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[hang]

Homework Statement

the question asks for the angle when the kinetic energy is one forth of the maximum kinetic energy. At first I equate :

gravitational potential energy + elastic potential energy + 1/4 maximum rotational energy= maximum rotational energy

then I assume that sin(theta)~theta and cos(theta)~ 1

However, the g value (gravity) just get cancelled and my answer is wrong which is as expected since I did not include g.

Where have I gone wrong? when the rotational energy is maximum, or when angular speed is maximum, do the system still has potential energy?

Relevant Equations

energy and moment of inertia

 

[BvU]

Hi,

Where have I gone wrong?

Hard to say if you don't post the steps you take to find your answer ...

$\$

 

[hang]

this is my working, sorry for the mess

 

[haruspex]

View attachment 281584
this is my working, sorry for the mess

You approximated $\cos(\theta)$ as 1, thereby throwing away a $\theta^2$ term, but kept the $\theta^2$ term from $\sin^2(\theta)$.

 

[hang]

You approximated $\cos(\theta)$ as 1, thereby throwing away a $\theta^2$ term, but kept the $\theta^2$ term from $\sin^2(\theta)$.

I approximated sin(theta) as theta ; and cos(theta) as 1. So shouldn't the sin^2 (theta) become (theta)^2 ?

So is my assumption for the energy correct? maximum rotational speed happens when potential energy is 0

thank you

 

[haruspex]

I approximated sin(theta) as theta ; and cos(theta) as 1. So shouldn't the sin^2 (theta) become (theta)^2 ?

That's not the right way to think about making approximations. You can't make them that independently.
If your equation is a sum of terms equating to zero then you need to chop each term off at the same power. You have a $\sin^2(\theta)$ term, not a $\sin(\theta)$ term. You chopped it off at $\theta^2$, so you need to do the same with the $\cos(\theta)$ term.
Note that it changes the answer.

 

[hang]

That's not the right way to think about making approximations. You can't make them that independently.
If your equation is a sum of terms equating to zero then you need to chop each term off at the same power. You have a $\sin^2(\theta)$ term, not a $\sin(\theta)$ term. You chopped it off at $\theta^2$, so you need to do the same with the $\cos(\theta)$ term.
Note that it changes the answer.

thank you so much, you are right. My approximation was wrong. And I also realized that I forgot to square the rotational speed in the equation. I really appreciate your help