"Anguler velocity calculation with varying acceleration"

Question

As you can see from the image we have inverted pendulum stick fixed to ground by the point "K".Rotation of the stick around is only movement.System has no gravity and friction.Only "F" force is applied to the point "P".The system is exposed this force for "t" amount of time.I want to calculate the change of "ϑ" in arbitary choosen time.

To solve this problem first i accepted "F" is always right angled(90 degree) to point "P" through motion. I calculated angular velocity like this:

F=m*a 5=1*a a=5 (m/s^2)

After 1 second , velocity of the stick is :

v=a*t =5.1= 5(m/s)

avarage velocity =(First speed + Last speed ) / 2 = (0+5)/2 =2.5 (m/s)

avarage angular velocity w=v/r => 2.5 / 1 = 2.5 (1/s)

So angular velocity is 2.5 (1/s)

My first question is how to interpret this number. Does it mean stick angle changes 2.5 degree per second?

when i dont make the assumption that F is right angled to P through movement. Effective force (F1) becames F1=F* sin(ϑ) , which means F1 changes according to angle of "ϑ" and therefore ,"a" the accelaration of the stick is always changing.

In order to calculate "a" in time of "t1"

a(t1)=F*sin(ϑ+∂ϑ) / m

// initial ϑ angle + change of angle through time ∂ϑ

Which method can be used to solve this kind equation to calculate anguler speed? I come up with a way to solve this but couldnt progress further..

Answer 1

So you calculate the torque at K as $$ \tau_K = -F_1 \ell \cos \theta = -F \ell \sin \theta \cos \theta \tag{1}$$

Now you calculate the MMOI about K $$I_K = m \ell^2 $$

and you find the rate of change of angular velocity from $$ \tau_K = I_K \ddot{\theta}$$

$$ \ddot\theta = -\frac{F \ell}{I_K} \sin \theta \cos \theta \tag{2} $$

You solve this with a direct integral.

The left-hand side is

$$ \int \ddot \theta \,{\rm d}\theta =\int \left(\tfrac{\rm d}{{\rm d}t} \dot \theta \right){\rm d}\theta = \int_0^{\dot \theta} \dot \theta {\rm d} \dot \theta = \tfrac{1}{2} \dot \theta ^2 \tag{3}$$

and the right-hand side

$$ -\int_{\theta_0}^\theta C \sin \theta \cos \theta\, {\rm d}\theta = -\tfrac{1}{2} C \left( \sin^2 \theta - \sin^2 \theta_0 \right) \tag{4}$$

where $C = \tfrac{F \ell}{I_K}$ and $\theta_0$ is the angle at time zero.

This means the angular velocity as a function of time is $$\dot \theta = \sqrt{ C \left( \sin^2 \theta_0 - \sin^2 \theta \right) } \tag{5} $$

which only has a solution if $\theta \leq \theta_0 $.

You will receive the same answer if you balanced kinetic energy with potential energy. The speed is a function of angle only.

The last step is to find the time it takes to reach this speed.

$$ t = \int \frac{1}{\dot \theta} \, {\rm d} \theta = \int_{\theta_0}^{\theta} \frac{1}{\sqrt{C \left( \sin^2 \theta_0 - \sin^2 \theta \right)}} \,{\rm d} \theta $$

which according to Wolfram Alfa the solution involves the Elliptic Integral of the First kind $\mathrm{F}(x,k^2) = \int_0^{\pi/2} \frac{1}{\sqrt{1-k^2 \sin^2 x}}\,{\rm d}x$ and evaluates to

$$ t = \frac{ \frac{\sqrt{\cos^2 \theta - \cos^2 \theta_0}}{\sin \theta_0} \mathrm{F}(\theta, \, \csc^2 \theta_0)}{\dot \theta} \tag{6} $$