Practice Makes Perfect VI

48 Practice Makes Perfect VI

More practice!

Exercise 48.1: Springs and Pulleys

A spring of constant $k$ has an unstretched length $L_0$ . You hang two identical masses from the spring as shown in the setup below. What is the length of the spring with the masses hanging from it? I’ll help you to get to the answer.

Start by making your drawing, and then compare to mine.

Consider the string on the right. There is a tension force pulling on the spring. What is the force pair of that tension?

Use the information above to find the length of the spring with the masses hanging from it.

Now imagine the mass on the right is $M$ , while the mass on the left is $m$ , and that $M>m$ , so that the masses (and spring) will move. Will the tension on the right string be bigger or smaller than the tension on the left string? Justify your answer.

Find the length of the spring $L$ as the big mass descends.

In the next problem we’ll learn about how rockets work. This problem is a bit tougher than some of the other stuff we’ve done, but… rockets!

First, though, we will learn how to use infinitesimals. Consider a very small number, e.g. $x=0.001$ . If we square this number, we get $x^2=0.000001$ , which is much, much smaller. That is, the product of two tiny numbers is a super-duper tiny number.

Now, remembe infinitesimals like $dx$ or $dy$ are the tiniest tiny thing you can have. Consequently, if you take the product of two infinitesimals, the result is so super tiny that we can ignore it: i.e. $dxdy=0$ . This is a rule for how to use infinitesimals, and it follows from our original observation about multiplying tiny numbers.

Bottom line: the product of two infinitesimals is zero.

Exercise 48.2: Rocket Launch

We will take a (highly simplified) look at how rockets work.

A rocket works by igniting the fuel to generate tremendous pressure. This pressure shoots out the gasses produced by the burning of the fuel out of the rocket downwards at very high speeds.

A. Explain why shooting stuff down at high velocity makes the rocket go up.

Here’s a nice demonstration of this same principle using more pedestrian materials: a tricycle and a fire extinguisher. Original video can be found here.

B. Let $m(t)$ be the amount of fuel in the rocket at time $t$ , and $m_0$ be the total amount of fuel at ignition. The rocket begins to burn through the fuel at a constant rate $k$ . We will adopt the convention that $k$ is positive, i.e. the amount of fuel burnt in time $dt$ is $dm_{\rm burnt}=kdt$ . These are infinitesimals because we are only able to burn a tiny amount of fuel $dm_{\rm burnt}$ in time $dt$ .

Find an expression for $m(t)$ , and determine for how much time it takes for the rocket to burn through all of its fuel.

C. Burnt fuel is ejected from the rocket at a fixed speed $v_{\rm e}$ (the “e” is for “exit velocity”) from the point of view of the rocket. If the rocket is moving upwards with speed $v$ , what is the velocity of the burnt fuel as seen from the ground?

D. Let $M$ be the mass of the rocket (without fuel). After the rocket burns fuel in the instant $dt$ , it’s upwards velocity will have increased. Denote its final velocity $v+dv$ . Draw a picture of the rocket+fuel at time $t$ , and time $t+dt$ . You can check your answer with mine.

E. Use your drawing to set up the conservation of momentum equation taking into account that there is an external force: gravity! Remember what you learned about infinitesimals when setting up this equation. Simplify your answer to find an equation for $dv/dt$ .

Let’s take a second to look at the final answer for the acceleration,

$\large \frac{dv}{dt} \normalsize = \frac{kv_e}{M+m} - g$ .

The first term looks like the acceleration due to a constant force $kv_e$ , which makes sense since the thrust is constant. The second term is the acceleration due to gravity.

E. The above expression is in terms of $m(t)$ . But- we know what $m(t)$ is! Plug this back in, multiply both sides by $dt$ , and integrate to find $v(t)$ .

Again, let’s pause and look at our final answer,

$v(t)=v_e\ln \left( \large\frac{M+m_0}{M+m_0-kt}\normalsize\right) - gt.$

We see that the final speed is proportional to the exit velocity of the burnt fuel. This makes sense: if I shoot out the fuel at high speed, there is more recoil, and the rocket ends up moving faster.

F. Find an expression for the speed of the rocket after it burns all its fuel. Use the fact that $k=m_0/\tau$ to get rid of any $k$ ‘s in your solution.

Consider our final expression, $v_{\rm f} = v_e \ln( 1+m_0/M) - g\tau$ . To make the final speed large, we need a large number inside that logarithm, which means that $m_0/M$ is a large number: we need a lot of fuel! That’s why rockets have such ridiculously large fuel tanks. Note this also means that we can ignore the “1” in the logarithm, since $1+x\approx x$ when $x$ is a large number. Thus, the final speed of the rocket is

$v_{\rm f} \approx v_e \ln(m_0/M) - g\tau.$

G. To achieve Earth orbit, a rocket needs to move at a velocity of $\approx 8\ {\rm km/s}$ , or about 18,000 mph. The fully loaded space shuttle weighs $\approx 100\ {\rm tonne} = 10^5\ {\rm kg}$ . A typical fuel exit velocity is $v_e \approx 2.5\ {\rm km/s}$ , and the rockets fire for about 2 min. How much fuel does a rocket engine need to get the space shuttle to space?

This is not a bad estimate! Note that the fuel required to get the rocket up to space is weight roughly 40 times the weight of the rocket! In practice, the amount of fuel used is a bit less ( $\approx$ 1700 tonne). This is partly due to the fact that rockets jettison part of their tanks half way through launch, reducing the amount of mass that needs to go up.

This calculation demonstrates the fundamental problem with rockets: to go up, you need a lot of fuel, but that means you need to bring up that fuel with you, further increasing your need for fuel! It all add ups very quickly. Indeed, that’s why it is difficult to build jet packs out of machine guns.

48 Practice Makes Perfect VI

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