Chapter Review
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Chapter 11: Rotational Dynamics and Static Equilibrium Chapter Review |
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Chapter Review
In this chapter we continue the study of rotation by looking at the dynamics of rotational motion. Here is where we introduced rotational versions of concepts like force, linear momentum, and Newton's second law.
11-1 - 11-2 Torque and Angular Acceleration
If you want to cause a nonrotating object to rotate you must apply a force. However, not just any force would result in a rotation. Crucial to determining the rotation that results from the force are two things, (a) how the applied force is directed relative to the axis of rotation, and (b) the distance of the point at which the force is being applied from the axis. These two things combine with the magnitude of the force to form a quantity called torque. It is through the application of a torque that an object will begin to rotate. The magnitude of the torque t that results from the application of a force F is given by
t = rF sin q,
where r is the magnitude of a radial vector r from the axis of rotation to the point of action of the force, and q is the smallest angle between r and F. The SI unit of torque is the 
; when dealing with torque the is not called a joule.
For convenience, the above expression for torque is often looked at in two ways. The quantity Fsin(q) equals the component of F tangential to a circle of radius r centered on the axis of rotation, Ft; so, we can write t = rFt. Grouped the other way, the quantity rsin(q) equals the component of r perpendicular to the line of force, 
, called the moment arm (or lever arm) of the force; so, we can also write 
.
All of the above equations are for the magnitude of the torque only. Torque is a vector quantity and we'll briefly discuss how to find its spacial direction later. In our applications torque will be a one-dimensional vector so we can account for its direction with just an algebraic sign. The convention is as follows:
a torque is positive if its tendency is to rotate an object counterclockwise and negative if its tendency is to rotate an object clockwise.
Example 11.1 Teather Ball: A child is playing teather ball; she has a 2.0 m long cord with one end connected to a vertical pole. Attached to the other end of the cord is a light ball of mass 0.43 kg. As the ball whirls around in a horizontal circle, the string makes an angle of 35o with the pole. The child then hits the ball with a force of 3.6 N, that lies in the horizontal plane of the ball's motion, and makes an angle of F = 30o with the tangent to the ball's path (away from the pole). What magnitude of torque does the child's force apply to the ball about an axis through the pole?
Picture the Problem Figure (a) shows the ball connected to the pole by a cord of length L. Figure (b) is a top view showing the ball's circular path and the force F applied by the child.
Strategy To complete this calculation we must correctly determine the quantities that are relevant to calculating torque: r, F, and q.
Solution
| 1. In figure (a) r equals the horizontal distance between the ball and the pole's axis: |  |
| 2. Since the vector r is perpendicular to the tangent line, the angle q can be found from F: |  |
| 3. Directly calculate the magnitude of the torque: |  |
Insight We only wanted to calculate the magnitude of the torque in this problem. If we wanted to associate a sign or direction to this torque, according to the above picture, it would be negative because the torque contributes to a clockwise rotation of the ball.
Torque is the quantity that plays the role of force for rotational motion. Just as force causes translational acceleration, torque causes angular acceleration. The relationship between t and a is very similar to that between force and linear acceleration,
t = Ia.
Comparing this with F = ma, we can see how strong the similarity is by recalling that the moment of inertia I is the rotational quantity that acts like mass does for translational motion. The above equation is referred to as Newton's second law for rotation.
Physlet Illustration: Torque and Moment of Inertia | |
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| A mass (between 100 and 1000 g) is hung by a string from the edge of a disk-like pulley causing it to accelerate as shown. The time is shown in seconds and the angular velocity, omega, is given in radians/sec. Determine the mass of the pulley. Start | |
Hints
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Exercise 11.2 Engine Specifications: The engine specifications on a Pontiac Grand Am SE states that it supplies 225 
of torque at 4200 rpm. If this same torque is applied to a solid cylinder of mass 3.3 kg and radius 32 cm that rotates about its central axis, what angular acceleration would result?
Solution: The following information is given in the problem.
