Torque Question

[stitchattack]

A particle of mass m is attached to a rod of negligible mass and length L. The rod is attached to the ground at point O. If the rod makes an angle theta with horizontal as it falls, calculate the torque about point O.

A. Lmgsin(theta)
B. Lmgcos(theta)
C. mgsin(theta)
D. mgcos(theta)

The answer is B. I thought torque = rFsin(theta). The explanation states that "if the rod were vertical, what torque would you expect because of mass? None. So, as theta goes to 90, the trigonometric functions must go to zero. This is what cos(theta) does.." This doesn't really help me understand the problem.

Can someone explain why the answer is B? Also, how do you know when sin/cos (theta) is used? Thanks!

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[Lighthill-FFT]

A tricky question. The explanation was written by a physicist.

You are right, the formula for torque is technically the cross product between the force generated and the axis it acts on (L).

But we have to consider the way the force behaves as it falls. Why? Because cross products are ambiguous on which axis to choose as the primary axis to base calculations off of (horizontal vs vertical). It just says to take sin($/theta$). So we do a little thought experiment.

As the mass falls, does it experience more torque or less torque? Well, torque is sort of hard to picture, so lets picture force instead which is directly proportional to torque (the mg part of the answers). As the mass falls, force falls. As force falls, torque should fall (because the masses don't speed up if I push less hard). We want our torque to fall as it approaches the vertical.

We check how the two trig functions behave.

As defined, sin(theta) would be sin(90)=1 as it falls. The function (L,m,g are all constant, so its just some number) reaches a maximum at vertical. That doesn't sound right.

As defined, cos(theta) would be cos(90)=0 as it approaches vertical. Sounds right. Must be this.


Of course, it doesn't answer the question of how a sin magically becomes a cos. The answer comes with a redefinition of your reference point. If this were a research problem, I would redefine vertical as a reference point because it makes life so much simpler. Why would I measure from horizontal if vertical is all I care about?

From this reference frame, we see that the answer is obviously Lmgsin(theta), because according to this definition at vertical the angle measured is 0, sin(0)=0 and life is well. But because this is a question made up by people to see if you know what you are doing rather than blindly applying formulas, they told you to do it from a unintuitive reference frame. To convert from horizontal to vertical reference frames, we can draw a little diagram in our heads and realize sin(90-theta) is the correct equation in the reference frame I would have used. Using a trig identity, sin(90-theta)=cos(theta).

Regarding your question about sin vs cos, you use it when you do. There is no rule because it all depends on what you define as the standard.

 

[techfan]

I had trouble with this one also, would you mind drawing a diagram? I can't picture where you are taking the angles

 

[justadream]

@techfan

I could be wrong but here is my interpretation:

O======================== [MASS]
=
=
=
=
=
=
=


The bar is falling clockwise I believe.

 

[techfan]

But where are you measuring theta?

 

[stitchattack]

I'm also having a hard time picturing the angles. When the mass hits the ground, theta will be 0 so what happens then? If torque = Lmgcos(0) then that means torque is at max..is that what we're trying to justify? I'm still confused.

 

[Lighthill-FFT]

I drew a nice diagram and everything. Its too bad my scanner inst working. Enjoy this crappier diagram I drew in paint. Theta is the theta used in the question. Theta prime is the other reference frame.

Answering Stitchattack's question, it might be helpful to get a better grasp on what torque is exactly. Open the nearest door. Now close it. Notice you didnt push the door in the direction you wanted it to go (otherwise you would be pushing into the door). You pushed it perpendicular to the cross product between the edge of the door and the axis defined by the edge and the hinge of the door (the doors height and length). You pushed it against its width, unless you push doors really weird.

In the case of the question, the force provided by you is substituted by the force of gravity. Therefore, the more force the moment arm (not the mass) experiences the more torque is generated. As the ball goes down, less gravitational force is available because the moment arm is going down. At the extreme case (the arm is vertical), the moment arm experiences no force at all and therefore no torque.

