Torque on a free body

  • Thread starter tomz
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Hi everyone, I have a very simple and classic question here.

Suppose I have a 10cm,0.01kg ruler (uniform density) in space. And I apply a 1N force 2 cm from the left end of it. What happened at this instance? I guess centre of the ruler will start to accelerate as a=F/m=100ms^2. But what about the torque? which point should I consider as the pivot position. (Here I want what happen at the very first instance only). Is it the centre of mass? If it yes, why??

Thank you for answering!!
 

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  • #2
jbriggs444
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You can choose any axis you like. They will all work.

There are two choices that are attractive at first glance

1. Consider the point of application of your force as the axis of rotation.

This has the advantage that your force exerts no torque. You can now compute the linear motion of the ruler's center-of-mass (100 m/s^2 as before), compute how much angular momentum that represents, compute the moment of inertia of the ruler about this axis and deduce a rotational acceleration rate that keeps total angular momentum
constant.

2. Consider the center of mass as the axis of rotation.

This has the advantage that the linear motion of the ruler does not contribute to total angular momentum. You can now compute the linear motion of the ruler based on the force applied and separately compute the torque and moment of inertia of the ruler about its center and deduce a rotational acceleration rate based on that.

You should perform both calculations and assure yourself that the results match.

With any other axis you can cancel neither the torque applied by the forst nor the angular momentum contributed by the linear acceleration of the center-of-mass. Analysis is still possible but is somewhat more difficult. It will still produce the same result.

Two additional attractive choices are the two ends of the ruler (making it slightly easier to determine its moment of inertia).
 
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