Ask Uncle Colin: An elastic speed limit

Ask Uncle Colin is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.

Dear Uncle Colin,

I clumsily dropped a particle of mass $m$! Luckily, it's attached to a light elastic string with a modulus of elasticity of $3mg$ and natural length $a$. The other end of the string is attached to the point where I dropped the weight from.

When I say 'dropped', I mean 'propelled downwards with a speed of $\sqrt{3ga}$', of course. I'm worried it's going to go too fast! What's the fastest it goes?

-- Hands Off Our Kinetic Energy!

Hello HOOKE! There are a few ways to approach this. You hint at one of them in your name: you can work with the conservation of energy to find out when the velocity stops changing.

The total energy throughout the travel is made up of (potential) + (kinetic) + (elastic). I'll take the zero level for height as the point where the particle is dropped.

Initially, then, the energy is $E_0 = (0) + \left( \frac 12 m v_0^2 \right) + 0$, which comes to $E_0 = \frac 32 mga$.

Once the elastic becomes taut, let's write 'height' as $(a+x)$, where $x$ is the extension of the string; the velocity is unknown.

$E_t = -\left( mg(x+a) \right) + \left( \frac 12 mv^2 \right) + \left( \frac 12 \lambda \frac{x^2 }{a} \right)$. That needs some tidying up: $E_t = mgx + mga + \frac 12 mv^2 + \frac 32 mg \frac{x^2}{a}$

By conservation of energy, we have $E_0 = E_t$, so $\frac 32 mga = -mgx - mga + \frac 12 mv^2 + \frac 32 mg \frac{x^2}{a}$.

It's still a mess. Everything has a factor of $m$ we can get rid of, and it's probably good to multiply by 2 as well: $3ga = - 2gx - 2ga + v^2 + 3g \frac{x^2}{a}$

Now rearrange to get $v^2$ on its own: $v^2 = g\left( 5a + 2x - 3\frac{x^2}{a} \right)$

Differentiate with respect to $x$ and you get $v \diff vx = g\left( 2 - 6\frac{x}{a} \right)$, which has its extremum when $2 - 6\frac xa = 0$, or when $x = \frac 13 a$.

Putting this back into the $v^2$ equation, we get: $v^2 = g\left( 5a + \frac 23 a - \frac{a}{3} \right)$, or $v^2 = \frac {16}{3} ga$. The maximum speed is $\frac 43 \sqrt {3ga}$.

As I say, there are alternative approaches: you can find the maximum speed by looking for the point where the net force is zero; you can model the freefall part of the travel with SUVAT and then use either energy or a differential equation; I'm sure there are other approaches that didn't spring to my mind.

I hope your particle is all right after its trauma!

-- Uncle Colin

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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