 |
|

|
 |

10/21/09 - Rotational Motion
|
 |
 |
 |
  | Arc length, radius and angle in radians
|
 |
 |
 |
 |
  | We are all used to measuring angles in degrees and we know their are 360° in a circle.
|
 |
 |
 |
 |
  | Did you ever wonder why we divide the circle into 360 parts? It comes from the fact that there are 365 days in a year. So, one degree is about how much the earth moves daily in its orbit around the sun.
|
 |
 |
 |
 |
  | At some point it was decided that we should measure angles in some unit that did not depend specifically on the earth or the sun, but just depended on the properties of a circle.
|
 |
 |
 |
 |
  | You learned in trigonometry that angles can also be measured in radians by drawing the angle from the center of a circle to the edge of the circle and dividing the length of the arc formed by the sides of the angle by the radius of the circle: Angle in radians = (arc length)/radius or θ = S/R. If the arc length S equals the radius R, then θ = R/R = 1 radian. If the arc length is the whole circumference of the circle, then θ = (2πR)/R = 2π radians, so there are 2π radians in a circle. Hence 2π radians = 360° or π radians = 180°.
|
 |
 |
 |
 |
  | We can convert an angle from degrees to radians by multiplying it by π/180°.
|
 |
 |
 |
 |
  | We can convert an angle from radians to degrees by multiplying it by 180°/π.
|
 |
 |
 |
 |
  | For many purposes it's simpler to work in radians than in degrees.
|
 |
 |
 |
 |
  | For example, if I know a laser beam is spreading out with an angle of 1 milliradian = 0.001 radian, I can estimate the width of the beam at any distance simply by multiplying the angle by the distance. In other words, S = Rθ. So, at a distance of 100 m, the length of a soccer field, the beam would have a width of S = (100m)(0.001) = 0.1 m. At a distance of a mile, the beam would have a width of S = (1 mile)(0.001) = 0.001 mile ≈ 5 feet.
|
 |
 |
 |
 |
  | Tangential velocity, radius and angular frequency in radians/second
|
 |
 |
 |
 |
  | An object moving in a circular path has an angular position which is constantly changing in time. The rate of change of the angular position with time is related by the circle's radius to the speed with which the object is moving on the edge of the circle. In other words, if S = Rθ, then ∆S/∆t = R(∆θ/∆t) or v = Rω. The rate at which the object moves along the arc is the tangential velocity v (usually measured in meters/second), and the rate at which the angle changes is called angular velocity (usually measured in radians/second) and is represented by the Greek letter omega ω.
|
 |
 |
 |
 |
  | Using this formula we can easily find how fast (in meters/second) the earth is moving in its orbit around the sun. The angular velocity of the earth around the sun is (2π radians)/(seconds in a year) = (2π)/(3.16x107) ≈ 2 x 10-7 radians/second. The distance R of the earth from the sun is 1.496 x 1011 m, so v = Rω = (1.496 x 1011m)(2 x 10-7radians/second) ≈ 3 x 104 m/s = 30,000 m/s ≈ 60,000 miles/hour.
|
 |
 |
 |
 |
  | Using the same formula, we can find the angular velocity ω if we know v and R. If we know a car is going around a curve of radius R = 100 m at a speed of 20 m/s, then its angular velocity is ω = v/R = (20m/s)/(100m) = 0.20 radians/s.
|
 |
 |
 |
 |
  | Tangential acceleration, radius and angular acceleration in radians/second2.
|
 |
 |
 |
 |
  | If an object is changing its speed as it goes around a circle, then it has angular acceleration. Since a = ∆v/∆t we can write a = ∆v/∆t = R(∆ω/∆t) = Rα, where the rate of change of angular velocity ∆ω/∆t is denoted by the Greek letter alpha α.
|
 |
 |
 |
 |
  | Suppose an object has an angular acceleration of α = 0.2 radians/s2 as it goes around a circle of radius R = 10m. What is its tangential acceleration in m/s2? a = Rα = (10m)(0.2radians/s2) = 2 m/s2, meaning it changes its speed by 2m/s every second.
