# Center of Mass/Torque

So, I have recently learned about 'Centers of Mass' and have discovered I was completely ignorant to the fact that the point where a force is applied will determine whether the force is a rotational force or a normal force.

Naturally, I have a few questions.

Q1: I have two objects on the ground, that are say at-least 2 cm apart, and center of mass is between the two objects. Does that mean if I applied a net force at the centre of mass the system of objects would move in that direction?

Q2: I have attached an image, in the image, we are looking at 3 objects with masses $5$$kg, 2$$kg$ and $3$$kg in red dots respectively all sitting on a horizontal axis. We are defining \bar{x} as the location of the center of mass from the point A and the total weight of this system is W=10g N Now, in the next line, finding the torque about A and defining clockwise as positive.$$10g\bar{x} = 5g(2) + 2g(3) + 3g(6) $$Now, I don't understand how this is so because, I thought that force applied at the centre of mass would move the object(s) in the direction in which the force was applied. So, basically there would be no torque and the weight of the system of objects would just move the system downwards, since weight acts downwards. So 10g\bar{x} = 0 ## 3 Answers Yes, for (1), you're correct. However, that's assuming that the two objects are joined by some kind of light rod. In that case, applying a force at the center of mass would cause some translational motion. Regarding (2), your calculation for the center of mass is that the center of mass is 3.4m from A. Using the formula for torque,$$τ = r * f$$The 3kg mass is 2.6m from the center of mass, the 2kg mass is 0.4m away from the center of mass and the 5kg mass is 1.4m away from the center of mass. Hence,$$τ_{net} = 3g * 2.6 - 2g * 0.4 - 5g * 1.4 = 0$\$

In your calculation, you were taking moments around the point A. What your calculations actually mean is that if all three masses were joined by a rod that was fixed to point A (hence the entire system can only rotate around A), there would be a net torque, which is true.

For Q1, you cannot apply a force to a point in space. If you apply the force to either of the masses, the center of mass will accelerate in the direction of that force. For Q2, the equation, τ = Iα, normally applies to a rigid body rotating about a fixed axle. In a more general system, the sum of torques determines the rate of change of the angular momentum, L. The, L, can be calculated as the sum of two parts: one part associated wth the motion of the center of mass, and a second part associated with rotation about the center of mass. In your problem, the angular momenta are associated with the linear motion of each mass relative to point, A. This does respond to the sum of the externally applied torques about point, A.

For your second question, it is the torque about the center of mass that matters in terms of purely translational motion or coupled translational and rotational.

In your situation the origin is (arbitrarily) chosen as the point of summation of torques to get $$5g(2) + 2g(3) + 3g(6) = 10g \overline{x}$$

The above leads to finding $$\overline{x} = 3.4$$ (the center of mass), and now if you sum the torques about $$\overline{x}$$ you get

$$5g (2-3.4) + 2g (3-3.4) + 3g (6-3.4) = 0 g$$

Your first question answer is yes if the two masses are rigidly attached. But why?

It comes down to the definition of momentum. The net force applied on a rigid body only affects the motion of the center of mass because Newton's laws equates forces to changes in momentum, and momentum is defined only by the motion of the center of mass.

There is a corollary statement, that a body with a pure torque applied (zero net force) will rotate about the center of mass. In some ways, this is an alternative definition of the center of mass. The point attached to a body frame which does not move if an arbitrary torque vector is applied.