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Chapter 8

Chapter 8. Rotational Motion. Introduction. Introduction. Until now, objects have been treated as point particles The point particle treatment is the correct way to describe translational motion The point particle approximation does not allow us to analyze rotational motion

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Chapter 8

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  1. Chapter 8 Rotational Motion

  2. Introduction Introduction • Until now, objects have been treated as point particles • The point particle treatment is the correct way to describe translational motion • The point particle approximation does not allow us to analyze rotational motion • Need to learn how to use Newton’s Laws to describe and analyze rotational motion • The size and shape of the object will have to be taken into account

  3. Describing Rotational Motion Section 8.1 • To deal with rotational motion, rotational quantities need to be defined • Angular position • Angular velocity • Angular acceleration • These will be analogous to the translational (linear) quantities of position, velocity and acceleration

  4. Coordinate System • Need to identify the rotation axis • In A, the rod is fixed at one end and the rotational axis is the z- axis • The angle θ is measured with respect to the x-axis • For the CD in B, the z-axis is the axis of rotation • The angular position, θ, is specified by the angle the reference line makes with the x-axis Section 8.1

  5. Radian • The end of the rod sweeps out a circle of radius r • Assume the end of the rod travels a distance s along the circular path • At the same time, the rod sweeps out an angle θ Section 8.1

  6. Radian, cont. Section 8.1 • The distance s and angle θ are related by • θ is measured in radians • Angles can also be measured in degrees • 360°= 2 π rad • Both measure one complete circle

  7. Angular Velocity Section 8.1 • Angular velocity describes how the angular position is changing with time • Denoted by ω • For some time interval, Δt, the average angular velocity is

  8. Instantaneous Angular Velocity Section 8.1 • The instantaneous angular velocity is • The instantaneous angular velocity equals the average angular velocity when ω is constant • SI unit is rad/s • May also see rpm (revolutions / minute)

  9. Angular Velocity, Direction • Since angular velocity is a vector quantity, it must have a direction • If θ increases with time, then ω is positive • Therefore, a counterclockwise rotation corresponds to a positive angular velocity • Clockwise would be negative Section 8.1

  10. Angular Acceleration Section 8.1 • Angular acceleration is the rate of change of the angular velocity • Denoted by α • The average angular acceleration is • The instantaneous angular acceleration is • SI units is rad/s² • The instantaneous angular acceleration equals the average angular acceleration when α is constant

  11. Angular Acceleration and Centripetal Acceleration Section 8.1 • Angular acceleration and centripetal acceleration are different • As an example, assume a particle is moving in a circle with a constant linear velocity • The particle’s angular position increases at a constant rate, therefore its angular velocity is constant • Its angular acceleration is 0 • Since it is moving in a circle, it experiences a centripetal acceleration of ac = v2/ r • This is not zero, even though the angular acceleration is zero • The centripetal acceleration refers to the linear motion of the particle • The angular acceleration is concerned with the related angular motion

  12. Angular and Linear Velocities Compared • When an object is rotating, all the points on the object have the same angular velocity • Makes ωa useful quantity for describing the motion • The linear velocity is not the same for all points • It depends on the distance from the rotational axis Section 8.1

  13. Period of Rotational Motion Section 8.1 • One revolution of an object corresponds to 2 π radians • The object will move through ω / 2π complete revolutions each second • The time required to complete one revolution is the period of the motion • Denoted by T with

  14. Connection Between Linear and Angular Velocities Section 8.1 • The linear velocity of any point on a rotating object is related to its angular velocity by v = ωr • r is the distance from the rotational axis to the point • Technically, this is the relationship between the speeds, since direction has not been taken into account • When a point is farther from the axis of rotation (rA > rB , for example), then its linear velocity will be greater • vA > vB • The angular velocities of both points are the same

  15. Connection Between Linear and Angular Accelerations Section 8.1 The relationship between linear and angular velocities can be used to determine the relationship between linear and angular accelerations Analysis indicates the angular acceleration and the linear acceleration of a point a distance r from the rotation axis is given by a = α r

  16. Torque and Newton’s Laws for Rotational Motion Section 8.2 A connection between force and rotational motion is needed Specifically, how forces give rise to angular accelerations The approach will be similar to looking at forces and linear motion

