How Do You Properly Approximate Terms in an Equation?

In summary, it is important to consider all terms when making approximations and to ensure that each term is chopped off at the same power. In this case, the incorrect approximation of sin(theta) as theta and cos(theta) as 1 led to an incorrect answer. It is also important to remember to include all necessary terms in the equation.
  • #1
hang
7
1
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
1618504889799.png
1618504923431.png
 
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  • #2
Hi,

hang said:
Where have I gone wrong?

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

##\ ##
 
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Likes hang
  • #3
1618509989849.png

this is my working, sorry for the mess
 
  • #4
hang said:
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)##.
 
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  • #5
haruspex said:
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
 
  • #6
hang said:
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.
 
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  • #7
haruspex said:
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
 

FAQ: How Do You Properly Approximate Terms in an Equation?

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