Unclear on interpretation of equation - oscillation maybe?

[SeventyOne]

Equation: angular acceleration = 3 (angular displacement) - 100
At start, when displacement is 0, the initial angular velocity is 20 rad per second clockwise.
I expect that the (-100) term will cause the angular velocity to decrease to 0 but since displacement is increasing during this process, the angular acceleration will be becoming less negative due to the first term in the equation which depends on the displacement value.
When the velocity reaches zero, the displacement is at its maximum. The acceleration is still negative therefore the body will then spin in the opposite direction, anticlockwise.
If I'm correct so far, the displacement will now decrease.
As the displacement decreases, the angular acceleration becomes more negative making the velocity faster anticlockwise. This further decreases the displacement...and then what?

 

[Chestermiller]

Equation: angular acceleration = 3 (angular displacement) - 100
At start, when displacement is 0, the initial angular velocity is 20 rad per second clockwise.
I expect that the (-100) term will cause the angular velocity to decrease to 0 but since displacement is increasing during this process, the angular acceleration will be becoming less negative due to the first term in the equation which depends on the displacement value.
When the velocity reaches zero, the displacement is at its maximum. The acceleration is still negative therefore the body will then spin in the opposite direction, anticlockwise.
If I'm correct so far, the displacement will now decrease.
As the displacement decreases, the angular acceleration becomes more negative making the velocity faster anticlockwise. This further decreases the displacement...and then what?

How is the angular acceleration related kinematically to the time derivatives of the angular displacement?

 

[SeventyOne]

Sorry, but I don't know.

 

[person_random_normal]

Understand that - acceleration is cause and velocity is effect

 

[Fred Wright]

I think what Dr. Chestermiller is asking you is: Do you know how to translate the word problem into a differential equation?
hint: angular acceleration $= \frac {d^2 \theta}{dt^2}$ and angular displacement $= \frac {d \theta}{dt}$.
Can you solve:
$$\frac {d^2 \theta}{dt^2} -3\frac {d \theta}{dt}=100$$ and apply the initial condition?

 

[SeventyOne]

No, I don't know how to do that. I've taken a brief look on youtube and there are a lot of videos there on how to solve differential equations. So far, I haven't found one similar to this equation. I'll keep looking. Thanks very much.

 

[SeventyOne]

It seems that I have a second order non-homogenous differential equation with initial values.
The videos seem to have equations of the form
Ay'' +By' + Cy = f(x)
Mine is
y'' - 3y' = 100
I've found the first part of the solution to be
y= -20/3 + 20/3 e^3x
and the particular solution to be
-100/3 x
This is all totally new to me and I'm not sure that I have any of this correct, especially the particular solution. My thinking on that was that y'' = 0 and y' = constant, therefore if y'' - 3y' = 100 then y = -100/3 x
So the general solution is
y= -20/3 + 20/3 e^3x -100/3 x
θ(t) = -20/3 + 20/3 e^3t -100/3 t
I put this into my graphing calculator
at t= 0, θ=0 then θ goes negative
there's a turning point around x= 0.159 and at t=0.318, θ=0
after this, θ becomes more and more positive.
This isn't at all what I expected for the motion of the rotating body. I was expecting what I described above and then, as displacement returned to zero and then became negative, that the velocity would continue to increases until the equipment broke apart.
I'm confused. I think I haven't solved the equation correctly. Can anyone help?

 

[SeventyOne]

Using this website
http://www.wolframalpha.com/widgets/view.jsp?id=17adb9cfd6c67a920240e6db6fd8e8a1
y''-3y'+0y = 100
y(0)=0
y'(0)=20
The solution is
y = 1/9 (-300x+160e^3x -151)
Not the same as my solution. Taking this solution as correct, the graph of y versus x shows the exponential increase of y with increasing x. Again, not what I was expecting for angular displacement versus time. I'm totally stuck.

 

[Fred Wright]

I made a typo in my previous post. The equation should read:$$\ddot {\theta}-3\dot {\theta}=-100$$ where the dots indicate time differentiation and $\theta$ is the angle. The first thing you want to do is solve the homogeneous equation $$\ddot {\theta_h}-3\dot {\theta_h}=0$$ and write $$\frac {d \dot {\theta_h}} {dt}=3\dot {\theta_h} \Rightarrow \int \frac {d \dot {\theta_h}} {\dot {\theta_h}}=3 \int dt$$$$log(\dot {\theta_h})=3t +C \Rightarrow \dot {\theta_h}=Ae^{3t}$$
where $A=e^C$. Integrate $\dot {\theta_h}$ to get$$\theta_h=\frac{A} {3}e^{3t}+B$$where B is a constant indicating the initial angular displacement, which we can take to be 0. We seek a solution which is the sum of the homogeneous part and the particular part $$\theta=\theta_h + \theta_p$$ From the right hand side of the original eqn. we suspect that $\theta_p$ is a polynomial in t. Let's guess that $\theta_p = -100t$ and plugging $\theta_h + \theta_p$ into the equation we see that indeed $$\theta=\frac{A} {3}e^{3t} -100t$$You can now apply the initial condition $\dot {\theta}=20$ at $t=0$ to solve for A, i.e. $A=20$. We see that the rotation reverses direction at $$e^{3t}=15t$$ which must be solved graphically for t.

Salaam,
Fred

 

[SeventyOne]

Fred, can I just ask you this first? Why is the equation θ'' - 3θ' = -100 and not θ'' - 3θ = -100 ?
Thank you.

 

[Fred Wright]

Fred, can I just ask you this first? Why is the equation θ'' - 3θ' = -100 and not θ'' - 3θ = -100 ?
Thank you.

You are right! I misread angular displacement to be angular velocity--sorry, I feel like a fool. The approach, however, is the same. Solve the homogeneous equation $$\ddot {\theta_h} - 3\theta_h =0$$ and guess the solution: $\theta_h=Ae^{\sqrt {3}t}$ which satisfies the homogeneous equation. Now guess the particular solution $\theta_p=-\frac {100} {3}$ and $$\theta=Ae^{\sqrt {3}t}-\frac {100} {3}$$ which satisfies the general equation. Apply the initial condition to solve for A.
Evidently $\theta$ increases exponentially with time and never changes direction.

 

[SeventyOne]

Using θ'' -3θ = -100, the solution I get is
θ(t)=-22.44e^-√3t - 10.893e^√3t + 100/3
This fits with what I had expected to happen to the rotating body. So is this correct?

 

[Fred Wright]

I think you are correct. That's the answer I get.

 

[SeventyOne]

Thank you SO much for your help. I really appreciate it.