Finding torque by integration of weight

[Cooojan]

Homework Statement

Hi. I ve got a problem, where I have to show torque by integrating the weight of the rod over the whole it's length.

Homework Equations

[/B]
Result, what I am suppose to get is:

$\tau_{rod} = \frac{mgb}2$

The Attempt at a Solution

[/B]
When I try to integrate, I am only getting:

$\tau_{rod} = \frac{mgb^2}2~~~~$

I'm sure I'm doing some silly mistake. But why can't I do this:

$\frac{d \tau_{rod}}{db} = mg~b~~~~~~⇒~~~~~~ \int \,d \tau_{rod}= mg \int b\,db~~~~~~⇒~~~~~~ \tau_{rod} = \frac{mgb^2}2~~~~$??

I only get the right answer by doing this:

$\frac{d \tau_{rod}b}{db} = mg~b~~~~~~⇒~~~~~~ \int \,d \tau_{rod}b= mg \int b\,db~~~~~~$

but I don't really understand why should I be doing anything like that, and this seems to me like completely wrong approach.

Im sorry if this seems very stupid :/

 

[kuruman]

If you have mass element $dm$ at distance $x$ from the the pivot, then the element contributes torque $d\tau = dm~x$. The next thing you have to do is add up (integrate) all such elements over the length of the rod. Note that both $dm$ and $x$ are variables. You need to invent a linear density and cast $dm$ in terms of $dx$.

 

[haruspex]

I'm sure I'm doing some silly mistake

b is the whole length of the rod. That is a bound on the integral, but not the variable of integration. Create a separate variable for that to avoid confusion.
In particular, you may find you are confusing mass with mass-per-unit-length.

 

[Cooojan]

If you have mass element $dm$ at distance $x$ from the the pivot, then the element contributes torque $d\tau = dm~x$. The next thing you have to do is add up (integrate) all such elements over the length of the rod. Note that both $dm$ and $x$ are variables. You need to invent a linear density and cast $dm$ in terms of $dx$.

I'm so cofused right now xDD

 

[Cooojan]

b is the whole length of the rod. That is a bound on the integral, but not the variable of integration. Create a separate variable for that to avoid confusion.
In particular, you may find you are confusing mass with mass-per-unit-length.

True! Thank you guys. I feel like my head is about to explode )))))

 

[Cooojan]

Is that suppose to be an obvious solution to this? Because I feel so dumb right now xD
Are there any video tutorials on this kind of problems, because I can't find any simmilar ones. Even tho it should be a common kind of problem, I guess... I've really been sitting with this problem for too long now... And this is just a first part of the exercise ://///

 

[Cooojan]

OK! I think I got this now... Please confirm if I am doing the right thing...

lin.density = $\frac mb$

$dF=g \int_0^b \,dm$

$g \int_0^b \,dm = \frac {mg}{b} \int_0^b \,dx$

$g \int_0^b \,dm =g~dm= dF$

$g~dm~x = d\tau$

$d\tau=\frac{gm}{b}\int_ 0^b x\,dx$

$d\tau=\frac{gm}{b}[\frac{x^2}{2}]_0^b$

$\tau=\frac{gmb^2}{2b}=\frac{gmb}{2}$

 

[kuruman]

$F=g \int m'\,dm$

Imagine cutting up the rod into length elements $dx$. If $m'$ is the linear density, then what is mass element $dm$? If I give you length $dx$, how much mass is in that length?

Also, we are not looking for a force, we are looking for a torque. Please reread post #2.

 

[Cooojan]

Imagine cutting up the rod into length elements $dx$. If $m'$ is the linear density, then what is mass element $dm$? If I give you length $dx$, how much mass is in that length?

Also, we are not looking for a force, we are looking for a torque. Please reread post #2.

Since we don't deal with volume here, but only length, so lin.density should be $=\frac mb$..
And as I understand, the mass per length $dx$ must be $\frac{m}{dx}$.. And torque is just the the force time radius times $sin\theta$,
so in this case, I thought I could multiply both sides with that particular $x$, to get that particular $\tau$

Im sorry if that's annoying...

 

[kuruman]

Since we don't deal with volume here, but only length, so lin.density should be $\frac {m}{b}$.

That is correct.

And as I understand, the mass per length $dx$ must be $\frac{m}{dx}$

Incorrect. We already decided what the linear density is. Linear density is also known as mass per length. Units might be kg/m.

Perhaps you can clarify these ideas in your head if you answer this "real-life" situation: Given a salami of length b = 1.0 m and mass m = 2 kg, what is the mass, dm, of a slice of length dx = 3×10^-3m?

 

[Cooojan]

It all suddenly became very clear, when u put it in salami terms...
Especially as I started getting seriously hungry, after breaking my head with this one.
That is one great salami, with fantastics mass of $dm$ per $dx$ unit length.
Thank you very much :)

 

[kuruman]

t all suddenly became very clear, when u put it in salami terms...

I'm glad. Can you finish the problem now?

 

[Cooojan]

yes)) it's done :)