APC PHYSICS CHAPTER 11 Mr. Holl Rotation Student Notes 11-1 Translation and Rotation All of the motion we have studied to this point was linear or translational. Rotational motion is the study of spinning bodies such as wheels, gears, jet engines, helicopter blades, planets and spinning athletes such as figure skaters and gymnasts. 11-2 Rotational Variable In this chapter we deal with rigid bodies rotating around a fixed axis, called the axis of rotation. The diagram to the right shows two masses moving in rotational motion around a fixed point or axis of rotation. All of the formulas you have learned for translation have corresponding formulas that deal with rotation, and every variable has a corresponding rotational variable. In general, Greek letters are used to represent rotational quantities. 1) Angular Position - where an object is located angularly as measured from a reference line. The variable is è and is measured in radians. Remember, è = s/r where s is the translational distance (arc length) from the reference point and r is the radius of the path followed. One complete revolution is 2ð rad. è is not reset at zero each revolution, so if you have gone around the circle 3 times your location is 6ð rad. 2) Angular Displacement - As you move between two points your angular location changes just as your linear position changed in translational problems. Äè = è 2 - è1
3) Average Angular Velocity - The average rate of change in angular displacement. The symbol used is lowercase omega - ù. 4) Instantaneous Angular Velocity - rate of change of angular displacement at any instance, or averge ù as Ät approaches zero. The unit for both average and instantaneous angular velocity is radians/second, abbreviated rad/s. 5) Average Angular Acceleration - Average rate of change of angular velocity. 6) Instantaneous Angular Acceleration - Rate of change of angular velocity at any given instant. 2 The unit for angular acceleration is rad/s.
Example 11-1) The diagram to the right shows the overhead view of a child on a merry-go-round. If the ride is rotating at a constant 12 rev/min: a) What is its angular velocity in rad/s? b) What is the child s angular displacement in 4 sec? c) How far has the child actually traveled? Example 11-2) The angular position of a reference line on a spinning wheel is given by the equation: è = 3 t - 27t+ 4, where t is in seconds. a) Find ù(t) and á(t) d) At what times is the angular velocity zero? e) Describe the wheels motion when t 0.
3 Example 11-3) A child s top is spun with an angular acceleration of á = 5t - 4t where the coefficients are compatible with seconds and radians. At t= 0 the top has an angular velocity of 5 rad/s, and a position of è = 2 rad with respect to some reference line. a) Obtain an expression for the angular velocity ù(t) of the top. b) Obtain an expression for the angular position of the top, è(t). c) After 2 seconds find the angular acceleration, angular velocity and final angular displacement of the top. 11-3 Angular Quantities as Vectors When dealing with linear vectors we expressed directions as positive and negative, with angular quantities we use clockwise and counterclockwise about the axis. When describing rotational vectors we establish the direction of ù using a right-hand rule. When your fingers are aligned in the direction of the rotation, your thumb will point in the direction of the angular velocity vector. The vector defines the axis of rotation. ex) When looking down at a playing CD, what is the direction of the velocity vector? ex) In which direction is the object to the right rotating if the velocity vector is in the +x dir.?
11-4 Rotation with Constant Angular Acceleration Just as with translational motion we had a group of formulas which were useful as long as the acceleration was constant, the same group can be derived for angular motion. These are: Example 11-4) 2 A grindstone has a constant angular acceleration of á = 0.35 rad/s It starts from rest with an arbitrary horizontal line at è = 0. o a) What is the angular displacement è of the reference line at t= 18 s? b) What is the wheel s angular velocity at t= 18 s?
Example 11-5) Lets assume that the grindstone in 11-4 does not start at rest but has an initial speed of ù o = -4.6 rad/s. a) What will the grindstone be doing over the first few seconds? b) At what time will the grindstone be at rest? c) How many revolutions will the grindstone make before coming to rest? Example 11-6) During the analysis of a helicopter engine you determine that the rotor s velocity changes from 320 rev/min to 225 rev/min during the 1.5 minutes of observation. a) What is the average angular acceleration of the rotor? b) Assuming that the rotational acceleration remains constant, how long will it take the rotor to stop? c) How many revolutions will it make before coming to rest. from the original 320 rev/s?
