# Torque in a rotating rod

by aligass2004
Tags: torque
 P: 236 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/f.../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.
 HW Helper P: 4,124 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...
 P: 236 Is it Tnet = mL - w? The w represents the mass hanging.
HW Helper
P: 4,124
Torque in a rotating rod

 Quote by aligass2004 Is it Tnet = mL - w? The w represents the mass hanging.
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.
 P: 236 So is it Tnet = W - mL?
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P: 4,124
 Quote by aligass2004 So is it Tnet = W - mL?
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?
 P: 236 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.
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P: 4,124
 Quote by 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.
:) 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...
 P: 236 The torque due to the weight of the rod would be T = (2kg)(9.8)(.325m) = 6.37
HW Helper
P: 4,124
 Quote by aligass2004 The torque due to the weight of the rod would be T = (2kg)(9.8)(.325m) = 6.37
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...
 P: 236 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.
HW Helper
P: 4,124
 Quote by 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.
lol. nahh.. you won't need me.
 P: 400 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?

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