Lecture 12

Rotational Dynamics : Torques

You have been told that an object is in equilibrium (i.e it has no acceleration) when the net force acting on it is zero. Obviously you were told a lie, since it is easy to think of many situations where a body is accelerating even if no net force is present (think e.g. of a water sprinkler with two jets spurting in opposite directions: the two jets exert equal and opposite forces, but the sprinkler is rotating...). The reason for this ambiguity is that, as we mentioned in the previous lecture, until now we had only considered translations, and ignored the possibility of rotations. How do then forces come into play when dealing with rotations, therefore angular, as opposed to linear, accelerations?

When a body is free to rotate around a point (or, to be more precise, around an axis), it is a well known fact that the force inducing rotations will be the more effective the larger is the distance between the point where the force is applied and the axis of rotation (compare what happens if you try swinging a door by pushing near the hinges or near the handle). Also, you will be much more effective in opening the door if the force is perpendicular rather than parallel to the door. We formalize these intuitive concepts introducing the following quantities:

• given a force, we call line of action the direction along which the force is acting
• given a force (with its line of action) and an axis of rotation, we call lever arm the perpendicular distance between line of action and axis of rotation.
• finally we define as torque $\tau$ the product of force magnitude times lever arm, $\tau~=~Fl$.

By convention, torques are positive when they would a counter-clockwise rotation. Be clear that torques are not absolute quantities, but their numerical value depends upon the choice of axis of rotation.

The fact that when dealing with rotations what matters is not just the force but rather the torque has obviously been applied since antiquity: just think of wrenches, levers, etc...

If a torque is what causes rotations, then we can immediately guess that

a body will be in complete equilibrium (i.e. no accelerations) when be that both the net force and the net torque are zero

The first condition guarantees the absence of linear, i.e. translational, accelerations, while the second implies no angular, i.e. rotational, accelerations. This apparently innocuous and unambiguous statement has vexed Physics students (including pre-meds) over the years, as it allows to attack the most complicated problems of equilibrium....

Operationally, the procedure to solve an equilibrium problem is the following :

1.

identify all and only the forces acting on the object under exam

2.

impose the condition of zero net force, i.e. both the sum of the x and y (and z) components must be = 0

3.

choose an axis of rotation (which one? it doesn't matter, if the body is in equilibrium then the torques must be zero w.r.t. any possible axis) and compute the torques wrt the axis.

4.

impose the condition of zero net torque

Another fact to keep in mind is the intuitive knowledge that, for a body of extended dimensions, one can assume that all of its weight is applied at a given point, which we call center of gravity. How can we determine its location? If a body has a symmetrical shape and is homogeneous, the center of gravity will obviously be the center of symmetry. If the body has irregular shape or its made of inhomogeneous materials, one can use the following procedures:

1.

if the body can be thought of as being made of separate parts, whose weights and center of gravity positions are known, then, with respect to an arbitrary origin, we will have

Wx_cg =  W₁x_cg1+W₂x_cg2+W₃x_cg3+....+W_nx_cgn

where W is the combined total weight. Why is that? Because, in the expression above, the right hand side represents the sum of all the individual torques (wrt an arbitrary axis located at x=0), i.e. the total torque, while the left hand side also represents the total torque, under the assumption that all the weight is applied at the center of gravity....

2.

the center of gravity of a body of arbitrary shape can be found by suspending the body from two different points, and marking in each case the vertical line going through the suspension point. The center of gravity must be located on these lines, since it represents the condition of zero torque because of zero lever arm...

In the very good approximation that the force of gravity does not change over the extension of an even large object, then the center of gravity coincides with the center of mass.