|
Angular Acceleration Returning back to angular acceleration, let's consider an object traveling in a circle as pictured below.  We know that the object has centripetal acceleration, because it is constantly changing direction. Its vector is shown below.  But, what happens if the object does not maintain a constant speed? There has to be an acceleration, but it is not centripetal, because that is only the acceleration of direction, not speed. Since this acceleration affects the linear speed, it will point along with the linear speed as shown below. It is called the tangential acceleration because it is the change of the tangential velocity.  Again, this could mean speeding up, or slowing down. In the latter case, the vector arrow would be pointing in the opposite direction of the tangential velocity, but still tangent to the circle. This would be analogous to riding your car in a circle and either stepping on the gas to speed up, or breaking to slow down all the while keeping the same radius of the circle. You may be wondering what this has to do with angular acceleration. Remember that angular acceleration, or alpha, is given by,  We also know that omega is equal to v/r, from the v = (r)(omega) equation. Substituting that relationship into the alpha equation, and dropping the delta signs for convenience, we have,  Notice in this equation you have a v/t, which is acceleration. In fact, since that v is the linear or tangential speed, it is tangential acceleration. So, it turns out that alpha and tangential acceleration are related by,  Rearranging gives us a nice neat equation.  There is also a nice relationship between centripetal acceleration and omega. Since,  We can square the v = (r)(omega) equation and substitute into the v squared in the centripetal acceleration equation and get,  The r on the bottom cancels one of the r's on top and gives,  These two simple relationships allow you to solve problems with ease that might otherwise be difficult. For example, consider a merry-go-round with radius of .50-m that has an instantaneous angular velocity of 4.0 rad/s while undergoing an angular acceleration of 2.0 rad/s^2. We can find the total linear acceleration using the new equations. To find the centripetal acceleration, we plug in the numbers to the equation and get, Similarily, the tangential acceleration is, Since these are vectors, we can not just simply add them together to find the total. We have to use vector addition.  The picture above shows the two vectors, at and ac, added together to make a. Using the pythagorean theorem to solve for a we have,  This equals about 8.1 m/s^2. We do need to find the angle since this is a vector, but we will just stop with the magnitude. Angular Kinematic Equations It should not come as a surprise that there is a list of angular kinematic equations. So far, every linear equation has had an angular equation. All we have to do is replace the linear components with their angular counterparts. Recall the kinematic equations from chapter one. Below is a table that has the linear equations on the left, and the angular equations on the right. Notice how each equation compares to each other.  There is no difference between solving problems involving the linear quantities and solving problems dealing with angular quantities. The same rules and methods apply to both situations. Let's consider a problem using the angular kinematic equations. A motor begins to rotate from rest with an angular acceleration of 20.0 rad/s^2. Find the motor's angular speed 6.0 seconds later, and how many revolutions were made during that time interval. Since we know the motor started from rest, the initial omega is zero. Knowing alpha and time, we can use the first equation in the list.  Finding the revolutions, we use the angular displacement equation and get,  That answer is in radians, but we want to know the revolutions. So, using the conversion factor, the answer in revolutions would be about 57 revolutions. Let's continue this same problem and assume the motor begins to accelerate at - 15.0 rad/s as soon as it reaches the 6.0 second mark. Find how long it will take to slow to a stop. This time omega final is zero, and omega initial is the omega found in the first part of the problem, which is 120 rad/s. So, the equation becomes, Solving for t gives 8.0 seconds. Now let's say the motor continues to have that same acceleration for an additional 3.0 seconds. Find the final omega, and angular displacement. Using the same equation as before we will have,  To find the total angular displacement, we need to start all the way back when the motor began with the negative acceleration, which would mean we need the initial omega at that point, which is 120 rad/s, and the total time would be the first 8.0 seconds to slow to a stop, and the additional 2.0 seconds after. This makes the equation,  This is certainly less than the first amount of revolutions found at the first of the problem. How is that possible? Well, you have to remember that this is angular displacement, which is how many rotations or radians the object has ended up with respect to the starting position. So, in the process of speeding up and slowing down, the motor would have a positive angular displacement. However, when the motor continued to have the negative acceleration, it now turned in the opposite direction, which made omega negative and the displacement negative. As a result, the 72 revolutions means the motor ended up a positive 72 turns from the original position. It is positive because the motor did not turn enough in the negative direction to completely completely undo the turns made during the speeding up and slowing down process. To understand this concept, let's say the motor didn't continue for 2.0 seconds, but for 12.0 seconds. So, after the 8.0 seconds of slowing down, the motor continues with the negative acceleration for another 12 seconds. Solving for the angular displacement would entail,  Here is an example where the motor not only went back to the original position, but continued to turn in the negative direction, and ended up about 95 revolutions from the original position after 20 seconds of having a negative acceleration. Torque Remember that Newton's Second Law says any net force applied to an object will produce a linear acceleration. Can this Law apply to angular acceleration? Consider Newton's First Law. It applies to linear motion, but, it can also be applied to angular motion. The First Law is all about the net force being zero on an object. So, if a wheel is at rest, and the net force on that wheel is zero, will it rotate? Of course not. Likewise, if the wheel is rotating with a constant omega, and the net force is zero on the wheel, will the wheel's omega increase or decrease? How could it? Therefore, an object will only change states of rotation when acted upon by an outside influence. Sound familiar? Taking this further, an object will sustain angular acceleration when under an outside influence, which is the essence of the Second Law. In the linear sense, the outside influence is force. In the angular sense, the outside influence is referred to as Torque . Torque is the influence that causes an object to change states of rotation. Although torque is the angular equivalent to force, there are some fundamental differences. In the linear sense, it does not matter which direction the net force is applied, the object will always accelerate in the same direction as the net force. This is not necessarily true in the angular sense. Consider the picture below.  