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Chapter 9: Torque and Rotation PowerPoint Presentation
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Chapter 9: Torque and Rotation

Chapter 9: Torque and Rotation

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Chapter 9: Torque and Rotation

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  1. Chapter 9: Torque and Rotation 9.1 Torque 9.2 Center of Mass 9.3 Rotational Inertia

  2. Chapter 9 Objectives Calculate the torque created by a force. Solve problems by balancing two torques in rotational equilibrium. Define the center of mass of an object. Describe a technique for finding the center of mass of an irregularly shaped object. Calculate the moment of inertia for a mass rotating on the end of a rod. Describe the relationship between torque, angular acceleration, and rotational inertia.

  3. Chapter 9 Vocabulary • angular acceleration • center of gravity • center of mass • center of rotation • lever arm • line of action • moment of inertia • rotational • equilibrium • rotational inertia • torque • translation

  4. Inv 9.1 Torque Investigation Key Question: How does force create rotation?

  5. 9.1 Torque Motion in which an entire object moves is called translation. Motion in which an object spins is calledrotation. An object can both rotate and translate.

  6. 9.1 Torque A torque is an action resulting from a force that causes objects to rotate. Torque is notthe same thing as force. Torque is created by force, but torque also depends on the point of application and the direction of the force. A door pushed at its handle will easily turn and open, but a door pushed near its hinges will not move as easily.

  7. 9.1 Torque The point or line about which an object turns is its center of rotation. A force applied far from the center of rotation produces a greater torque than a force applied close to the center of rotation. Doorknobs are positioned far from the hinges to provide the greatest amount of torque for a given force.

  8. 9.1 Torque Torque is created when the line of action of a force does not pass through the center of rotation. The line of action is an imaginary line that follows the direction of a force and passes though its point of application.

  9. 9.1 Torque To get the maximum torque, the force should be applied in a direction that creates the greatest lever arm. The lever arm is the perpendicular distance between the line of action of the force and the center of rotation

  10. 9.1 The torque created by force • The torque created by a force is equal to the lever arm length times the magnitude of the force. • Torque is usually represented by the lower case Greek letter “tau,” t. • When calculating torque, if the line of action passes through the center of rotation, the lever arm is zero, and so is the torque, no matter how large a force is applied.

  11. 9.1 Calculating Torque Lever arm length (m) t = r x F Torque (N.m) Force (N)

  12. 9.1 Calculating torque • If more than one torque acts on an object, the torques are combined to determine the net torque. • If the torques make an object spin in the same direction, clockwise or counterclockwise, they are added together.

  13. 9.1 Units of torque • The units of torque are force times distance, or newton-meters. • A torque of 1 N⋅m is created by a force of 1 newton acting with a lever arm of 1 meter. • Torque is always calculated around a particular center of rotation.

  14. Calculating a torque A force of 50 newtons is applied to a wrench that is 30 centimeters long. Calculate the torque if the force is applied perpendicular to the wrench so the lever arm is 30 cm. • You are asked to find the torque. • You are given the force and lever arm. • Use: t = r F. • 4. Solve: t = (-50 N)(0.3 m) = -15 N⋅m

  15. 9.1 Rotational Equilibrium When an object is in rotational equilibrium, the net torque applied to it is zero. Rotational equilibrium is often used to determine unknown forces. What are the forces (FA, FB) holding the bridge up at either end?

  16. 9.1 Rotational Equilibrium • For the bridge not to move up or down, the total upward force must equal the total downward force. • This means FA + FB= 1,250 N.

  17. 9.1 Rotational Equilibrium • If we choose the left end of the bridge, and set the total of the remaining torques to zero, the force on the right support (FB) is calculated to be 400 newtons. • Since the total of both forces must be 1,250 N, that means the force FA on the left must be 850 N. • For the bridge to be in rotational equilibrium, the total torque around any pointmust also be zero.

  18. Solve a rotational equilibriumproblem • You are asked to find the boy’s lever arm. • You are given the two masses and the cat’s lever arm. • Use: t = r F, and F = mg • Solve for the cat: t = (2 m)(4 kg)(9.8 N/kg) = +78.4 N⋅m • Solve for the boy: t = (d )(50 kg)(9.8 N/kg) = -490 d • For rotational equilibrium, the net torque must be zero. • 0= 78.4 – 490d d = 0.16 m. The boy must sit 16 cm from center. A boy and his cat sit on a seesaw. The cat has a mass of 4 kg and sits 2 m from the center of rotation. If the boy has a mass of 50 kg, where should he sit so that the see-saw will balance?

