G11-1. Solving Net Torque Problems
Solving a net torque problem is similar to solving a net force problem, but there are some differences. For both types of problems, you have to find the sum of something, either forces or torques. For net force problems, though, you set Fnet = ma, while for net torque problems, you set τnet = Iα. Here are some other differences:
We start with a method for calculating net torque and then show how to solve for the angular acceleration.
Calculating Net Torque
Example. A circular disc is rotated about an
axis O through its center by the application of two forces. A force of
magnitude 11 N is exerted at a distance of 0.34 m from the axis and at an
angle of 58° from a radial line extending from the axis through the point of
application Q of the force. A second force of magnitude 15 N is
exerted at a distance of 0.26 m from the axis and at an angle of 119° from a
radial line extending from the axis through the point of application P of
the force. Determine the net torque on the disc about its center and
which way the net torque accelerates the disc.
Calculating Angular Acceleration
Once the net torque is known, one calculates the angular acceleration using α = τnet/I, where I is the moment of inertia of the rotating, rigid body. We provide the following complete, example solution.
The situation is that of a thin rod of uniform density hinged to a level table at one end. See Figure 1 below. The mass of the rod is M and its length is L. Now consider Figure 2. We select the hinge at point A for the axis of rotation. Note that we don't have to select the physical axis of rotation as the point about which to calculate torques. However, it's convenient to do so in this situation. That's because the hinge exerts forces on the rod both to support it and to keep it from sliding horizontally. Since these forces act through the axis of rotation, they have 0 moment arms and hence exert 0 torque. Thus, the only force that exerts a torque is the weight, Mg, of the rod. We take this force as acting vertically from the center of mass. The moment arm of this force is the perpendicular distance, AB, from the hinge to the line of action of the force. This distance is AB = rcosθ, where r is the distance from the hinge to center of mass. You may wonder why we use the cosine function here rather than the sine function. The reason has to do with the way the angle θ is defined. With the angle shown in the diagram, the perpendicular distance from the axis of rotation to the line of application of the force is the side of the right triangle adjacent to θ. Thus, we use the cosine.
We can now write an expression for the torque produced by the weight of the rod. Since the rod is of uniform density, we can say that the center of mass is at the geometric center of the rod, thus, r = L/2 and AB = (L/2)cosθ. The torque exerted by the weight of the rod is therefore τMg = -Mg(L/2)cosθ. The minus sign indicates that the torque produces a clockwise rotation.
Since the weight is the only force exerting a torque on the rod, then τnet = τMg = -Mg(L/2)cosθ. We can now solve for the angular acceleration.
α = τnet/I
= -Mg(L/2)cosθ / I
= -Mg(L/2)cosθ / (ML2/3)
We substituted the expression from Table 10-1 of the text for the moment of inertia of a thin rod with axis at one end. Now let's do some checks. We've already addressed the sign. The units on the right are (m/s2)/m = 1/s2, which are the units of angular acceleration. Consider special values of the angle next. When θ = 90o, α = 0. This makes sense, because the rod would be vertical and in equilibrium. (Note that the equilibrium is unstable, because just a small motion would allow it to start rotating.) When θ = 0o, α = -(3/2)(g/L). The angular acceleration is greatest just before the rod hits the table. That makes sense, because the moment arm of the weight is greatest at that point. But does a magnitude of (3/2)(g/L) make sense? Consider the tangential acceleration at that point. This is vertically downward and given in magnitude by |at| = L|α| = (3/2)g. This says that the acceleration of the end of the board is 50% greater than g. Does it make sense that the acceleration of a falling object can be greater than g? The answer is yes, assuming that the object is part of a rotating, rigid object as in this situation. There are other points of the object with tangential accelerations less than g. For example, the center of mass has |aCM| = (L/2)|α| = (3/4)g or 75% of g. This suggests that there's a point on the rod that has a tangential acceleration of g. We leave it as an exercise for you to determine where that point is.
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