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Q & A: torque and Newton's laws

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Most recent answer: 01/08/2014
Is it possible to derive the torque equation r*F = I*alpha from Newton's second law? Every source I check makes sense with one particle: F = ma, so rF = mar = m*r^2*alpha = I*alpha. But when they extend this to more than one particle, I get a bit confused. For example, if we take two particles at different points along a massless rod that's pointed perpendicular to an axis about which it is rotating, then I is just (m_1)*(r_1)^2+(m_2)*(r_2)^2. But if we then apply a force F at distance r_2 (for example) from the axis of rotation, the earlier derivation no longer works, since r_1 does not equal r_2. Is there some detail that gets glossed over in introductory textbooks that would allow a derivation for a body of many particles?
- Nathan (age 23)
Boston, MA, USA

To boil your question down a little further, what it's really getting at is whether Newton's Third Law (momentum conservation) implies the conservation of angular momentum. Unless angular momentum were to be conserved, you could get angular acceleration without any external forces. 

It turns out that momentum conservation is not sufficient. Say you had two objects exerting forces on each other. Newton's Third Law says those forces are equally strong and point in opposite directions.  Unless those directions are along the line connecting the centers of the two objects, the two torques (about the center of mass) won't be zero and they won't cancel.  But the "along the line connecting the centers " rule is not one of Newton's three laws. So it's an extra law, a particular case of a general additional law, the conservation of angular momentum.

Mike W.

(published on 01/08/2014)

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