11-5 Relating Angular and Linear Variables In the diagram to the right suppose that the two masses are traveling with a constant and equal angular velocity. This means that their relative positions will remain constant because each will pass through the same angular displacement è in the same time interval. But the linear distance traveled by the outer mass will be significantly larger than that of the inner mass due to the fact that it is traveling a long a circular path of larger radius! We can use the relationship between è and arc length to convert angular quantities into linear quantities. 1) Position or Distance 2) Speed è = s/r therefore s = èr where s is the actual distance traveled Differentiating the equation for distance we get but ds/dt is the linear speed, and dè/dt is the angular velocity so: v = ùr 3) Period To this point we have always found the period, the time it takes to make one revolution, using the formula T = 2ðr/v, but if v = ùr than r = v/ù. If we plug this back into the equation we get: 4) Acceleration a) Tangential (linear) - This is the rate at which the object is speeding up or slowing down. If we differentiate v = ùr with respect to time we get Since dv/dt = a and dù/dt =á we get: a = ár
b) Radial or Centripetal Acceleration - When an object is traveling along a circular path it also has a centripetal or radial (along the radius) component of acceleration. We know 2 that we can find this using a = v /r. But we also now know that v = ùr so: r This is the acceleration responsible for the change in direction rather than the change in speed. Example 11-7) The diagram to the right shows a centrifuge used to train astronauts to get use to high accelerations and g forces. The radius of the device is r =15 m. a) What constant angular velocity must the centrifuge have if the rider is to experience 11g? b) What is the tangential acceleration of the astronaut if the centrifuge goes from rest to this angular velocity in 2 minutes? 11-6 Moment of Inertia As we ve seen so far, there seem to be angular quantities that correspond with the linear quantities we dealt with in the first 10 chapters. This is also true when it comes to mass. We defined mass (inertial mass) as a measure of how difficult it was to accelerate the object. The greater the mass, the harder (more force needed) it is to accelerate the object. When dealing with rotation there is also a similar quantity. It is called rotational Inertia, or more often, moment of inertia. The greater the moment of inertia of an object, the harder it is to make it accelerate angularly (á).
The moment of inertia (I), is determined by not only the objects mass, but the location of the mass (r). The farther the mass is from the axis of rotation, the greater its moment of inertia. For a single point mass (like a ball at the end of a string, or a coin on the edge of a turntable) the moment of inertia is found using: 2 I = mr If several masses are located in the system then the moment of inertia is: Example 11-8) The diagram to the right shows a 5 kg mass attached to a massless rod with a length of 60 cm. The axis of rotation is at the other end of the rod. What is the moment of inertia of the system? Example 11-9) 3 children of masses shown are sitting on a massless merry-go-round. Determine the moment of inertia of the system. Calculating Moment of Inertia For any rigid body the moment of inertia is found using the formula on the previous page. If we choose an infinitely small unit of mass (again our differential mass), dm, we can replace the with an integral to get: This equation allows us to find the moment of inertia for any standard shape object.
Example 11-10) The small, thin rod shown to the right has a length of L and a mass of M. What is the moment of inertia of the rod is the axis of rotation is placed in the center of the rod. What is the rod were to be rotated around one end? Example 11-11) Find the moment of inertia for a disk with a mass of M, a thickness of L and a radius of r. Fortunately for us we do not have to run this integration every time. The work has already been done for most rigid solids and the results listed below. I will provide you with this list for all quizzes and tests and the AP gives you one too.
Rotational Inertia for Some Objects
The Parallel Axis Theorem If you know the rotational inertia of a body about any axis that passes through its center of mass, you can find its rotational inertia about any other axis parallel to that axis with the parallel axis theorem. Where M is the mass, I cm is the moment of inertia through the center of mass and h is the perpendicular distance to the new axis. Example 11-12) Using your information from example 11-10, find the moment of inertia of the rod if it is rotated ½ way between the center and either end. Example 11-13) In an exciting physics contest a race down a hill is set between a ring (hoop), a disk (cylinder), and a sphere all of equal mass and radius. Place your bets! Who will win the race???? 11-7 Rotational Kinetic Energy A rotating object, such as a lawn mower blade certainly has kinetic energy even though the blade as a whole has no translational velocity. If we look at any little piece of the blade it has a mass, and a speed. Add these all up and we should have the total kinetic energy of the blade due to its rotation. 2 2 2 K = ½m1v 1 + ½m2v 2 + ½m3v 3 +... or K = ½m v i = 1 to n i 2 i The problem is that the v i is not the same for all particle. We can get around this by expressing the velocity in angular terms. v = ùr therfore K = ½m (ùr ) 2 i 2 If we rearrange and move the r over with the mass and pull the ½ out of the sum we get K = ½ m r 2 ù 2 But we have just seen that m r = I so i i i 2 2 i i K = ½ Iù Example 11-14) The giant Ferris Wheel at Great Adventure has a mass of 12,000 kg and a diameter of 40 m. It rotates at a rate of 1.2 RPM. Assuming that the mass is concentrated on the outer edge (like a hoop) find: a) The moment of inertia of the wheel.