Notice how on the wrench, the force is acting down, but the acceleration is pointing into the page, and perpindicular to the applied force. What about the torque? What direction does it point? It has to point in the direction of alpha right? The answer is yes. Here is where force and torque are not exactly the same quantities, in the sense that velocity is displacement over time and omega is angular displacement over time. Torque is not technically a force although it does depend on force. It also depends on one more thing: the lever arm. The lever arm is the distance between the applied force and the axis of rotation as shown in the example below.  It turns out that the torque on an object is equal to the force times the lever arm. Below is the equation where torque is represented by the greek letter tau.  As a result of this equation, the unit of torque is not Newtons, but Newton-meters, because force is in newtons and lever arm is measured in meters. (The more familiar unit of torque is the foot-pound). This is why torque is not technically a force. But, it is still the angular counterpart of force which we will see more clearly later. Right now, let's examine the equation above a little further. Obviously torque is directly related to force and the lever arm. So, doubling the lever arm while keeping the same force doubles the torque and vice versa. You could produce any size of torque that you wanted provided you could extend the lever arm indefinitely (more on this in chapter ten). This is such a useful relationship and its applications are found everywhere. In fact, most people have probably utilized this concept without even realizing it. Have you ever tried opening a door when the door knob was actually located in the middle instead of at the edge? Whoever designed those doors must have done that for aesthetic purposes, because it certainly doesn't make the door easier to open. In fact, it makes it quite worse. Since the lever arm is cut in half, you have to use twice the force to open the door! A cool way to feel the consequences of torque is to push a door open with one finger: first at the very edge farthest from the hinges, and then at a point close to the hinges. After pushing close to the hinges with one finger, you will have a new appreciation for torque. Consider the situation below.  In this situation, the force is acting at an angle, which leads to the question: what is the lever arm now? You can answer this two different ways with two different lever arms. First let's consider the force's components.  Notice that the Fx component is parallel to the wrench, whereas Fy is perpindicular. As a result, the Fx part of the force does not contribute to the torque, because it by itself would not produce a rotation. However, the Fy part would definitely produce rotation. The point is, in order for a force or it's component to produce torque, it must be perpindicular to the lever arm. Therefore, the lever arm in this situation would still be same as the original.  (You don't extend the lever arm to where the Fy component is pointing in the picture because that is just a representation of Fy. Fy acts at the point of contact of the resultant force.) That is one way to determine torque of a force that is not in and of itself acting perpindicular to the lever arm. The other way requires a different approach, but produces the same result. Since only a force perpindicular to the lever arm produces a force, you could extend an imaginary line from the force and another line from the center of rotation such that they meet at a 90 degree angle.  Now you can just take the entire force and multiply by the new lever arm, and you will get the same torque. Honest. Let's consider an example to make sure. Say the angle between the force and wrench is 30 degrees, and the force is 50 Newtons. If the distance from the force to the center of rotation is 10 cm, the picture would become,  Notice how the angle of the force is equal to the angle inside the triangle. That is due to the opposite interior angle rule of geometry. As a result, you can easily find the other lever arm (the leg of the triangle opposite to the 30 degree angle), since you know the hypotenuse (the 10 cm given above). This is given by, Where L is the new lever arm. Since the sine of 30 is .5, L is 10 times .5 or 5 cm. Now, to calculate the torque using the Fy method, we need to find Fy, which is given by, Which gives 25 Newtons. Therefore, the torque would be 25 times 10 or 250 Nm.  Doing it the other way would require multiplying the resultant force, 50 N, and the new lever arm, 5 m. This indeed yields 250 Nm!  Newton's Second Law Since the Second Law says F = ma, all we need to do is substitute the angular counterparts for F, m, and a to derive the angular equivalent of the Second Law. The result gives,  Moment of inertia, I, replaces m, alpha replaces a, and torque replaces F. Even though torque is technically not just a force, it still replaces force in the equations, which is why it was mentioned earlier that torque is the angular equivalent of force. Let's consider the following example that ties this all together. A 3.0 kg grindstone 15 cm in radius is turning at 150 rad/s. When the motor is turned off, a chisel is used to help slow it down. If the chisel makes an angle of 45 degrees with respect to the grindstone, and the force is 5.0 N, how long will it take the grindstone to slow down? The picture below illustrates this problem.  Since we want to know how long it will take to make the stone stop we will use,  We know the final and initial angular velocities, and we know time, but we don't know alpha. Therefore, we must find alpha by using the angular Second Law equation. Finding the torque, we need to determine the component that is perpindicular to the axis of rotation. This would be the the Fx labeled in the picture above. To find Fx, we need to use the equation 5cos(45), which equals approximately 3.5 N. So, the torque is -3.5 times .15, which equals -.53 Nm. It is negative because it acts in the opposite direction of rotation. Since the grindstone is a disc, the moment of inertia is given by 1/2mr^2. Therefore, we have, As a result, alpha is found by, Now we use the omega equation and set the final omega to zero and get, Solving for t yields a time of 9.4 seconds. |