Torque of a uniform rod being pivoted.

[monkeysmine]

Homework Statement

A uniform rod 1.1 m long with mass 0.7 kg is pivoted at one end, as shown in Fig. 9-46, and released from a horizontal position.
Find the torque about the pivot exerted by the force of gravity as a function of the angle that the rod makes with the horizontal direction. (Use theta for θ.)
http://www.webassign.net/fgt/9-43.gif

Homework Equations

T = rFsin(theta)

The Attempt at a Solution

My attempt has consisted of the following:
1st attempt:
Using the basic formula, and then applying it to obtain ~7.6sin(theta), trying the negative version as well.

I think realized that sin might not be valid to use due to the nature of the direction of application of the force and tried 7.6cos(90-theta), which also did not work.

2nd attempt:
T = angular acceleration * moment of inertia
I solved the moment of inertia to be ~.2823 (mL^2/3).
I defined omega as d/(dt)[int(g)sin(theta)/r].

I then plugged in the values and came up with .2823d/(dt)[int(9.8)sin(theta)/1.1]

Any help would be appreciated.

 

[PhanthomJay]

You don't need to get into angular accelerations and moment of inertias per attempt 2. The problem is just asking for the torque as a function of theta, per your first attempt equation. You have a couple of errors. In calcualting the torque, where is the gravity force (weight force) applied, and using the cross product definition of torque, T=rf sin alpha, what is the correct angle to use for 'alpha'?

 

[monkeysmine]

I only tried 2) because 1) did not work.

I crossed .5Lcos(theta)i - .5Lsin(theta)j + 0k with 0i -mgj + 0k

 

[PhanthomJay]

I only tried 2) because 1) did not work.

I crossed .5Lcos(theta)i - .5Lsin(theta)j + 0k with 0i -mgj + 0k

That cross product formula looks OK, but it's way too easy to mess up the calculation when you do it that way. Instead, try using use T=r X F = rF sin alpha , where r is magnitude of the radius (.5L), F is the magnitude of the weight, and alpha is the angle between the radius and weight vector. alpha = ??

 

[monkeysmine]

Well I double checked my work on a calculator and I'm positive that my cross product was right. I did realize that an error I made was that R is actually r/2 since r goes from the pivot to the center of mass, but this still does not give me the correct answer.

 

[monkeysmine]

I got T = 3.8cos(90-theta) as a replacement to 3.9sin(theta). I just used my last "guesses" on this problem though, so I will talk to my teacher tomorrow about it. I am positive that one of my 50 answers were right (30+ variations on rounding and truncating) of 3.773cos(90-theta) and I also tried +/- 3.773cos(90-theta).

WebAssign is glitchy anyways, and often requires strange sig. figs or the like.
Thank you for the prompt help!

 

[PhanthomJay]

I got T = 3.8cos(90-theta) as a replacement to 3.9sin(theta). I just used my last "guesses" on this problem though, so I will talk to my teacher tomorrow about it. I am positive that one of my 50 answers were right (30+ variations on rounding and truncating) of 3.773cos(90-theta) and I also tried +/- 3.773cos(90-theta).

WebAssign is glitchy anyways, and often requires strange sig. figs or the like.
Thank you for the prompt help!

Well, sig figures aside, none of your answers are correct. I inadvertently gave you the correct answer, so I've deleted it.

 

[monkeysmine]

Well, sig figures aside, none of your answers are correct. I inadvertently gave you the correct answer, so I've deleted it.

3.773 = .5 * 1.1 * .7 * 9.8, so actually, I don't understand why you are saying my answers are incorrect.

 

[PhanthomJay]

That 3.773 (call it 3.8) is fine...it's the cos(90- theta) , (or sin theta), part of the equation, that's wrong. Theta is not the angle between the force and position vectors. I don't know how you got sin theta, even using your cumbersome cross product equation.

 

[monkeysmine]

...Okay sorry I don't know what I was thinking yesterday I meant I tried cos(theta) and cos(90-theta)...

 

[PhanthomJay]

Okay, I gather you are saying that the (magnitude of) the torque is either 3.8cos theta or 3.8 cos (90-theta); one of those answers is correct for the magnitude of the torque; the other is not. But in any case, the answer may be off by a minus sign in front of the 3.8. Usually, standard convention using the 'right hand rule' assumes that clockwise torques are negative. In that case, you need the minus sign.

 

[monkeysmine]

Yeah, I meant my original thought was cos(theta) for some reason I kept writing sin. It was probably just kind of late. Apparently Webassign demanded that we round and truncate making the answer 4cos(theta) (shady if you ask me) but hey.

 

[PhanthomJay]

Yes, shady, but correct I guess. Since the least number of significant figures in the given variables is 1 (0.7 has one sig figure), then the answer can have only one sig figure, hence the 4. That's the rule. But while 3.773 would be wrong, I would think that 3.8 ought to be accepted by webassign. I'm surprised they accepted the plus sign. I think a human teacher, instructor, tutor, professor, homework helper, or the like, would be a bit more forgiving, depending on how strict they are.