TORQUES AND ROTATIONAL MOTION

Just as net forces cause changes in translational motion, net torques cause changes in rotational motion. The logic behind setting these problems up runs about the same as the translational problems. There are, obviously, many ways in which something can rotate. Mainly, things can rotate at a constant rate (constant angular velocity), a changing rate (constant angular acceleration), or with changing angular acceleration. Fortunately, you will only be held accountable for the constant angular velocity and constant angular acceleration.

Constant Angular Velocity

For a wheel, person, or llama to rotate at the same rate, the torques must cancel each other out or not exist whatsoever. Letís start by looking at something that doesnít rotate. In this case, the net torques are acting to cancel each other out. When I was a kid, playgrounds used to have something called a teeter-totter, that looked like the object shown at the right. If both people had the same mass, the teeter-totter would balance when each person sat equidistant from the pivot. However, if the people had different masses, you had to adjust the placement of the pivot to compensate for the difference. The reason the adjustment was needed is that balance is achieved when the torques balance.

Torques are more than just force. In the simplest case, like the one shown above, they are a product of force and its distance from the axis of rotation. Since a torque also considers distance, then a small force may well be capable of producing a large torque. In the teeter-totter illustration, letís assume that the longer side is twice the shorter side in pictures B and C.



If you choose the pivot as the axis of rotation, the FBD for the torques acting on the beam in the following picture would appear as shown.



The torque equation for the second picture would look like this initially:

tnet = t1 - t2

The negative sign means that the second torque would cause the beam to rotate in the opposite direction. However, since it isnít rotating and we now know that torque can be computed as force times distance, then this equation becomes:

0 = F1d1 - F2d2

0 = m?g(2x) - (50)gx

And the mass of the unknown object can be found by doing a little algebra

2m?gx = 50gx

m? = 25 kg

The torques in picture C could be analyzed in much the same way as B to find the mass of the unknown object. In fact, the FBD diagram and initial equations would appear exactly the same.



The only difference crops up when values are substituted for F and d. The solution follows:

tnet = t1 - t2

0 = F1d1 - F2d2

0 = 50g(2x) - m?gx

And the mass of the unknown object can be found by doing a little algebra

m?gx = 100gx m? = 100 kg

Notice that in either case, the smaller force was able to produce an equal torque by acting on the beam at a distance that is twice as far from the pivot.

Questions:

What other force must be present to prevent motion from occurring in pictures B and C that is not drawn in the FBD?

How big would it have to be in each case?

Why wasnít it drawn in?

The teeter-totter problem is a good starting point, but it is tremendously simplified. In reality, the beam is going to have mass, the forces may not act perpendicular to the beam, and for most of the problems you will encounter the beam or object is going to rotate. Even so, the basic approach will remain the same:

1. Draw FBDs as required

2. Setup equations for net force and torque

3. Substitute appropriately, and

4. Solve

Since we all want 5s on the AP exam, letís take a look at the things we didn't consider initially.

First, let's look at the angle between the beam and the force. Remember that we said a torque results from a force being applied on an object at some distance away from the axis of rotation and directed neither toward nor away from the axis of rotation. In the earlier problem, the forces were pushing on the beam such that no part of the force went toward or away from the axis of rotation (they acted perpendicular to the beam). However, if we allow the beam to turn, as shown at the below, part of the force is directed toward or away from the axis of rotation.





In other words, if you drew a line from where the force is applied to the pivot, only part of that force is acting perpendicular to that line. The rest of the force is acting along the beam. Consequently, the only part of the force that can change the way the beam rotates is that which acts perpendicular to the beam because the other part acts to slide the beam along the pivot.

Let's take a closer look at the force acting on the lower left end of the beam. Notice we can break down that vector into components (parts) that run parallel and perpendicular to the beam.



Since the parallel part goes through the axis of rotation, it doesn't contribute to the torque. The only part that contributes is the perpendicular part, which can be quantified as F1sin θ. As a result, we can improve upon the equation used earlier (torque = force times distance). A better (more universal) way of calculating the magnitude (size) of a torque is:

τ = Frsinθ, where

r is the distance from where the torque is applied to the axis of rotation

θ is the reference angle between the force vector and a line drawn between the axis of rotation and the point of the application of the force.



Although this equation is all you need to compute the size of a torque, the best equation will give the direction of the torque as well. Mathematically, a torque is best described by using an operation called the cross product. Thus, a torque equation becomes:



A cross product is a mathematical operation for multiplying two vectors. It yields a third vector that is perpendicular to the two that created it. If you happen to remember your 3-dimensional vector math from last year, then you compute a torque as follows:



If not, we will cover this operation in greater detail later. It looks more confusing than it is.

