# What are the steps to the approximation for this falling pendulum problem?

I am trying to understand the math here. How long would it take for an upright rigid body to fall to the ground? But I am lacking in some basic mathematical concept/techniques.

In several answers, an estimate is given that the angular acceleration can be approximated as

$$\ddot{\theta}=\dfrac{g}{h}\theta \tag{1}$$

Which can be approximated as

$$t=\sqrt{\dfrac{h}{g}} \ln\left(\dfrac{\theta}{\theta_0}\right) \tag{2}$$

Could someone kindly explain the steps/method to go from equation 1 to equation 2?

• Are you asking how long it takes once the "fall" begins? Because in classical mechanics, the rigid body will never leave its upright position until disturbed. There's a standard "gedanken" in introductory quantum mech that asks for the Heisenberg-based probability of the object remaining upright for X amount of time. Aug 31, 2022 at 14:10
• The solution of equation (1) is $\theta=\theta_0\,e^{ \sqrt{\frac g h}\,t}$ solve this equation for t you obtain Eq. (2)
– Eli
Sep 2, 2022 at 17:03
• Thanks @Eli Basically I am trying to understand how to get that $\theta_{t} =\theta_{0}e^{\sqrt{\frac{g}h}t}$. Sorry for my question was not very clear. Anyway after more studying on how to solve homogenous ODE, and running some simulated values I conclude that the answer to that linked question may not be accurate. The solution for $\theta_{t}$, I believe, should be $$\theta_{t} =\frac{\theta_{0}}2e^{\sqrt{\frac{g}h}t} + \frac{\theta_{0}}2e^{-\sqrt{\frac{g}h}t}$$ Sep 4, 2022 at 3:21

This is not a complete solution but may help you analyse it.

My assumption is that you have a rigid body on the point of toppling and you want to know when it will have rotated 90 deg to hit the surface.

Torque (or moment) and Moment of Inertia (MOI, or Second moment of mass) are kind of analogous with force and mass and we have the relationship

Torque = I x Alpha

where I is the MOI and Alpha is angular acceleration. MOI is highly dependant on the geometry of the object, and needs to be calculated 'about the pivot point'.

So to get to an expression for theta you need to integrate twice from the expression for Alpha

Now the torque is not constant. If the centre of mass (m) is at height h then the torque is equal to (h cos theta x m x g)

Once you have the expression for Theta in terms of h, m, and I, then solve for theta = 90 deg.

Due to my poorly written question, I unfortunately did not get a satisfactory answer. After much studying and some experiments, my own solution differs from the linked question's solution. Here is my solution.

Given the equation, find $$t$$, the time it takes to go from $$\theta_0$$ to $$\theta_t$$, with intial condition $$\theta'_0 = 0$$ $$\theta''=\frac{g}hθ\tag{1}$$

Part 1 solve the differential equation

substitute $$\theta=e^{rt}, \theta''=r^2e^{rt},$$ and get $$r^2e^{rt}=\frac{g}he^{rt}$$ $$\Rightarrow r=\sqrt{\frac{g}h}$$

solution for $$\theta$$ $$\theta_t = C_{1}e^{rt} + C_{2}e^{-rt} \tag{1a}$$ differentiate both sides $$\theta'_t = rC_{1}e^{rt} - rC_{2}e^{-rt}\tag{1b}$$

solve $$(1a)$$ and $$(1b)$$ using intial value $$\theta'_0 = 0$$, $$C1 = C2 = \frac{\theta_0}2$$

The solution for the differential equation is $$\theta_t=\frac{\theta_0}{2}(e^{\sqrt{\frac{g}h}t} + e^{-\sqrt{\frac{g}h}t})\tag{1c}$$

Part 2, solve for t

For larger values of $$t$$, the $$e^{-rt}$$ term has a much smaller contribution to the sum. $$\theta_t\approx\frac{\theta_0}{2}e^{\sqrt{\frac{g}h}t}$$ $$t\approx\sqrt{\frac{h}g}ln|\frac{2\theta_t}{\theta_0}|\tag{2a}$$

However, if $$\sqrt{\frac{h}g}t$$ is small, the $$e^{-rt}$$ term cannot be ignored. One way to approximate, is to use the first few terms of the Taylor series of the exponential function. $$e^x=1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ...$$ then $$e^x + e^{-x}=2(1 + \frac{x^2}{2!} + \frac{x^4}{4!} + ...)$$ $$\Rightarrow e^x + e^{-x}\approx2(1 + \frac{x^2}{2!} + \frac{x^4}{4!})\tag{2b}$$

let $$y=x^2=(\sqrt{\frac{g}h}t)^2\tag{2c}$$ subsitute $$(2b)$$ and $$(2c)$$ into $$(1c)$$ $$\theta_t\approx\theta_0(1 + \frac{y}{2!} + \frac{y^2}{4!})$$ $$\frac{\theta_0}{4!}y^2 + \frac{\theta_0}{2!}y + (\theta_0-\theta_t) = 0\tag{2d}$$

solve for $$y$$ using quadratic formula, and finally $$t=\sqrt{\frac{h}g}\sqrt{y}$$