The Second Condition for Equilibrium

We define the moment of inertia, or rotational inertia, as  mr^{2} for a point mass, where  m is the mass of the object being rotated and  r is the radius from the pivot point to the end of the mass.

However, the moment of inertia for a distribution of mass that makes a shape, such as a rotating cylinder or sphere, applies different equations. For example, for a solid sphere rotating about a central axis going through the core of the sphere, the moment of inertia is  I_{\mathrm{solid\ sphere}}=\frac{2}{3}mr^{2} where  m is the mass of the entire sphere and  r is the sphere's radius.

In Section 6.1, we said that an object in equilibrium must have no rotational acceleration. We can restate this by saying that an object in equilibrium must have a torque of zero. When no torque is acting on a system, no rotational acceleration is given to the system, and it remains in equilibrium.

As you read, pay attention to Figure 9.6, which shows the torque on a rotating plank of wood secured at a pivot point at one end. This diagram shows how the direction of the force impacts the rotation of the plank of wood.

When the force is perpendicular to the length of the plank of wood, the plank experiences torque, and it rotates. When the force is parallel to the length of the plank of wood, it does not experience a net force and therefore does not rotate or experience torque. When the force is at an angle other than 90° from the length of the plank, the plank experiences less torque than if the force was at 90° from the plank's length.

Torque

The second condition necessary to achieve equilibrium involves avoiding accelerated rotation (maintaining a constant angular velocity). A rotating body or system can be in equilibrium if its rate of rotation is constant and remains unchanged by the forces acting on it. To understand what factors affect rotation, let us think about what happens when you open an ordinary door by rotating it on its hinges.

Several familiar factors determine how effective you are in opening the door. See Figure 9.6. First of all, the larger the force, the more effective it is in opening the door - obviously, the harder you push, the more rapidly the door opens. Also, the point at which you push is crucial. If you apply your force too close to the hinges, the door will open slowly, if at all. Most people have been embarrassed by making this mistake and bumping up against a door when it did not open as quickly as expected. Finally, the direction in which you push is also important. The most effective direction is perpendicular to the door - we push in this direction almost instinctively.

In the figure, six top views of a door are shown. In the first figure, a force vector is shown in the North West direction. The perpendicular distance of the force from the point of rotation is r. In the second figure, a force is applied in the opposite direction at the same distance from the hinges. In the third figure, a smaller force in applied at the same point. In the next figure, a horizontal force is applied at the same point. In this case, the perpendicular distance from the hinges is shown as r sin theta. In the next figure, force is applied at a distance near the hinges. In the final figure, the force is shown along the direction of hinges toward the handle of the door.

Figure 9.6 Torque is the turning or twisting effectiveness of a force, illustrated here for door rotation on its hinges (as viewed from overhead). Torque has both magnitude and direction. (a) Counterclockwise torque is produced by this force, which means that the door will rotate in a counterclockwise due to \mathbf{F}. Note that r_{\perp} is the perpendicular distance of the pivot from the line of action of the force. (b) A smaller counterclockwise torque is produced by a smaller force \mathbf{F} / acting at the same distance from the hinges (the pivot point). (c) The same force as in (a) produces a smaller counterclockwise torque when applied at a smaller distance from the hinges. (d) The same force as in (a), but acting in the opposite direction, produces a clockwise torque. (e) A smaller counterclockwise torque is produced by the same magnitude force acting at the same point but in a different direction. Here, \theta is less than 90^{\circ}. (f) Torque is zero here since the force just pulls on the hinges, producing no rotation. In this case, \theta=0^{\circ}.


The magnitude, direction, and point of application of the force are incorporated into the definition of the physical quantity called torque. Torque is the rotational equivalent of a force. It is a measure of the effectiveness of a force in changing or accelerating a rotation (changing the angular velocity over a period of time). In equation form, the magnitude of torque is defined to be

\tau=r F \sin \theta.

where \tau (the Greek letter tau) is the symbol for torque, r is the distance from the pivot point to the point where the force is applied, F is the magnitude of the force, and \theta is the angle between the force and the vector directed from the point of application to the pivot point, as seen in Figure 9.6 and Figure 9.7. An alternative expression for torque is given in terms of the perpendicular lever arm r_{\perp} as shown in Figure 9.6 and Figure 9.7, which is defined as

r_{\perp}=r \sin \theta

so that

\tau=r_{\perp} F.