  17. Torque • Torque is the product of an applied force and the distance it is applied from the support point • Denoted τ • The point P is called the pivot point • Since the object can rotate around that point Section 8.2

  18. Lever Arm and Torque Section 8.2 • The lever arm is the distance between the pivot point and where the force acts • When the force is perpendicular to a line connecting its point of application to the pivot point, the torque is given by τ = F r • Torque in rotational motion is analogous to force in translational motion • Torque is the product of the force and the distance to the pivot point • SI unit is N. m

  19. Torque and Angular Acceleration • Only forces with a component perpendicular to the rod can contribute to the angular acceleration • Newton’s Second Law for rotational motion is written as Στ = I α • I is called the moment of inertia of the object Section 8.2

  20. Newton’s Second Law – Rotational Section 8.2 • The Στ is the vector sum of all the torques acting on the object • The moment of inertia, I, plays the same role that mass did in translational motion • For a ball and massless hinged rod, I = m r2

  21. Analogy with Translational Motion Section 8.2 • Torque plays the role of force in rotational motion • Torque depends on the magnitude of the force and where the force is applied relative to the pivot point • There may be multiple forces acting on the system, all involving a single pivot point

  22. Analogy with Translational Motion, 2 Section 8.2 • The motion of inertia enters into rotational motion in the same way that mass enters into translational motion • For an object composed of many pieces of mass located at various distances from the pivot point, the moment of inertia of the object is • The moment of inertia depends on the mass and on how that mass is distributed relative to the axis of rotation

  23. Analogy with Translational Motion, 3 Section 8.2 Newton’s Second Law for translational motion can be used to derive Newton’s Second Law for rotational motion For rotational motion, Newton’s Second Law becomes Στ = I α

  24. Linear and Rotational Comparison Section 8.2

  25. Torque and Lever Arm: Generalized • In general, the force that produces a torque does not have to be applied in a perpendicular direction • Assume the force acts at an angle ϕ with respect to the rod • Only the perpendicular component contributes to the torque Section 8.2

  26. Torque and Lever Arm: Generalized, cont. Section 8.2 • The perpendicular component of the applied force is Fapplied sin ϕ • The torque is Fapplied r sin ϕ • This is a general definition of torque • It can be used when the force is not directed perpendicular to the lever arm • When ϕis 90°, sin ϕ = 1 and Fapplied = F| • When ϕ is 0°, sin ϕ = 0 and Fapplied = 0 • A force applied parallel to the lever arm cannot cause an object to rotate

  27. One Way to Think About Torque • Since the perpendicular component of the force is Fapplied sin ϕ, the torque can be expressed as Fapplied r sin ϕ • The angle can be found by extending the radius line beyond the point where the force acts • The angle is between the force and this radius line Section 8.2

  28. Another Way to Think About Torque • You can also use the perpendicular distance from the pivot point to where the force is acting to calculate the torque • rperpendicular = r sin ϕ • This gives the general expression for the lever arm • Therefore, τ = Fapplied r sin ϕ Section 8.2

  29. Ways to Think About Torques, Final • Using the idea of the perpendicular force or the perpendicular distance (lever arm) will give the same results • For example: • ϕ = 90° gives maximum torque • ϕ = 0° give zero torque Section 8.2

  30. Torque and Direction – Mass at End • For a single rotation axis, the direction of the torque is specified by its sign • A positive torque is one that would produce a counterclockwise rotation • A negative torque would produce a clockwise rotation Section 8.2

  31. Torque and Direction – Distributed Mass • We could imagine breaking the clock hand up into many infinitesimally small pieces and finding the torques on each piece • A more convenient approach is to use the center of gravity of the hand Section 8.2

  32. Center of Gravity Section 8.2 • For the purposes of calculating the torque due to the gravitational force, you can assume all the force acts at a single location • The location is called the center of gravity of the object • The center of gravity and the center of mass of an object are usually the same point

  33. Rotational Equilibrium Section 8.3 Equilibrium may include rotational equilibrium An object can be in equilibrium with regard to both its translation and its rotational motion Its linear acceleration must be zero and its angular acceleration must be zero The total force being zero is not sufficient to ensure both accelerations are zero

  34. Example: Equilibrium • The applied forces are equal in magnitude, but opposite in direction • Therefore, ΣF = 0 • However, the object is not in equilibrium • The forces produce a net torque on the object • There will be an angular acceleration in the clockwise direction Section 8.3

