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Torque in a rotating rodby aligass2004
Tags: torque 
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#1
Oct1807, 10:25 PM

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.../p1325alt.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
Oct1907, 12:27 AM

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... 


#3
Oct1907, 12:38 AM

P: 236

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



#4
Oct1907, 12:40 AM

HW Helper
P: 4,124

Torque in a rotating rod
try to fix the expression. 


#5
Oct1907, 12:44 AM

P: 236

So is it Tnet = W  mL?



#6
Oct1907, 12:48 AM

HW Helper
P: 4,124

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


#7
Oct1907, 12:51 AM

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.



#8
Oct1907, 12:55 AM

HW Helper
P: 4,124

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
Oct1907, 12:59 AM

P: 236

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



#10
Oct1907, 01:01 AM

HW Helper
P: 4,124

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
Oct1907, 01:06 AM

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.



#12
Oct1907, 01:06 AM

HW Helper
P: 4,124




#13
Feb2410, 04:21 PM

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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