Gravitational torque about a point on a rod

  • Thread starter cdbowman42
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1. A 2.0 kg uniform, horizontal rod is 1 m in length. What is the gravitational torque about the point A that is 25 cm from the left end of the rod.


2. T=Fr
F=mg
T=mgr


3. The center of gravity of the rod is directy at its center, so is therefore 25 cm from the point A. I thought the torque would be the product of the amount of mass on that side of the center of gravity (3/4 or the 2 kg rod since it is uniform), acceleration due to gravity, and the distance of the point to to center of gravity. The answer in the back of the book says -2.1 N*m (Assuming clockwise motion is negative), but I get 3.675 N*m. Where am I going wrong?
 

Answers and Replies

  • #2
gneill
Mentor
20,925
2,867
The book's answer would appear to be incorrect. Then again, so does yours! :smile:

Something's gone awry in your calculations. Perhaps you should present them in detail.
 

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