  35. Rotational Equilibrium, cont. Section 8.3 • For an object to be in complete equilibrium, the angular acceleration is required to be zero • Στ = 0 • This is a necessary condition for rotational equilibrium • All the torques will be considered to refer to a single axis of rotation • The same ideas can also be applied to multiple axes

  36. Problem Solving Strategy – Equilibrium Section 8.3 • Recognize the principle • The object is in translational static equilibrium if its linear acceleration and linear velocity are both zero • The object is in rotational static equilibrium if its angular acceleration and angular velocity are both zero • Sketch the problem • Show the object of interest along with all the forces that act on it • Include a set of coordinate axes

  37. Problem Solving Strategy, 2 Section 8.3 • Identify the relationships • Find the rotation axis and the pivot point • Calculate the torque from each force • Determine the lever arm for each force • Calculate the magnitude of the force using τ = F r sin ϕ • Determine the sign of the torque • If the force acting alone would produce a counterclockwise rotation, the torque is positive • If the force acting alone would produce a clockwise rotation, the torque is negative • Add the torques from each force to get the total force • Be sure to include the sign of each torque

  38. Problem Solving Strategy, 3 Section 8.3 • Solve • Solve for the unknowns by applying the condition for rotational equilibrium • Στ = 0 • If necessary, apply the condition for translational equilibrium • ΣF = 0 • Check • Consider what your answer means • Check that your answer makes sense

  39. Problem Solving Reminders Section 8.3 • There are some new considerations to remember • Draw the entire object to show where the forces are acting on it • You can no longer draw the object as a point • Decide where to put the pivot point • There is often a natural choice for the pivot point • Sometimes there can be more than one plausible choice • For an object in rotational equilibrium, any spot may be chosen to be the pivot point without affecting the final answer • To simplify the equations, remember that forces whose lever arms are zero will not contribute to the torque

  40. Rotational Equilibrium Example: Lever • Use rotational equilibrium to find the force needed to just lift the rock • We can assume that the acceleration and the angular acceleration are zero • Also ignore the mass of the lever • mlever << mrock Section 8.3

  41. Lever Example, cont. Section 8.3 • Three forces are present • The force from the rock on the lever • The force the person applied to the lever • The force the of support on the lever • Choosing this point to be the pivot point produces a zero torque from this force • The force exerted by the person can be less than the weight of the rock • If Lperson > Lrock • The lever will amplify the force exerted by the person

  42. Amplification of Forces in the Ear • The incus is supported by a hinge that acts like a lever • The ratio of the lever arms is about 3 • The lever amplifies the forces associated with a sound vibration by the same factor Section 8.3

  43. Example: Tipping a Crate • We can calculate the force that will just cause the crate to tip • When on the verge of tipping, static equilibrium applies • If the person can exert about half the weight of the crate, it will tip Section 8.3

  44. Moment of Inertia Section 8.4 The moment of inertia of an object composed of many pieces of mass is The moment of inertia of an object depends on its mass and on how this mass is distributed with respect to the rotation axis The definition can be applied to find the moment of inertia of various objects for any rotational axis SI unit of moment of inertia is kg ·m2

  45. Moment of Inertia, cont. • The value of I depends on the choice of rotation axis • In the two examples, m and L are the same • Their moments of inertia are different due to the difference in rotation axes Section 8.4

  46. Various Moments of Inertia Section 8.4

  47. Rotational Dynamics Section 8.5 Newton’s Second Law for a rotating system states Στ = Iα Once the total torque and moment of inertia are found, the angular acceleration can be calculated Then rotational motion equations can be applied For constant angular acceleration:

  48. Kinematic Relationships Section 8.5

  49. Example: Real Pulley with Mass • Up to now, we have assumed a massless pulley • Using rotational dynamics, we can deal with real pulleys • The torque on the pulley is due to the tension in the rope • Apply Newton’s Second Laws for translational motion and for rotational motion Section 8.5

  50. Problem Solving Strategy – Rotational Dynamics Section 8.5 • Recognize the principle • Find the total torque • Find the moment of inertia • Use Newton’s Second Law to find the angular acceleration • Sketch the problem • Show all the objects of interest • Include all the forces that act on the objects • Include coordinate axes for translational motion

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