6.6: Torque
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- Nov 1, 2024
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Rotational Analog to Force
We have developed multiple analogs between linear and rotational motion so far: velocity, acceleration, kinetic energy, and mass. We have seen Newton's 2nd Law which relates force to mass and acceleration, which tells us how the linear motion of an object changes due to external forces. Now we want to develop the analog of the 2nd Law with will relate "rotational force", known as torque,
Torque is a vector since angular acceleration is a vector, and rotational inertia is a scalar. Let us examine which variables torque depends on by thinking about its units:
The last equality in the above equation comes from definition of Newtons:
Figure 7.5.1: Torque on a Point Mass
Focusing on magnitude only for now and applying Newton's 2nd Law for linear motion to the mass and using the connection between linear and angular acceleration,
For a point mass the rotational inertia from the pivot is
Comparing the above result with Equation
This result is consistent with the argument in Equation
Figure 7.5.2: Direction of Torque on a Point Mass
Thus, cases 1 and 2 generate a torque but in the opposite directions based on Equation
Figure 7.5.3: Vector Direction In and Out of the Page
What we can conclude from the three cases in Figure 7.5.2 that whether a force generates a torques depends of its orientation relative to the pivot. We can define a force position vector,
Figure 7.5.4: Direction of Force and Torque
Equation
The equation above only give the magnitude of the torque, while the right-hand rule determines the direction. Mathematically, a cross product contains both the magnitude and direction of torque:
But in this class, we will typically use equation
Figure 7.5.5 shows the position vector
Figure 7.5.5: Position Vector and Moment arm
However, there are cases where using the position vector to calculate the magnitude of torque is not the simplest method geometrically. Thus, another method with the moment arm,
Combing the above equation and Equation
The two methods described by Equations
What we have learned by describing torque in this section so far is that the location of the force is essential in order to determine torque. Thus, when drawing a force diagram on a selected system, we can no longer collapse the extended object(s) into a point, which we did when calculating forces. For torque we need to draw the entire object and draw forces at the locations of their application. This is known as the extended force diagram, which needs to contain all the force vectors at their locations, the pivot or the point of reference from which torques are calculated, and all the distances marked from the pivot to each force. The extended force diagram contains all the information that a "collapsed" free-body diagram contains, so it can be used for calculating the net force as well as the net torque.
Example
Shown below is a top view of a square shape anchored at the lower left corner with 5 forces acting on it. The length of each side of the square is 0.4 m. Forces 1-4 at at one of the corners of the square, while force 3 acts right at the center, as shown below. The magnitudes of the forces are given as:
a) Calculate the net torque on the square. Express your answer in terms of magnitude and direction.
b) Calculate the net force on the square. Express your answer in terms of magnitude and direction.
- Solution
-
a) The following diagram shows a convenient choice of using either position vector,
or the moment arm to solve for the torque due to each force.Force
is parallel to the position vector . Thus, there is no perpendicular component which results in zero torque due to this force:Force
has the same position vector as ( ), but there perpendicular component of this force to the position vector, :The magnitude of the torque due to force 2 is then:
Using the right hand-rule, fingers point up and curl toward
in the clockwise direction resulting in the thumb pointing into the page. Thus, the torque due to points into the page.For
it is simpler to use the moment arm method since you do not need to calculate any angles, as labeled in the figure . If you were to use the position vector, you would need to draw a vector pointing from the pivot to the location of , and then calculate component of which is perpendicular to . Also, you would need to calculate the length of , which is half of the diagonal of the square. But using the moment arm gives a simple result to the magnitude of the torque due to force 3:Using RHR, we find that
results in a clockwise rotation, so the torque point out of the page.For
it is again simpler to use the moment arm to calculate torque. Using as marked in the figure we find:Using RHR, we find that force 4 results in a clockwise rotation, so the torque point out of the page.
Force
is parallel to the position vector, since makes exactly a angle with the corner of the square, as does . Since the two vectors are parallel there is no perpendicular component of the force to the position vector which results in zero torque:Let us define a convention that vectors pointing out of the page are positive, and those pointing into the page are negative. Putting everything together the net torque is:
The magnitude of net torque is
and it point into the page.b) To find the net force we need to break down the forces by components. In the x-direction:
In the y-direction:
The magnitude is:
And the angle is:
pointing in the northeast direction.
