Dimensional Analysis: Inverse Cosine

[raddian]

Homework Statement

For the following dimensional equation, find the base dimensions of the parameter f:

M M^-3 = a cos( f L )

Homework Equations

M represents mass, a represents acceleration due to gravity, in terms of mass * length over seconds squared [[M * L]/[t²]] where L represents length and t represents time.

For example, solving for k in the equation:

ML² = k L t M²

results in k = L M^-1 t^-1

The Attempt at a Solution

The answer, which is given, is L^-1,

So I got

cos^-1(M^-2 a^-1) = f L

the output of the cos^-1 function results in a dimensionless unit of measure (radian/degree).

Therefore,

c = f L

where c is a constant/dimensionless quantity,

therefore f = L ^-1

The problem is that I cannot understand how the inverse cosine function works in this sense. Can its input be "dimensionally inequivalent", I can't find the word for it.

 

[berkeman]

Homework Statement

For the following dimensional equation, find the base dimensions of the parameter f:

M M^-3 = a cos( f L )

Homework Equations

M represents mass, a represents acceleration due to gravity, in terms of mass * length over seconds squared [[M * L]/[t²] where L represents length and t represents time.

For example, solving for k in the equation:

ML² = k L t M²

results in k = L M^{-1 t}

The Attempt at a Solution

The answer, which is given, is L^-1,

So I got

cos^-1(M^-2 a^-1) = f L

the output of the cos^-1 function results in a dimensionless unit of measure (radian/degree).

Therefore,

c = f L

where c is a constant/dimensionless quantity,

therefore f = L ^-1

The problem is that I cannot understand how the inverse cosine function works in this sense. Can its input be "dimensionally inequivalent", I can't find the word for it.

Why are you taking the inverse cosine? It looks like you are given a simple cos() equation to analyze...

BTW, it looks like there is a typo in your latex for this line: "results in k = L M-1 t" -- the t appears in the exponent in your equation instead of in a denominator...

 

[raddian]

Why are you taking the inverse cosine? It looks like you are given a simple cos() equation to analyze...

BTW, it looks like there is a typo in your latex for this line: "results in k = L M-1 t" -- the t appears in the exponent in your equation instead of in a denominator...

Correct, I will edit that. And I didnt use latex btw, I used what and is called.

I am using inverse cosine so I can "remove" f and L from inside the parenthesis. I didnt think of another way.

 

[berkeman]

Correct, I will edit that. And I didnt use latex btw, I used what and is called.

I am using inverse cosine so I can "remove" f and L from inside the parenthesis. I didnt think of another way.

No need to remove or simplify anything in this case. What are the units for the argument for any trig function?

 

[raddian]

No need to remove or simplify anything in this case. What are the units for the argument for any trig function?

Degrees/radians?

 

[berkeman]

Degrees/radians?

Well, good thought, but both of those are dimensionless. There are no degrees or radians in the MKS or cgs systems of units, I believe.

Think about what a trig function is defined as in terms of triangles. cos() = what over what?

 

[raddian]

Well, good thought, but both of those are dimensionless. There are no degrees or radians in the MKS or cgs systems of units, I believe.

Think about what a trig function is defined as in terms of triangles. cos() = what over what?

A length over a length--in essence, a ratio.

 

[berkeman]

A length over a length--in essence, a ratio.

correct-a-mundo! Which is why the argument to trig functions is dimensionless. Makes sense?

 

[raddian]

correct-a-mundo! Which is why the argument to trig functions is dimensionless. Makes sense?

Yes absolutely. BUT (and there's a big but lol)

In this question, cos() = M^-2 a^-1.

The right side of the equation is not a ratio... it has units.. ):

Btw, M^-2 a^-1 = M^-2 M^-1 L^-1 t² = M^-3 L^-1 t²

 

[berkeman]

Yes absolutely. BUT (and there's a big but lol)

In this question, cos() = M^-2 a^-1.

The right side of the equation is not a ratio... it has units.. ):

Btw, M^-2 a^-1 = M^-2 M^-1 L^-1 s² = M^-3 L^-1 s²

No the right side cannot have units, because cos() does not have units. So what do the units of a have to be?

And the problem had asked you for the units of f, which only depend on what else is in the argument for the cos() funstion.

 

[berkeman]

Actually now I see they are saying a is acceleration, so the original equation does not look dimensionally correct to me... Is that the actual question?

 

[raddian]

So what do the units of a have to be?

AHHH THIS IS IT! After looking at my question, variables with unknown units are italicized.

I thought a was acceleration! Thanks very much!I guess we are stuck again. Ill try to take a picture and post it. Is that allowed?

 

[berkeman]

Ah, sweet!

 

[raddian]

Actually now I see they are saying a is acceleration, so the original equation does not look dimensionally correct to me... Is that the actual question?

I think we are stuck again... this was my original problem. I will take a picture and host it on this site.

 

[raddian]

Theres the equation see attached

 

[berkeman]

No attachment yet...

 

[raddian]

Sorry I forgot to press upload, its there now

 

[raddian]

I just want to add, I want to ask my prof. but I'm nervous, I'm a freshman, and I don't know when would be appropriate, since class time = lecture time, not homework help time... Is that what office hours are for?

 

[berkeman]

I just want to add, I want to ask my prof. but I'm nervous, I'm a freshman, and I don't know when would be appropriate, since class time = lecture time, not homework help time... Is that what office hours are for?

Yes, office hours, or if there is a TA for the class, you can ask them during their office hours.

Can you zoom out on the pic to show the whole problem statement? I need to leave work now, but will check in later tonight from home to see if I can help any more with the question.

 

[raddian]

updated attachment

 

[berkeman]

That looks fine. What do you think the new issue is? You found the dimensions of f and you can find the dimensions of a if you want.

 

[raddian]

If a equals an unknown dimension (whose units may equal to M^-2) then I completely understand.

If a means acceleration, then I think there is a problem. Cos() will not equal a ratio if a equals acceleration.

I can kinda sleep better tonight lol.

 

[berkeman]

If a equals an unknown dimension (whose units may equal to M^-2) then I completely understand.

Perfect.

On a personal note -- understanding dimensional analysis and carrying units along in my equations was one of the most important things I learned in my first year of undergrad. What a powerful tool. Especially in working with big complicated equations and simplifications, making sure all the units keep matching has helped me avoid so many typo mistakes. And it's helped me be sure that I've set up my initial equations correctly. Good stuff.

 

[raddian]

Pro tip