservation seems to be broken, that it's only temporary. Whatever energy gets borrowed will soon be returned, and in that sense, over any longer duration of time we find the same amount of energy to be present. This "borrowing" of energy does not fit into our current discussion and will have to wait until we discuss modern physics.[br][br]For now we will consider cases where the energy of a given system is being conserved. For instance, if we consider the ball/earth system from the previous section, what if the ball falls under the influence of gravity but also experiences drag? The drag is due to the atmosphere which is not part of the two-body ball-earth system. If you would like to argue that the atmosphere is part of earth and earth is part of the ball/earth system, I understand your confusion. However, we are considering the earth only for its contribution to the gravitational potential energy, not for all its other forms of energy. In other words, if anything other than an exchange from gravitational potential energy to kinetic energy of the ball occurs, it is from outside the system.[br][br]I want to consider in this section only cases without any influence from non-conservative forces. In other words, no friction of any kind. While this type of problem is tough to find representation of in the macroscopic world, they do exist. Problems in space are pretty good examples, and slow-moving low friction systems in general are not too bad either. What's nice about such cases is that the math leads to very simple conclusions. Ignoring non-conservative work terms means:[br][center][math]\sum W_{NC}=\Delta K + \Delta U \\[br]\Delta K + \Delta U = 0. [br][/math][/center]

Consider the problem of a ball falling a certain distance subject to gravity without any drag. Back when we did kinematics, we had to do two integrals to figure out how far a ball would fall after a given time had elapsed. It required further algebra to relate that to the distance fallen. Using this new equation of energy conservation we can see the algebra is trivial:[br][center][math]\Delta K + \Delta U = 0 \\[br]K_f-Ki+U_f-U_i = 0 \\[br]\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2 + mgy_f-mgy_i = 0 \\[br]\text{Supposing for simplicity that $v_i=0m/s$ and $y_f = 0m$, we get: } \\[br] \frac{1}{2}mv_f^2 = mgy_i \\[br]v_f=\sqrt{2gy_i}.[br][/math][/center]

[color=#1e84cc][i] EXAMPLE: Consider the rollercoaster above. Let's assume that the rollercoaster in the diagram is moving at [math]10m/s,[/math] and is 7.0m above the bottom dip in the track. Let's find out how fast the rollercoaster will be going as it arrives at the bottom of the dip. To be fair, we need to neglect the length of the rollercoaster since what will matter in our equation is the height. With a long rollercoaster this would be the average height of the several cars. So I will assume that the average height will have descended by 7.0m by the time it arrives at the dip. We will also neglect all dissipative forces. [br][br]SOLUTION: Since there are no dissipative forces, [math]\Delta K +\Delta U=0.[/math] Applying that equation, we get [math]K_f-K_i+(U_f-U_i)=0.[/math] This gives [math]\frac{1}{2}m(v_f^2-v_i^2) +mg(h_f-h_i)=0.[/math] Notice that [math]h_f-h_i = -7.0m,[/math] or that the rollercoaster is losing height. So we get [math]v_f = \sqrt{2g(h_i-h_f)+v_i^2}=15.5m/s.[/math] [/i][/color]

[color=#1e84cc][i]EXAMPLE: Suppose Tarzan (mass m) holds onto a vine of length L that makes an angle [math]\phi[/math] with respect to vertical, and swings downward. As he passes the bottom point, what is the tension in the vine? Assume that there is no appreciable air drag.[br][br]SOLUTION: This problem includes elements from Newton's laws, circular motion and energy conservation. From the current section we should expect that as he falls, there is a loss of gravitational potential energy in exchange for an increase in kinetic energy - or he speeds up. So [math]\Delta K +\Delta U = 0.[/math] The initial kinetic energy is zero, so [math]\frac{1}{2}mv_f^2 = -mg\Delta y=mgh.[/math] [br][br]Tarzan's path is circular, so as he swings, he is undergoing centripetal acceleration toward the center of the circular path. The magnitude of that acceleration is [math]a_c=\frac{v^2}{r}.[/math] [br][br]Lastly, at the bottom of the swing the centripetal acceleration is upward and is due to the sum of the forces, which are upward tension and downward gravitational force. Therefore we may write (if we call upward the +y direction) [math]\sum F_y = F_T-mg=ma_c=\frac{mv^2}{r}[/math] where positive is upward. [br][br]To piece these all together we need an expression for [math]\Delta y[/math] in terms of [math]\phi.[/math] Notice that [math]\Delta y = L-L\cos\phi[/math] if you draw the diagram like a unit circle where L (the length of the vine) is the radius. [br][br]Substituting this into the energy conservation equation and writing the kinetic energy in terms of centripetal acceleration gives [math]F_T=2mg(1-\cos\phi)+mg.[/math] Notice that for a vine this term has a maximum where [math]\phi=\pi/2[/math] since that's a horizontal vine. This leads to a max tension of [math]F_T=3mg.[/math] Tarzan would have to have a tremendous grip to hang on from that initial swing angle since he'd be supporting 3 times his own weight! [/i][/color]

I went to the circus with my two young sons as it passed through town a few years back. During the show there was an act in which a man rode a motorcycle around a vertical loop that was perhaps 5-7 meters in diameter. What was interesting, and made for a spectacle, was that the top segment of the loop was missing - perhaps as much as 45 degrees of the circular segment. He rode around and went inverted perhaps 10 times, and then exited the loop without mishap. [br][br][color=#1e84cc]EXAMPLE: Suppose as a variation on that act that I wish to ride my bicycle down a long ramp and simply coast under the influence of gravity, enter the loop and go around just once before exiting. How high should I start on the ramp (in terms of the loop's radius R) in order to be successful? We will suppose that non-conservative forces like air drag and friction are negligible, even though in reality they certainly are not.[br][br]SOLUTION: [br]First let's discuss getting over the top of a circular loop with the top part missing. To avoid a serious crash, the normal force at the top (contact between bike and track) must go to zero. That means that only the gravitational force is present. Since the track is circular, the gravitational force [math]F_g=mg[/math] must be producing centripetal acceleration given by [br][br][center][math]F_g=ma_c = m\tfrac{v^2}{r}. \\[br]mg=m\tfrac{v^2}{r} \\[br]\text{Solving for $v^2$ since I'll need it below, gives me:} \\[br]v^2=gr. [/math][/center][br][br]If I ignore non-conservative forces like air drag and rolling resistance, I can write [br][br][center][math]W_{nc}=\Delta K + \Delta U_g \\[br]0= \Delta K + \Delta U_g. \\[br]\text{Expanding the kinetic energy term and the potential energy, this gives:} \\[br]0=\tfrac{1}{2}mv_f^2-\tfrac{1}{2}mv_i^2+mg\Delta y. \\[br]\text{Since $v_i=0$, plugging in the $v^2$ expression from above gives: } \\[br]0=\tfrac{1}{2}mgr-0+mg\Delta y.[br]\text{ Solving for the height descended, from top of ramp to top of loop, we get: } \\[br]\Delta y = -\tfrac{1}{2}r. \\[br]\text{So we need to start half a radius above the top of the loop or 5/2r above the ground.}[br][/math][/center][/color]