|
 |
 |
 |
 |
  | Suppose a pilot accelerates at 5 m/s2 as he flies around a circular arc of radius 1000 m? What is his angular acceleration? α = a/R = (5m/s2)/(1000m) = 0.005 radians/s2.
|
 |
 |
 |
 |
  | For each of the rotational quantities there is a linear analog:
|
 |
 |
 |
 |
  | Angle θ in radians is analogous to displacement D in meters.
|
 |
 |
 |
 |
  | Angular velocity ω in radians/second is analogous to linear velocity in meters/second.
|
 |
 |
 |
 |
  | Angular acceleration α in radians/second/second is analogous to linear acceleration in meters/second/second.
|
 |
 |
 |
 |
  | This means that just as we have the following relationships among the linear quantities:
|
 |
 |
 |
 |
  | The slope of the displacement-time graph is the linear velocity.
|
 |
 |
 |
 |
  | The slope of the velocity-time graph is the linear acceleration.
|
 |
 |
 |
 |
  | The area under the linear acceleration-time graph is the change in velocity.
|
 |
 |
 |
 |
  | The area under the velocity-time graph is the change in displacement in meters.
|
 |
 |
 |
 |
  | We similarly have the following relationships among the rotational quantities:
|
 |
 |
 |
 |
  | The slope of the angle-in-radians-time graph is the angular velocity.
|
 |
 |
 |
 |
  | The slope of the angular-velocity-time graph is the angular acceleration.
|
 |
 |
 |
 |
  | The area under the angular-acceleration-time graph is the change in angular velocity.
|
 |
 |
 |
 |
  | The area under the angular-velocity-time graph is the change in angle in radians.
|
 |
 |
 |
 |
  | A force that acts at some distance from a point of rotation produces a torque. The magnitude of the torque depends on both the magnitude of the force and the distance from the point of rotation. Torque τ = R Fperpendicular.
|
 |
 |
 |
 |
  | We can increase the torque either by increasing the force F or by increasing the distance from the point of rotation or both.
|
 |
 |
 |
 |
  | R is referred to as the "moment arm".
|
 |
 |
 |
 |
  | Only the component of the force at right angles to the moment arm causes rotation, so if the angle between the moment arm and the force is θ, we can say that the torque τ = R F sin(θ). In the diagram below, the angle between the moment arm and the force is 90°, so the force produces the maximum torque.
|
 |
 |
 |
 |
  | In the drawing below, the angle between the moment arm and the force is 45°, so only the horizontal component of the force (F cos 45°) produces a torque.
|
 |
 |
 |
 |
  | Torque is the rotational analog of force.
|
 |
 |
 |
 |
  | That means that just as an unbalanced force produces a linear acceleration, an unbalanced torque produces a rotational acceleration.
|
 |
 |
 |
 |
  | For an unbalanced force, the linear acceleration is inversely proportional to the mass:
|
 |
 |
 |
 |
  | For an unbalanced torque, the rotational acceleration is inversely proportional to the moment of inertia:
|
 |
 |
 |
 |
  | α = torque/(moment of inertia) = τ/I
|
 |
 |
 |
 |
  | The moment of inertia I of an object is a combination of the mass of an object and the way it's distributed. Experimentally it's been found that mass close to the point of rotation is easier to accelerate than mass farther away from the point of rotation. For example, rotating a baseball bat from the small end is harder than rotating a baseball bat from the large end. The moment of inertia is actually the sum (or integral) of a bunch of slices of a mass times the distance squared they are from the point of rotation: I = Σ r2∆m.
|
 |
 |
 |
 |
  | Our textbook has a table of moments of inertia for different common distributions of mass on p. 262.
|
 |
 |
 |
 |
  | Read pp. 240-260 of the text.
|
 |
 |
|


 |
 |
 |