In the first part of the figure, a hockey stick is shown. At a point A near the bottom, a nail is fixed. A force is applied at a point near the holding grip of the hockey stick. A quarter circular arrow shows that the stick rotates in the counterclockwise direction. The perpendicular distance between the pivot point and the force vector direction is labeled as r-perpendicular, and the angle between the direction of force and the line joining the pivot A to the point of application of force is given as theta. In the second part of the figure, the pivot point is near the top of the stick and the point of application of the force is about the same as that in the first part of the figure. An upward quarter circle arrow shows that the stick rotates in the clockwise direction.

Figure 9.7 A force applied to an object can produce a torque, which depends on the location of the pivot point. (a) The three factors \mathbf{r}, \mathbf{F}, and \theta for pivot point \mathrm{A} on a body are shown herer is the distance from the chosen pivot point to the point where the force \mathbf{F} is applied, and \theta is the angle between \mathbf{F} and the vector directed from the point of application to the pivot point. If the object can rotate around point \mathrm{A}, it will rotate counterclockwise. This means that torque is counterclockwise relative to pivot A. (b) In this case, point B is the pivot point. The torque from the applied force will cause a clockwise rotation around point \mathrm{B}, and so it is a clockwise torque relative to \mathrm{B}.


The perpendicular lever arm r_{\perp} is the shortest distance from the pivot point to the line along which \mathbf{F} acts; it is shown as a dashed line in Figure 9.6 and Figure 9.7. Note that the line segment that defines the distance r_{\perp} is perpendicular to \mathbf{F}, as its name implies. It is sometimes easier to find or visualize r_{\perp} than to find both r and \theta. In such cases, it may be more convenient to use \tau=r_{\perp} \mathbf{F} rather than \tau=r F \sin \theta for torque, but both are equally valid.

The SI unit of torque is newtons times meters, usually written as \mathrm{N} \cdot \mathrm{m}. For example, if you push perpendicular to the door with a force of 40 \mathrm{~N} at a distance of 0.800 \mathrm{~m} from the hinges, you exert a torque of 32 \mathrm{~N} \cdot \mathrm{m}(0.800 \mathrm{~m} \times 40 \left.\mathrm{N} \times \sin 90^{\circ}\right) relative to the hinges. If you reduce the force to 20 \mathrm{~N}, the torque is reduced to 16 \mathrm{~N} \cdot \mathrm{m}, and so on.

The torque is always calculated with reference to some chosen pivot point. For the same applied force, a different choice for the location of the pivot will give you a different value for the torque, since both r and \theta depend on the location of the pivot. Any point in any object can be chosen to calculate the torque about that point. The object may not actually pivot about the chosen "pivot point".

Note that for rotation in a plane, torque has two possible directions. Torque is either clockwise or counterclockwise relative to the chosen pivot point, as illustrated for points B and A, respectively, in Figure 9.7. If the object can rotate about point A, it will rotate counterclockwise, which means that the torque for the force is shown as counterclockwise relative to \mathrm{A}. But if the object can rotate about point \mathrm{B}, it will rotate clockwise, which means the torque for the force shown is clockwise relative to B. Also, the magnitude of the torque is greater when the lever arm is longer.

Now, the second condition necessary to achieve equilibrium is that the net external torque on a system must be zero. An external torque is one that is created by an external force. You can choose the point around which the torque is calculated. The point can be the physical pivot point of a system or any other point in space-but it must be the same point for all torques. If the second condition (net external torque on a system is zero) is satisfied for one choice of pivot point, it will also hold true for any other choice of pivot point in or out of the system of interest. (This is true only in an inertial frame of reference). The second condition necessary to achieve equilibrium is stated in equation form as

\text{net} \, \boldsymbol{\tau}=0

where net means total. Torques, which are in opposite directions are assigned opposite signs. A common convention is to call counterclockwise (ccw) torques positive and clockwise (cw) torques negative.

 


Source: Rice University, https://openstax.org/books/college-physics/pages/9-2-the-second-condition-for-equilibrium
Creative Commons License This work is licensed under a Creative Commons Attribution 4.0 License.

Last modified: Monday, November 1, 2021, 12:52 PM