The Center of Mass
You may have realized by now that modeling objects as point particles is a rather drastic oversimplification, even though it is often very useful. We have now seen instances when the extended geometry of a non-point object becomes important. Focusing on just one point of an object can describe perfectly adequately the translational motion of that object, but it does not tell us anything about the object’s rotation. It turns out that we can consider all of the forces acting on the object as if they acted at one point, the center of mass, as far as translation is concerned. That is, if we are concerned only about an object’s translation, it doesn’t matter where the forces act on the object. We can consider them all to act at a single point! This is truly a great simplification. The special point where we consider the forces to act is called the center of mass.
The center of mass is defined as a geometric average position of all the masses. For example, the center of mass of two point particles of equal mass is exactly at the midpoint between the two masses. If the masses are unequal, the center of mass would be closer to the heavier mass. Mathematically the center of mass can be described with an equation which expresses the average based on the relative weights of the masses. Figure 7.5.6 below shows a generalized system of two points masses,
For the two masses shown in Figure 7.5.6, the center of mass is written as:
From the equation we see that when
In nature most objects are not made up of a collection of point masses, but rather of a continuous mass. Mathematically, this means that we need to turn the summation to an integral over small mass elements, which can become rather complicated for an object in three-dimension with mass which is not uniform.
Torque due to Gravity
As we have just established in order to calculate torque on an object due to a force, you need to know the exact location where that force act. Gravity is a special case since it acts everywhere on a given object. And gravity can certainly cause torque, if you attempt hold a long and heavy stick at one end, it will want to rotate down due to torque cause by gravity. Thus, we want to figure out where to place the force of gravity on an extended object so we can calculate the amount of torque generated by gravity relative to some reference point.Consider the example in Figure 7.5.7 below, where a massless rod with two point masses at each end is balanced by a fulcrum, the pivot point where an object is balanced. The left picture shows an extended force diagram. Since the two masses are point masses, the force due to gravity on each mass is located exactly at the location of each mass.
Figure 7.5.7: Balanced Rod with Two Point Masses
Relative to the fulcrum about which the rod pivots, the two weights generate torque in opposite directions. The right mass causes a torque into the page, while the left mass generates a torque out of the page. To balance torque, in addition to the direction being opposite, their magnitudes need to be the equal,
Imagine that we did not have knowledge of the two point masses at the end of this rod, but instead just moved the fulcrum along the rod until it became balanced in this particular location. Since we now view this object as having some total mass (rather than a sum of two point masses) but with unknown distribution of that mass and there is no net torque once the fulcrum balances the object, the logical location of the weight of the object is at balance point. At the fulcrum which is the pivot, the position vector to the weight force is zero, so there is zero torque due to the weight. If we placed the weight vector anywhere other than the fulcrum, then it would generate a torque relative to the pivot, which is not consistent with the object being balanced. In other words, we can replace the two weights at the two point masses in the left picture with just one weight at the fulcrum of magnitude,
Derivation
We want to show that the result in Equation
The center of mass relative to x=0 is given by:
We can rewrite
and
Plugging in the expression for
and
Now let us multiple the equation for
which is the exact equation that we obtained from balancing torque. Thus, the fulcrum was balanced at the center of mass of the two point object.
It turns out that the center of mass is the same as the center of gravity (where you can support the object and it will not rotate) as long as the gravitational force is uniform. Near the surface of the Earth, for all objects of ordinary size, the gravitational force can certainly be considered uniform. Thus, for all problems we consider, the center of mass and center of gravity are the same point.
If objects are constrained to rotate about a particular axis, such as a rod mounted by a hinge to a vertical wall the object is fixed at a specific location, the torques are typically computed about that axis. If there is no constraint, torques should be computed about the center of mass, the point about which the object will rotate. To see this, imagine you exert two forces of equal magnitude and opposite direction at different locations along the object. The different locations of force will result in a non-zero net torque, so the object will start rotating. Equal and opposite forces result in zero net force, so the center of mass cannot accelerate. If the object is rotating about the center of mass, it means that the center of mass is stationary, which is consistent with zero net force. If the object was rotating about any other reference point, that would imply that the center of mass is rotating about that point. In that case the velocity of the center of mass would be changing since its direction is changing, which can only happen if there is net force. Thus, the object must rotate about its center of mass.
Work
We are familiar with the concept of work as a way that the energy of a system is changed. In terms of force and distance, work is done to translate a system from position
where the dot product of the two vectors reminds us that it is only the components of force and displacement in the same direction that contribute to the integral.
A similar expression holds for the work done by a torque to rotate a system from angle
The energy of a particular system can be changed by the process of a force exerted by an outside object doing work on an object in the system and/or by a torque exerted by an outside object doing work on an object within the system. In either case, the work can be positive (increases the energy of the system) or negative (decreases the energy of the system).