At this moment, you only need to know how to calculate the magnitude as previously stated, Frsinθ, and be able to find the direction using . . . THE RIGHT HAND RULE!!!!
The right hand rule works for any cross product and is used whenever two vectors produce a third vector that is perpendicular to both.

Example:

Find the direction of the torque generated by F1 in the drawing shown below:





Since the equation is,



the steps proceed as follows:

Step 1: Straighten your right hand and place the bottom of your hand at the pivot and point your finger toward F1. Your hand represents the displacement vector, r.

Step 2: Turn your hand so that you can bend your fingers, naturally, in the direction of the force, F1.

Step 3: Stick out your thumb. Your thumb represents the direction of the torque.

Alas, we are getting ahead of ourselves. For the time being, I only want you to concern yourselves with setting up torque problems. Which leaves us with one more thing to cover for tonight.

Constant Angular Acceleration

The approach to solving these problems is exactly the same as constant angular velocity, with one exception. The net torque will not equal zero. Otherwise, the two setups are identical.



The pulley shown above illustrates two forces, each exerting a torque on it. To simplify the problem we are assuming that the force, F3, from the pulley's axle prevents translation, but provides on torque. Thus, the only forces exerting torques are F1 and F2. Since they act against each other the net torque equation would appear as:

tnet = t1 - t2

tnet = F1r1sinq1 - F2r2sinq2

In this case each angle is 90į, so the equation simplifies to:

tnet = F1r1 - F2r2

We can go one step further by substituting the rotational equivalent of ma, which is Iα. The "I" in rotation stands for "moment of inertia" and equates to mass in translation. The "α" in rotation stands for angular acceleration and equates to acceleration in translation. Therefore, by analogy to force problems the final net torque equation may be written as:

Ianet = F1r1 - F2r2

All problems follow the same format, the trick is drawing the FBD correctly, identifying the axis of rotation, and setting up the equations.

The ladder problem:








The pulley problem:








Draw a FBD, net force and net torque equations for the beam suspended by the string in the picture below.

The beam and pivot problem:




Check your answer

You can weigh a rigid object by balancing it with a mass hung from one end. If a meter stick balances at the 20 cm mark with 250 grams hung from one end, what is the mass of the meter stick?


Problems:

1. How do torques differ from forces in terms of what they do?
Need a hint?


Check your answer?

2. What do torques cause?
Need a hint?


Check your answer?

3. Explain why it is easier to curl 50 lbs, than lift 50 lbs with your arm fully extended.
Need a hint?


Check your answer?

4. What do you know about the forces and torques on an object that is not moving?
Need a hint?


Check your answer?

5. A 5N force is applied to a stick, 3 m from the pivot. At what distance from the pivot would a 2 N force have to be applied to prevent the stick from rotating?
Need a hint?


Check your answer?

6. What is the rotational equivalent of mass?
Need a hint?


Check your answer?

7. When does something have a net angular acceleration?
Need a hint?


Check your answer?

8. Corey balances a meter stick at the 25 cm mark when he puts a 100 g mass at the 100 cm mark and a 500 gram mass at the 0 cm mark. What is the mass of the stick?
Need a hint?


Check your answer?

9. A 50 kg ladder leans against a wall as shown below. What is the minimum force of friction needed to prevent the ladder from sliding. Assume that the force of friction from the wall is negligible.




Need a hint?


Check your answer?

10. An Atwood's machine consists of a pulley and two masses as shown below and to the right. If the masses are 10 kg and 5 kg from left to right, and the acceleration of the masses is 1 m/s2, what is



a) The tension in the left string

b) The tension in the right string

c) The moment of inertia of the pulley, given it has a radius of 0.2 m.

(Hint: To solve this problem you need to realize that a = αr)


Need a hint?


Check your answer?

11. A wheel that has a moment of inertia of 50 kg-m2 is pulled on by two ropes as shown below. One rope is wrapped around the inner radius of 0.1 m and the other is wrapped around the outer radius of 0.4 m. Determine the angular acceleration of the wheel.




Need a hint?


Check your answer?

12. Why do some problems have tensions the same on either side of a pulley and others, like number 10, have different tensions on either side of the pulley?


Need a hint?


Check your answer?

13. Why is there a ďsinθĒ in the torque equation?
Need a hint?


Check your answer?

14. What is the equation for the net force that causes circular motion? What is the name for this force?
Need a hint?


Check your answer?

15. A 10 gram marble rolls around a cone at a radius of 0.2 m at a constant height. Assuming friction is negligible, determine the speed of the ball.




Need a hint?


Check your answer?


Questions? Send comments to Like I care
Main Home Page
AP Physics Home Page