Given:  , M = 3.3 kg, r = 32 cm, w = 4200 rpm |
Find: a |
We know that the relationship between torque and angular acceleration is t = Ia. Therefore, if we can determine the moment of inertia for the cylinder we can use it to determine a. From table 10-1 in the text we can see that the moment of inertia for a solid cylinder about an axis through its center is
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Substituting this into Newton's second law for rotation and solving for a yields
.Inserting numerical values and converting produces the final result of
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Notice that to answer the question of this problem we did not need the angular velocity. For the engineers working on this car the angular velocity is important because it is at this angular speed that the result applies. You should always try to understand the relevance of a calculated result; otherwise, you will greatly diminish its usefulness to you.
Practice Questions
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11-3 - 11-4 Static Equilibrium and Balance
In many applications, especially the design of buildings and other structures, you want objects to be balanced so that they are not falling down or tipping over on people. This means that we want the objects to be in static equilibrium. We have seen examples for which an equilibrium of forces means that an object will not translate; however, to have complete static equilibrium the object must not rotate either. For the object to be nonrotating, we must have an equilibrium of the torques on the object. The conditions for static equilibrium then are that both the net force and the net torque on an object be zero:
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When doing an analysis for equilibrium, it is useful to remember that here is a place where the center of mass concept is often very useful. The weight of an object can exert a torque about an axis of rotation that doesn't pass through the center of mass. Therefore, as far as torque is concerned, we can treat rigid objects like point masses whose mass is located at the center of mass. This will be illustrated by example below.
Example 11.3 Hanging the lights: A horizontally suspended 2.7 m track for fluorescent lights is held in place by two vertical beams that are connected to the ceiling. The track and lights jointly weigh 15 N and each beam is attached 60 cm from opposite ends of the track. What upward force does each beam apply to the track?
Picture the Problem The picture on the right shows the light (at the bottom) attached to a track which hangs from the ceiling by two vertical beams.
Strategy To solve this problem we will apply the equilibrium conditions. The first step is to draw a free-body diagram of the track (shown on the right) with the left beam applying a force labeled F1 and the right beam's force as F2. The axis for calculating torques is located at the left beam. In this example we will calculate torque as force times lever arm.
Solution
| 1. Write out the torque due to F1: |  |
| 2. Write out the torque due to F2: |  |
| 3. Write out the torque due to gravity: |  |
| 4. Apply the equilibrium condition for torque: | |
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| 5. Apply the equilibrium condition for force: | |
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Insight Each beam contributes equally to supporting the track. Once you become more used to this type of analysis you'll be able to solve a problem like this without any calculations. The symmetry of the problem shows that F1 and F2 have equal lever arms, about an axis throught the center of the track. Since they apply torques of equal magnitude about this axis, they must also apply forces of equal magnitude. Since they share the 15 N equally, they must each be 7.5 N.
Example 11.4 Decorating: A decorative ornament is hung from the end of an 82.3 cm long horizontal beam which is attached to a wall as shown below. The beam weighs 12.1 N and the ornament weighs 7.62 N. (a) What is the tension in the cord? (b) What are the horizontal and vertical components of the force that the wall directly applies to the beam?
Picture the Problem The picture on the left shows the ornament being supported by a horizontal beam attached to a vertical wall. The beam is supported by a cord attached to its end. The picture on the right is a partial free-body diagram of the beam (the unknown force exerted on the beam by the wall is not shown). The lever arm for the tension T is also shown on the diagram as 
.
Strategy Fwall is unknown at this stage so it was not explicitly shown in the diagram on the right. Therefore, we chose the axis to be at that location to simplify the equation for the net torque. Let's first apply the torque condition and see where it leads us. (When needed I will choose up as the +y direction and right as the +x direction)
Solution
Part (a)
| 1. Write the equilibrium condition for the torques: |  |
| 2. From the geometry of the problem the lever arm for the tension can be written in terms of L: |  |
| 3. Dividing through by L the torque equation becomes: |  |
| 4. Solving for T gives: |  |
Part (b)
| 5. Apply the equilibrium condition for the x-components of the forces: |  |
| 6. Apply the equilibrium condition for the y-components of the forces: | |
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Insight The signs of the results in part (b) tell us that Fwall,x is to the right and Fwall,y is upward. We could have discovered these facts without calculations however by looking carefully at the equilibrium situation. This will be discussed in more detail in the Other Useful Tips section below. Also, the length of the beam does not enter the calculations.
Practice Quiz
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