If we define things from the reference point of theta' in my diagram, we see that at the bottom theta'=0 -> sin(0)=0 and all is well. But thats not what the question asked. It asked for theta. At the bottom, theta=90 and sin(90)=/= 0. We have to find a formula to convert from theta to theta prime.

Imagine a hypothetical case where theta was 30 degrees. theta' is therefore 60 degrees, or 90-30. Therefore, to convert from theta to theta', we just subtract theta from 90. Therefore, we rewrite theta in terms of theta': sin(90-theta)=sin(theta')=cos(theta).

EDIT:If anyone is still having problems, I'll take the time to fix my scanner. Let me know!

 

[The Brown Knight]

draw it out. angle between rod and attached particle is 90-theta
you'll find that the torque is T = Lmg sin (90 - theta)
using high school trigonometry: sin (90-theta) = sin90 cos(theta) - cos90 sin(theta) = (1)(cos theta) - (0)(sin theta) = cos (theta)

 

[techfan]

@Lighthill-FFT I'm trying to figure out if you measure theta from the point you're pushing from or from the origin. I think I have a problem with cross products (maybe I've used the right hand rule so much that I don't understand the actual definition). Does your drawn out diagram show how theta' got "converted" (I don't remember all of the proper geometric vocabulary) to theta?

 

[Lighthill-FFT]

Cross-products don't matter much in this specific question; you just have to know that the force is produced on the third perpendicular axis. Thats just math gobbledygook for this simple idea:

Put a 3D coordinate system on the door, The direction that doesn't involve the door (idealizing the door to a 2D shape) is the direction of torque. If you want a detailed explanation of torque, I can try to explain it to you.

Theta has nothing to do with the direction of torque. Theta is defined by the question: its the angle between the pendulum and the horizontal. Ill give you another example that hopefully helps clear some things up.

Your arm can act effectively as a pendulum (your arm defined as from the shoulder down). You can move it in a circle (in an ideal world of course) by moving your arm up an down. Where is the torque generated? Its generated in the direction demonstrated by moving your arm toward and away from your chest (like a giant hugging motion, I guess).

Likewise, when you move your arms in that giant hugging motion, you generate torque in the plane above/below your movement. The right hand rule tells you the direction the torque is pointing, but the point of this example is torque is produced perpendicular to the movement plane.

Theta' to Theta just depends on how things are defined. Theta' is defined so that at the bottom the right hand rule works as it should. Theta is defined so that just using the formula gives you the wrong answer.

 

[Czarcasm]

A particle of mass m is attached to a rod of negligible mass and length L. The rod is attached to the ground at point O. If the rod makes an angle theta with horizontal as it falls, calculate the torque about point O.

A. Lmgsin(theta)
B. Lmgcos(theta)
C. mgsin(theta)
D. mgcos(theta)

The answer is B. I thought torque = rFsin(theta). The explanation states that "if the rod were vertical, what torque would you expect because of mass? None. So, as theta goes to 90, the trigonometric functions must go to zero. This is what cos(theta) does.." This doesn't really help me understand the problem.

Can someone explain why the answer is B? Also, how do you know when sin/cos (theta) is used? Thanks!

I'm not sure if everyone is over=thinking it, or I'm oversimplifying it, but for torque, you want the force perpendicular to r or the lever arm perpendicular to force - whichever is easier to find in any given scenario. In this scenario, force is pointing straight down to the ground, but r is slanted. For me personally, I find it easier to just find the lever arm. We're given the angle from the horizontal. So, the length of the rod perpendicular to force (lever arm), according to trig, is Lcos(theta). So collectively, torque would be: mgLcos(theta).

Also, it might help to realize that the center of mass is shifting relative to the pivot as the rod is tilting. Initially, it acts through the pivot. But as the rod tilts, the center of mass shifts and the lever arm increases proportionally ... up to a maximum, where the rod is flat on the ground.