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Given the stated system in the photo we are suppose to prove simple harmonic motion when given an initial displacement $x$

Problem statement

I first considered the total energy of the system which we will call $H$ $$H=\frac{1}{2}mv^2+\frac{1}{2}I_{0}\omega^2+2\left(\frac{1}{2}kx^2\right)$$

However if we take the point to consider as the contact region between the circle and the ground we can take away the translational kinetic energy term and it reduces to

$$H=\frac{1}{2}I_{1}\omega^2+2\left(\frac{1}{2}kx^2\right)=\textrm{constant}$$

We note that this equals a constant by the law of conservation of energy Then consider the total time derivative of $H$ And we note that differentiating a constant gives us 0

$$\frac{\mathrm dH}{\mathrm dt}=I_{1}\dot {\theta}\ddot {\theta}+2kx\dot {x}=0$$

We convert from polar form to generalised Cartesian coordinates

$$\frac{\mathrm dH}{\mathrm dt}=I_{1}\frac{\dot {x} \ddot {x}}{R^2}+2kx\dot {x}=0$$

$$I_{1}=\frac{7}{5}mR^2$$

Substituting I yield..... $$\frac{\mathrm dH}{\mathrm dt}=\frac{7}{5}m\dot {x} \ddot {x}+2kx\dot {x}=0$$ Doing some manipulation.... $$\frac{7}{5}m\dot {x} \ddot {x}=-2kx\dot {x}$$

We can then cancel out $\dot {x}$ from both sides

$$\frac{7}{5}m \ddot {x}=-2kx$$

I'll skip some steps and simply completing the manipulation we obtain

$$\ddot {x}=-\frac{10k}{7m}x$$

which resembles the form $a=-\omega^2x$

where $\omega=\sqrt{\frac{10k}{7m}}$


Most importantly, why does this method work? And what is this method called as I have never seen it before, I heard my friends briefly mentioned this is a Hamiltonian but it doesn't resemble as such Is this a valid method to approach mechanics problems by writing down the total energy and taking the time derivative as being equal to zero?

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It is a method that is generally used for conservative unidimensional problems (problems with only one degree of freedom, here your angle $\theta$ or cartesian coordinate $x$). You'll notice that it is equivalent to using Newton's second law in this case : let us write the total energy

$E = \frac{1}{2} m v^2 + V(x)$,

$V$ being potential energy. The problem is conservative so $E$ is a constant, we derivate with respect to time to get

$$0 = m v \dot{v} + \frac{\mathrm d V(x)}{\mathrm dt}$$

but $$ \frac{\mathrm d V(x)}{\mathrm dt} = \frac{\mathrm d x}{\mathrm dt} \frac{d V(x)}{dx} = v \frac{\mathrm d V(x)}{\mathrm dx}$$

and remember that the conservative force $F$ is linked to its potential energy by $F = - \frac{\mathrm d V(x)}{\mathrm dx}$. Simplifying by the velocity $v$, we get

$m \frac{\mathrm dv}{\mathrm dt} = F$ which is nothing but Newton's second law in one dimension.

To summarize : if you have a conservative problem, if it is in one dimension, use the conservation of energy method which is usually much easier to apply, and is equivalent to Newton's second law. If you are in more that one dimension (problems of satellites orbiting around planet, etc) you'll need to apply a vectorial theorem such as Newton's second law or the angular momentum theorem.

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  • $\begingroup$ Can one possibly apply this to generalised coordinates like how Euler-Lagrange equations can be applied to each coordinate $\endgroup$ – Russell Yang Feb 9 '16 at 14:32
  • $\begingroup$ Yes, this method is know as Hamiltonian mechanics. $\endgroup$ – Dimitri Feb 9 '16 at 14:35
  • $\begingroup$ And by conservative, would this method work of friction were involved in which it is non conservative. Could this method then work if I consider the work done by friction and differentiate that WRT time @Dmitri $\endgroup$ – Russell Yang Feb 9 '16 at 14:36
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    $\begingroup$ Yes it also would, but you need to know the precise expression for the dissipative force. The energetic theorem then needs to be applied as $\frac{dE}{dt} = F_d . v$ where $F_d$ is the sum of all dissipative forces. $\endgroup$ – Dimitri Feb 9 '16 at 14:39

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