Approximating a Simple Harmonic motion

[hms.tech]

Homework Statement

Homework Equations

F=ma

The Attempt at a Solution

I did the first three parts . The last part of this question is quite hard, i tried using Newton's 2nd law of motion but ... here is what happens :

T is the tension, as stated in the question .

so the equation of motion should be :

T cos($\varphi$) = m a
Since $\varphi$ = pi/2 -2θ
so using some simplification we get cos$\varphi$=sin(2θ)

Now, since θ is small sin(2θ) can be approximated to 2θ.

so : $\frac{2Tθ}{m}$ = a

for angular acceleration = α

so, α = $\frac{2Tθ}{mr}$
r= a meters (the radius)

final answer (which is wrong) α = $\frac{2Tθ}{ma}$

Note : a is acceleration
a is the radius given in the question
α is the angular acceleration
The correct answer given in the answer archives is :
α = -2Tθ (whoa...where did the -ve sign come from :S)

 

[TSny]

final answer (which is wrong) α = $\frac{2Tθ}{ma}$

Look's like a typing error in writing the denominator above.

α = -2Tθ (whoa...where did the -ve sign come from :S)

(Again, a typing error? The dimensions on each side do not match.)

Did you take into account the direction of the force T cos($\varphi$)? Is it in the positive or negative direction of motion?

 

[hms.tech]

Look's like a typing error in writing the denominator above.




(Again, a typing error? The dimensions on each side do not match.)

Did you take into account the direction of the force T cos($\varphi$)? Is it in the positive or negative direction of motion?

No there is no typing error, i just rechecked the solution archive, it is exactly as i wrote it !

I really can't say anything about its direction since i have not defined them yet.
Lets do that ! For the tension
Let the clockwise tangent be -ve and the tangent which points in the anti clockwise direction be positive !
Now let's check :
Tension is -ve for the first part of its journey (as shown in the diagram see 1st post) while at the same time the displacement (angular) is positive. Well that solves the PHYSICS of the problem. I really can't see , in all honesty, how does mathematics prove the exact same thing. in other words, i understand why there must be a -ve sign in the equation but i can't prove it using the mathematical model . Can u help me there ?

 

[TSny]

No there is no typing error, i just rechecked the solution archive, it is exactly as i wrote it !

Your final answer:

α = $\frac{2Tθ}{ma}$

cannot be correct because the right side does not have dimensions of angular acceleration. I think you meant to write $r$ instead of $a$ in the denominator.

Likewise, the equation you stated as the solution from the archives:

α = -2Tθ

cannot be correct for the same reason.

Regarding the signs, let $\textbf{u}_{\theta}$ be a unit vector tangent to the circle in the counterclockwise direction. Project Newton's second law $\textbf{F}_{net} = m\textbf{a}$ along $\textbf{u}_{\theta}$. You should find $\textbf{F}_{net}\cdot\textbf{u}_{\theta}=-Tsin(2\theta)$ which is negative for positive θ and positive for negative θ.

 

[hms.tech]

so, α = $\frac{2Tθ}{mr}$
r= a meters (the radius)

final answer (which is wrong) α = $\frac{2Tθ}{ma}$

Note : a is acceleration
a is the radius given in the question
α is the angular acceleration

Understandable, my presentation is confusing, that is why you misunderstood my approach.

The variable "r" has a value of "a" in this situation. (both of them are just lengths and have dimensions of length)

Thanks for the explaining about the mathematical model for the signs .

At least now you do agree that my solution is dimensionally consistent !

I would like you to check my entire solution as presented in the first post and confirm whether its true. If yes, then we would have to agree that this archive is faulty .

 

[TSny]

The variable "r" has a value of "a" in this situation. (both of them are just lengths and have dimensions of length)

Ah, my mistake! Thanks for setting me straight. (I was reading "a" as linear acceleration.)

At least now you do agree that my solution is dimensionally consistent !
I would like you to check my entire solution as presented in the first post and confirm whether its true.

Yes, your answer is correct except I would include the negative sign to show that the angular acceleration is negative when the angle is positive.

If yes, then we would have to agree that this archive is faulty .

Agreed.