b) The angular velocity of the wheel. c) The rotational kinetic energy of the wheel. 11-8 Torque When trying to make an object rotate we must apply a torque. A torque is the product of a force, and the distance the force is applied from the axis of rotation. The greater the force or the greater the distance (sometimes called the lever arm) the greater the torque produced. The diagram tot he right shows an overhead view of a door. If we grab the outer edge pull perpendicular the torque created is ô = rf Suppose I now pull not perpendicularly, but at an angle ö from the direction of r. Now we must resolve F into two components, F r (radial component) and F t (tangential component). The radial component creates no torque. the torque is created by the tangential component. This means that the torque we create is ô = r F t. Looking at the diagram we can see that F t = F sinö, so torque can be written as: ô = r F sinö where ö is the angle between the direction of the lever arm and the direction of the force. Another way to visualize this is to extend the direction of the force and until you can find the perpendicular distance to the axis. This r is known as the moment arm and
torque can be written as ô = r F. As shown in the diagram to the right, r = r sinö so no matter which way you look at it the torque formula comes out the same. The definition of torque ô = r F sinö should look familiar! It is our definition of a cross product, therefore we can also write torque as ô = r x F, and its direction is given by the right hand rule! Example 11-15) The diagram to the right shows two forces applied to a wheel with a radius of 20 cm. a) Determine the magnitude and direction of the torque created by each force. b) What is the net torque acting on the wheel? 11-9 Newton s Second law For Rotation (Rotational Dynamics) Newton s Second Law states where the (t) stands for translational. If ô = rf t then we can rewrite the formula for torque as F =ma t t ô = ma r t but at = ár giving ô = m(ár)r or ô = mr 2 á 2 Since mr is the moment of inertia, I we can again rewrite the formula as or more correctly ô = Iá ô = Iá The sum of the torques acting on an object (net torque) is always equal to the moment of inertia times the angular acceleration.
Example 11-16) If the wheel in example 11-15 is taken to be a solid disc with a mass of 6 kg and a radius of 40 cm, what will be its angular acceleration? Example 11-17) The diagram to the right shows a uniform disk with a mass M = 2.5 kg and a radius of R = 20 cm. The disk is mounted on a fixed horizontal axis. A block with a mass of 1.2 kg hangs by a massless rope wrapped around the disk. Find: a) The downwards acceleration of the mass. b) The angular acceleration of the disk. c) The tension in the rope.
Example 11-18) The Atwood s machine to the right shows two masses attached with a string that passes over a pulley that has a mass of 2 kg and a radius of 20 cm and a frictional toque of 8.0 mn (WOW - We ve been waiting all of our physics lives for a pulley with mass and friction!!!!) Determine: a) The acceleration of the system. b) The tension in the rope attached to m 1. c) The tension in the rope attached to m 2. 11-10 Work and Rotational Kinetic Energy When we did work accelerating an object linearly we found the work by using; W = F d = Fd cosè when the force was constant, and when the force was changing; W = F ds (s = distance) (s) Taking this concept to rotation we know that s = rè therefore ds = drè = r dè giving W = F r dè (è) But Fr = ô so the work done accelerating an object angularly becomes For Power, average power was rate at which work was done
Instantaneous power was force times instantaneous velocity. When dealing with rotation, power is torque times instantaneous angular velocity. WORK - ENERGY THEOREM (ô = Iá) Example 11-19) A bicycle wheel (hoop) has a mass of 8 kg and a radius of 50 cm. It starts at rest and the pedaler exerts a torque of 3.2 Nm to the wheel for 10 s. a) What is the moment of inertia of the wheel? b) What is the angular acceleration of the wheel? c) What is the angular velocity of the wheel after the 10 s? d) What is the angular displacement of the wheel after the 10 s? e) How much work was done accelerating the wheel? f) What is the average power of the pedaler? G) If the accelerating torque is now removed, and the wheel comes to rest after 30 seconds, what is the average frictional torque acting on the wheel?