# Homework Help: Torque in a rotating rod

1. Oct 18, 2007

### aligass2004

1. The problem statement, all variables and given/known data
How much torque must the pin exert to keep the rod from rotating if the rod has a length L=65cm and a mass m=2.0kg? Calculate this torque about an axis that passes through the point where the pin enters the rod and is perpendicular to the plane of the figure.

http://i241.photobucket.com/albums/ff4/alg5045/p13-25alt.gif

2. Relevant equations

3. The attempt at a solution
I haven't attempted the problem because I'm not quite sure where to start. I know that you have to find Tnet, and I know the mass and the length fit into the equation somewhere. I'm just not sure how to set the equation up.

2. Oct 19, 2007

### learningphysics

What is the total torque the rod and the 500g mass exert about the pin?

The pin needs to create a torque of the same magnitude but the opposite direction, to cancel the above torque...

3. Oct 19, 2007

### aligass2004

Is it Tnet = mL - w? The w represents the mass hanging.

4. Oct 19, 2007

### learningphysics

that's the idea, but there are mistakes... weight acts at the center of mass... take clockwise as positive... counterclockwise as negative for signs...

try to fix the expression.

5. Oct 19, 2007

### aligass2004

So is it Tnet = W - mL?

6. Oct 19, 2007

### learningphysics

no. take it step by step. what is the torque due to the 0.500kg hanging mass? take clockwise positive counterclockwise negative...

use the definition of torque... what is the force? what is the distance?

7. Oct 19, 2007

### aligass2004

T = -Fr and I think the F is just the weight of the mass. In case you couldn't tell, torque and I don't get along well.

8. Oct 19, 2007

### learningphysics

:) that's ok.

yes, that's the definition you need... here gravity acts downward... so the rod would turn clockwise due to gravity...

so we use a +...

torque due to the 0.500kg = 0.500*9.8*L

try to find the torque due to the weight of the rod... it acts at L/2...

9. Oct 19, 2007

### aligass2004

The torque due to the weight of the rod would be T = (2kg)(9.8)(.325m) = 6.37

10. Oct 19, 2007

### learningphysics

exactly. you're getting along better already. :)

so the total torque is 0.500*9.8*0.65 + 6.37...

so that's a clockwise torque about the pin... so the pin needs to exert a counterclockwise torque of the same magnitude...

11. Oct 19, 2007

### aligass2004

Got it. T = 9.555. I think I need you to sit beside me during my exam in a couple weeks.....just for the torque problems.

12. Oct 19, 2007

### learningphysics

lol. nahh.. you won't need me.

13. Feb 24, 2010

### jumbogala

Re: Torque

I am doing a very similar problem to this one. My question is, why do you have to pick the left end of the rod as your pivot point?

I tried using the center of mass and another point as pivot points and got different answers for the torque from the pin each time. How do you know which is the correct one?