Where's my mistake? (Pendulum with gravity)

[Efeguleroglu]

I have a pendulum and an object with radius "R" and mass "m". There are forces: constant gravitational acceleration and tension on the rope. I can write:
$$x=R sin(\theta) \ \ y=R cos(\theta)$$
$$\dot{x}=R\dot{\theta}cos(\theta) \ \ \dot{y}=-R\dot{\theta}sin(\theta)$$
$$\ddot{x}=R\ddot{\theta}cos(\theta)-R{\dot{\theta}}^{2}sin(\theta)\ \ \ (1)$$
$$\ddot{y}=-R\ddot{\theta}sin(\theta)-R{\dot{\theta}}^{2}cos(\theta)\ \ \ (2)$$
$$a=\frac{T}{m}=\frac{{\dot{x}}^{2}+{\dot{y}}^{2}}{R}={\dot{\theta}}^{2}R\ \ \ (3)$$
$$R\ddot{\theta}cos(\theta)-R{\dot{\theta}}^{2}sin(\theta)=\frac{T}{m}sin(\theta)={\dot{\theta}}^{2}Rsin(\theta)$$
Combine (1) and (3).
$$\ddot{\theta}cos(\theta)-{\dot{\theta}}^{2}sin(\theta)={\dot{\theta}}^{2}sin(\theta)$$
$$tan(\theta)=\frac{\ddot{\theta}}{2{\dot{\theta}}^{2}}\ \ \ (4)$$
Combine (2) and (3).
$$-R\ddot{\theta}sin(\theta)-R{\dot{\theta}}^{2}cos(\theta)=\frac{T}{m}cos(\theta)-g=R{\dot{\theta}}^{2}cos(\theta)-g$$
Find tan(θ) and equate to (4).
$$tan(\theta)=-\frac{2{\dot{\theta}}^{2}}{\ddot{\theta}}+\frac{g}{cos(\theta)R\ddot{\theta}}=\frac{\ddot{\theta}}{2{\dot{\theta}}^{2}}\ \ \ (5)$$
$$cos(\theta)=\frac{2{\dot{\theta}}^{2}g}{R({\ddot{\theta}}^{2}+4{\dot{\theta}}^{4})}\ \ \ and\ \ \ y=\frac{2{\dot{\theta}}^{2}g}{({\ddot{\theta}}^{2}+4{\dot{\theta}}^{4})}\ \ \ (6)$$
From (4).
$$cot(\theta)=\frac{2{\dot{\theta}}^{2}}{\ddot{\theta}}\ \ \ (7)$$
Combine (2) and (3).
$$-R\ddot{\theta}sin(\theta)-R{\dot{\theta}}^{2}cos(\theta)=\frac{T}{m}cos(\theta)-g=R{\dot{\theta}}^{2}cos(\theta)-g$$
Find cot(θ) and equate to (7).
$$cot(\theta)=\frac{g}{2sin(\theta){\dot{\theta}}^{2}{R}}-\frac{\ddot{\theta}}{2{\dot{\theta}}^{2}}=\frac{2{\dot{\theta}}^{2}}{\ddot{\theta}}\ \ \ (8)$$
$$sin(\theta)=\frac{\ddot{\theta}g}{(4{\dot{\theta}}^{4}+{\ddot{\theta}}^{2})R}\ \ \ and \ \ \ x=\frac{\ddot{\theta}g}{(4{\dot{\theta}}^{4}+{\ddot{\theta}}^{2})}\ \ \ (9)$$
So from (6) and (9).
$$x^{2}+y^{2}=g^{2}$$
But that doesn't make sense. There must be a mistake. How does
$$g=R$$
have to be hold?

 

[hutchphd]

Check the units of each expression you have written and you will find your error where they do not match. Do this always. I am terrible at Algebra.

 

[Efeguleroglu]

That's too embarassing. I forgot to square the denominator. Forgive me for wasting your time :(

 

[hutchphd]

That's too embarassing. I forgot to square the denominator. Forgive me for wasting your time :(

That's too embarassing. I forgot to square the denominator. Forgive me for wasting your time :(

Not a waste if you develop good habits! Glad I could help.

 

[Efeguleroglu]

How will I delete this thread?

 

[davenn]

How will I delete this thread?

it's left there ... some one else may learn from it

 

[Efeguleroglu]

I wrote
$$a_c=\frac{T}{m}$$
But it must be equal to
$$a_c=\frac{T+mgcos(\theta)}{m}$$
Actually I mistakenly used T as the centripetal force. It can easily be replaced. I just wanted to fix that.

 

[DennisN]

That's too embarassing. I forgot to square the denominator. Forgive me for wasting your time :(

Hi, @Efeguleroglu, I haven't checked the math, but I just want to say that "to err is human." I've done plenty of mistakes during the years, and I still do, though I hope the frequency has been reduced . And errors and failures are excellent learning opportunities!