- #1
cdbowman42
- 14
- 0
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?
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?