How do I differentiate m(L^2)/16?

[matt_crouch]

Homework Statement

i have been told to calculate DI/Dx i calculated I as being = m(L^2)/16

how do i differentiate this? I am confussed with what happens to the m is it a constant an just removed? same with the 1/16 i assume that is removed as a constant?

just need some clarification.

thanks

Homework Equations

The Attempt at a Solution

 

[danago]

Is that all you are told? If m and L are functions of x, then you will need to take that into consideration.

If m and L are just constants, then treat them like you would any other number.

EDIT: Hmm just re-read your question; did you mean you want to calculate dI/dL and not dI/dx?

 

[matt_crouch]

Well the question says that it passes a distance of x from the perpendicular axis ? does that mean i have to substitute a value for dx in?

 

[Mark44]

How about posting the problem exactly as it was given to you? That would save us a lot of time.

 

[matt_crouch]

OK.

Two point masses m1 and m2 are separated by a massless rod of length L.
a) write an expression for the moment of inertia I about an axis perpendicular to the rod and passing through it a distance x from mass m1
b)Calculate dI/dx and show that I is a minimum when the axis passes through the centre of mass of the system

 

[Mark44]

So m and L are not functions of x.
Now show us how you calculated I. If you haven't drawn a picture, now would be a good time to do that.

 

[matt_crouch]

i think i calculated I i drew a picture an since its two point masses i don't think i need to intergrate so therefore the Inertia is equal to the sum of the product individual mass and the distance from the axis squared.

so i then got [m1L^2]/32 +[m2L^2]/32 so simplify this to get
mL^2/16

 

[Mark44]

You are assuming that the two masses are equal, which is not given. Also, you have not used the information that the axis passes through a point a distance x from m1.

 

[matt_crouch]

should i replace L with X then? i don't really kno where to go

 

[Mark44]

To calculate the moment of inertia you need to know the mass and the length of the lever arm (i.e., the distance from the mass to the point for which you're calculating the moment of inertia).
What is the M.I. for mass m1?
What is M.I. for mass m2?

 

[matt_crouch]

so the M.I for mass one =m1x^2 ? that was using the equation I=mr^2 where r is the radial distance

so the M.I for mass two is = m2(L-x)^2

?

 

[Mark44]

Yes for both. The combined M.I. is I = m₁x² - m₂(L - x)², right? The two masses are on opposite sides of the axis, so their lever arms are acting in opposite directions.

Now, can you find dI/dx and finish off part b?

 

[matt_crouch]

ahh cheers thanks for the help ill give b a shot now. =]

 

[matt_crouch]

so b)

dI/dx = m.2x+x^2-[-2(L-x)]+(L-x)^2

not so sure if iv done that right :/

 

[Mark44]

so b)

dI/dx = m.2x+x^2-[-2(L-x)]+(L-x)^2

not so sure if iv done that right :/

No, not right. dI/dx will have both m₁ and m₂ in it.

To make your work more readable here, use the X² button for exponents and the X₂ button for subscripts. These buttons are in the advanced menu bar. If you don't see this menu bar, click the Go Advanced button just below the text entry area.

 

[matt_crouch]

sorry was supposed to put

m₁2x+x²-m₂(-2[L-x])+(L-x)²

 

[Mark44]

The first term is right (2m₁x) but the rest isn't, and I have no idea how you got what you show.

Start with I = m₁x² - m₂(L - x)², and show me your steps in getting dI/dx.

 

[matt_crouch]

Im not really sure i just used the product rule. I am not sure what happens to the mass or the length or if they are taken out as constants so need some help really

 

[Mark44]

The masses and length L are constants. You don't just "take them out."

I = m₁x² - m₂(L - x)²
dI/dx = d/dx(m₁x²) - d/dx(m₂(L - x)²)
= 2m₁x - m₂ d/dx ((L - x)²)

Use the chain rule to get the remaining derivative. If you don't know the chain rule, expand (L - x)² and differentiate that instead.

 

[matt_crouch]

so m₂d/dx(L-x)²
will that go to

m₂2(L-x)L ?

 

[Mark44]

so m₂d/dx(L-x)²
will that go to

m₂2(L-x)L ?

No. Why do you have that final L?

 

[matt_crouch]

chain rule? u differentiate the bracket then times by the deferential of the inside of the bracket since x goes to -1
should it be -L ?

 

[Mark44]

chain rule? u differentiate the bracket then times by the deferential of the inside of the bracket since x goes to -1
should it be -L ?

Yes, use the chain rule. To differentiate (L - x)², you get 2(L - x) times the derivative (not differential) of what's inside the parentheses. What is d/dx(L - x)?

What you said about x going to -1 makes no sense.

 

[matt_crouch]

is the derivative

dL/dx -1

 

[Mark44]

Yes, d/dx(L - x) = dL/dx - 1, but this should be simplified. What is dL/dx?

 

[matt_crouch]

i don't know :/

 

[Mark44]

What is dm₁/dx?
What is dm₂/dx?
What is dL/dx?

 

[matt_crouch]

im not sure how do i calculate them?

 

[Mark44]

What is d(2)/dx?
What is d(\pi)/dx?
What is the derivative of any constant?

 

[matt_crouch]

is the derivative 0 then ?

 

[Mark44]

Yes! Since dL/dx = 0, what do you get for dI/dx?
dI/dx = d/dx(m₁x² - m₂(L - x)²) = ?

 

[matt_crouch]

so dI/dx=2m₁+2m₂(L-x)

?

 

[Mark44]

Now the second term is correct, but the first one isn't. You had it right before.

Since you're having so much trouble with differentiation, I would advise you to go back and review the basic differentiation rules, especially the product rule and chain rule, and derivative of a constant rule.

 

[matt_crouch]

ahh ye just missed the x from the first term
thanks for all the help btw =]