Rotational Equilibrium Study Guide (page 3)
In this lesson, we will approach a more complex system, a rigid object that has constraints that do not permit a linear motion but allow circular motion. We will define moment of inertia and torque inertia, and we will apply the conditions of equilibrium when rotational motion occurs in different systems.
Similar to the linear motion where we defined inertia, an object rotating around a fixed axis will keep this motion unless acted upon by an external influence. The same will happen with an object at rest: It will keep the state of rest unless acted upon.
Whereas mass is a measure of linear inertia, the measure of rotational inertia is the moment of inertia, or I. This quantity, as inertia, is proportional to the mass of the object but is also related to how the mass is distributed around the axis of rotation. The same object rotating around two different axes will resist rotation differently, and therefore have a different moment of inertia.
Note that the moment of inertia depends on the distribution of mass around an axis. The further the mass is from the axis, the more rotational inertia the object has and the harder it is to change the state of current motion. In this chapter, we will address the behavior of rigid objects. These are objects that will keep their shape and form while under external influence.
An object continues to stay at rest or move in a uniform circular motion as long as there is no external influence to change that motion. This property is called rotational inertia.
If you consider a metal rod or a wooden dowel of mass m and length L rotating in one of the two situations shown by Figure 9.1, you will have different responses to rotation.
In this figure, the rod rotates around the axis passing through the middle of the rod, halfway between the ends. In this case, the moment of inertia can be calculated as shown to be I = m · L2/12.
If we consider now an axis passing through the end of the rod, as shown in Figure 9.2, the moment of inertia can be calculated and the result is larger: I = m · L2/3 hence it is harder to start rotating the object or stop the object if it is in rotation when the axis of rotation is as shown in Figure 9.2.
Other useful expressions for the moment of inertia are shown in Figures 9.3 through 9.6.
In our discussions, we have studied two different types of motion: linear motion and rotational motion. The difference between the two lies in the path that the points of the rigid body follow in their motion. While in linear motion, the path is made out of parallel lines; in rotational motion, there is a spinning of the object around an axis.
The conditions of equilibrium are different for these two types of motion. Imagine the following examples. In the first example, you apply two equal but opposite forces on the steering wheel of your car (or boat) in the manner shown in Figure 9.7.
What is the result of the forces? If the wheel is anchored at the middle, there is no action following the interaction with the object. There is no rotation of the steering wheel around the axis of rotation.
Now let's apply the forces as shown in Figure 9.8. The object will rotate around the axis of rotation. However, in terms of the net force, there is no difference in the two cases. They can both be represented in the same way:
Net F = 0
The difference in the two cases is in the point of application of the force. So, analog to force, we need another quantity that includes force but also includes the distance to the point of application of the force. We call this new quantity torque.
Torque is proportional to the applied force, F, and to the lever arm, I:
T = FI
- The pivot point is the point through which the axis of rotation passes.
- The line of action is an extension of the direction of the force.
- The lever arm is the length of the perpendicular dropped from the pivot point to the line of action.
- The unit of measurement for torque is N · m.
Consider a seesaw of length L and a child playing on it. Show the force, the level arm, and the pivot point. Find the torque acting on the end of the seesaw.
Figure 9.9 shows the line of action, the pivot point, the weight of the child playing on the seesaw, and the lever arm.
Because the board has a length L, the lever arm is And the torque is:
Value of Torque
The value of the torque in Figure 9.11 (a) is considered positive, or counterclockwise (CCW), whereas the second picture shows a negative, or clockwise (CW) torque.
Now consider that the child stepping up from the board and starting to push on it with a force equal to his weight and in the direction shown in the figure. Is the torque the same as before?
Figure 9.10 shows the line of action, the pivot point, the force of the child playing on the seesaw, and the lever arm. The lever arm in this case is:
T = F · l = F · (L/2) · cos α
T = m · g · (L/2) . cos α
In this case, the torque is smaller than when the child exerts her weight perpendicular to the plank.
In these two cases, the applied force has the same magnitude but different directions. Hence, the effect on the object will be different.
Another observation that we need to make about torque is that, similar to force, it is a vector: For the same values of the force and lever arm, we might end up with different directions of rotation. Let's look at the following example. In Figure 9.11(a), the rotation is counterclockwise (CCW), whereas in Figure 9.11(b), although the same value of force and lever arm was used, the object rotates clockwise (CW).
A rigid object is in rotational equilibrium when the net external force and the net torque are both zero.
Net F = 0
Net τ = 0
The two conditions have to be satisfied in the same time.
Equilibrium of a Rigid Object
We have seen that there are cases when one, two, or more forces can give a zero or nonzero torque, depending on the value of the lever arm. In the example with the rotating wheel, once the two forces are applied such that their extension goes through the center of rotation (and the lever arm is zero) and the forces are equal and opposite, there is no rotation and therefore no torque. This situation is a rotational equilibrium, and to establish the state, we need to talk about two conditions: net external force and net torque.
Two kids playing soccer hit the ball at the same time with equal and opposite forces directed along the same diameter of the ball. Does the ball start to spin around its own axis?
There is no data, so no conversion to SI is necessary.
F = –F
Net τ = ?
Because the line of action passes through the center of rotation, any thing perpendicular to it from the center would be zero. Hence, there is zero lever arm and zero torque for each of the forces.
Net F = F + ( –F) = 0 N
Net τ = F · 0 + F · 0 = 0 N · m
Consider the seesaw again, now with two children: One is 35 kg and the other 42 kg. Is this system in rotational equilibrium? The two sides of the seesaw are of equal length and the total length is 280 cm. If the seesaw is not in equilibrium, what has to be changed so that it equilibrates (keep masses constant)? Consider the seesaw to be of negligible weight.
Convert the quantities to SI. Next, draw a diagram and find the equations.
m1 = 35 kg
m2 = 42 kg
L = 280 cm = 2.8 m
Net F = ?
Net τ = ?
The net external force is the sum of the two weights and the normal force. Because the object does not have a linear motion, the three cancel each other out.
Net F = W1 + W2 – N = 0 N
Weight W1 determines a positive torque (CCW), whereas the second weight determines a negative torque.
The weight is perpendicular on the seesaw in both cases, so the lever arm is equal to half of the length of the seesaw.
τ1 = F · l = W1 · l = m1 · g · l = m1 · g · L/2
τ2 = – F · l = – W2 · l = – m2 · g · l = – m2 · g · L/2
Net τ = τ1 + τ2 = m1 · g · L/2 – m2 · g · L/2 (m1 – m2) · g · L/2 = –96 N · m
Net τ = –96 N · m
The minus sign tells us that the seesaw rotates clockwise, hence the system is not in rotational equilibrium.
In order for the seesaw to remain in rotational equilibrium, the net torque has to be zero. Although we cannot change the mass of the two children, we can change their position on the seesaw.
τ1 = F · x = W1 · x = m1 · g · x
τ2 = – F · (L – x) = – W2 · (L – x) = – m2 · g · (L – x)
NetT = τ1 + τ2 = m1 · g · x – m2 · g · (L – x) = [m1 · x – m2 · (L – x)] · g = 0 N · m
And we can solve for the unknown, x:
m1 · x – m2 · (L – x) = 0
(m1 + m2) · x – m2 · L = 0
(m1 + m2) · x = m2 · L
x = 1.5 m
The smaller weight has to have a longer lever arm to equilibrate the torque exerted by the second, larger weight.
Practice problems of this concept can be found at: Rotational Equilibrium Practice Questions
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