  19. 9.1 When the force and lever arm are NOT perpendicular • When the force and lever arm are not perpendicular, an extra step is required to calculate the length of the lever arm.

  20. Calculating a torque from an angled force • You are asked to find the force. • You are given the force and lever arm for one condition. • Use:  = r F. • Solve for the torque required to loosen the bolt •  = (50 N)(0.2 m) = 10 N⋅m • Calculate the torque with a force applied at 30 degrees: • 10 N⋅m = F × (0.2 m) cos 30o = 0.173 F • F = 10 N⋅m ÷ 0.173 = 58 N. It takes a larger force at 30o. A 20-centimeter wrench is used to loosen a bolt. The force is applied 0.20 m from the bolt. It takes 50 newtons to loosen the bolt when the force is applied perpendicular to the wrench. How much force would it take if the force was applied at a 30-degree angle from perpendicular?

  21. Chapter 9: Torque and Rotation 9.1 Torque 9.2 Center of Mass 9.3 Rotational Inertia

  22. Inv 9.2 Center of Mass Investigation Key Question: How do objects balance?

  23. 9.2 Center of Mass There are three different axes about which an object will naturally spin. The point at which the three axes intersect is called the center of mass.

  24. 9.2 Finding the center of mass If an object is irregularly shaped, the center of mass can be found by spinning the object and finding the intersection of the three spin axes. There is not always material at an object’s center of mass.

  25. 9.2 Finding the center of gravity The center of gravity of an irregularly shaped object can be found by suspending it from two or more points. For very tall objects, such as skyscrapers, the acceleration due to gravity may be slightly different at points throughout the object.

  26. 9.2 Balance and the center of mass • Objects balance because the torque caused by the force of the object’s weight is equal on each side.

  27. 9.2 Balance and the center of mass For an object to remain upright, its center of gravity must be above its area of support. The area of support includes the entire region surrounded by the actual supports. An object will topple over if its center of mass is not above its area of support.

  28. Chapter 9: Torque and Rotation 9.1 Torque 9.2 Center of Mass 9.3 Rotational Inertia

  29. Inv 9.3 Rotational Inertia Investigation Key Question: Does mass resist rotation the way it resists acceleration?

  30. 9.3 Rotational Inertia Previously inertia was defined as an object’s resistance to a change in its motion. Rotational inertia is the term used to describe an object’s resistance to a change in its rotational motion. An object’s rotational inertia depends not only on the total mass, but also on the way mass is distributed.

  31. 9.3 Rotational Inertia A rotating mass on a rod can be described with variables from linear or rotational motion.

  32. 9.3 Rotational Inertia To equate rotational motion to linear motion, the force is replaced by the torque about the center of rotation. The linear acceleration is replaced by the angular acceleration.

  33. 9.3 Linear and Angular Acceleration Angular acceleration (kg) a = a r Linear acceleration (m/sec2) Radius of motion (m)

  34. 9.3 Rotational Inertia The product of mass × radius squared (mr2) is the rotational inertia for a point mass where r is measured from the axis of rotation.

  35. 9.3 Moment of Inertia The sum of mr2 for all the particles of mass in a solid is called the moment of inertia (I). A solid object contains mass distributed at different distances from the center of rotation. Because rotational inertia depends on the square of the radius, the distribution of mass makes a big difference for solid objects.

  36. 9.3 Moment of Inertia

  37. 9.3 Rotation and Newton's 2nd Law If you apply a torque to a wheel, it will spin in the direction of the torque. The greater the torque, the greater the angular acceleration.

  38. 9.3 Rotation and Newton's 2nd Law • Force causes linear acceleration, torque causes angular acceleration. • Mass is a measure of linear inertia. • The moment of inertia (I ) is a measure of rotational inertia.

  39. Bicycle Physics • A modern bicycle is the most efficient machine ever invented for turning human muscle power into motion. • Bicycles work by a series of transformations from forces to torques and back.