lecture notes-Chapter8-posted

lecture notes-Chapter8-posted - 1 Chapter 8: Torque and...

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Unformatted text preview: 1 Chapter 8: Torque and Angular Momentum Rotational Kinetic Energy Rotational Inertia Torque Work Done by a Torque Equilibrium (revisited) Rotational Form of Newtons 2 nd Law Rolling Objects Angular Momentum 2 8.1 Rotational KE and Inertia For a rotating solid body: = = + + + = n i i i n n v m v m v m v m K 1 2 2 2 2 2 2 1 1 rot 2 1 2 1 2 1 2 1 For a rotating body v i = r i where r i is the distance from the rotation axis to the mass m i . ( 29 2 2 1 2 1 2 rot 2 1 2 1 2 1 I r m r m K n i i i n i i i = = = = = 3 The quantity = = n i i i r m I 1 2 is called rotational inertia or moment of inertia. Use the above expression when the number of masses that make up a body is small. Use the moments of inertia in table 8.1 for extended bodies. 4 Example: (a) Find the moment of inertia of the system below. The masses are m 1 and m 2 and they are separated by a distance r. Assume the rod connecting the masses is massless. r 1 r 2 m 1 m 2 r 1 and r 2 are the distances between mass 1 and the rotation axis and mass 2 and the rotation axis (the dashed, vertical line) respectively. 5 2 2 2 2 2 1 1 2 1 2 m kg 67 . = + = = = r m r m r m I i i i (b) What is the moment of inertia if the axis is moved so that is passes through m 1 ? Example continued: 2 2 2 2 2 1 2 m kg 00 . 1 = = = = r m r m I i i i Take m 1 = 2.00 kg, m 2 = 1.00 kg, r 1 = 0.33 m , and r 2 = 0.67 m. r 1 r 2 m 1 m 2 6 Example (text problem 8.4): What is the rotational inertia of a solid iron disk of mass 49.0 kg with a thickness of 5.00 cm and a radius of 20.0 cm, about an axis through its center and perpendicular to it? From table 8.1: ( 29 ( 29 2 2 2 m kg 98 . m 2 . kg . 49 2 1 2 1 = = = MR I 7 8.2 Torque A rotating (spinning) body will continue to rotate unless it is acted upon by a torque. hinge P u s h Q: Where on a door do you push to open it? A: Far from the hinge. 8 Torque method 1: = rF Hinge end F F || F r = the distance from the rotation axis (hinge) to the point where the force F is applied. F is the component of the force F that is perpendicular to the door (here it is Fsin ). Top view of door 9 The units of torque are Nm (not joules!). By convention: When the applied force causes the object to rotate counterclockwise (CCW) then is positive. When the applied force causes the object to rotate clockwise (CW) then is negative. 10 Torque method 2: F r = r is called the lever arm and F is the magnitude of the applied force. Lever arm is the perpendicular distance to the line of action of the force. 11 Line of action of the force Hinge end F r Top view of door Lever arm The torque is: sin sin r r r r = = sin rF F r = = Same as before 12 Example (text problem 8.12): The pull cord of a lawnmower engine is wound around a drum of radius 6.00 cm, while the cord is pulled with a force of 75.0 N to start the engine. cord is pulled with a force of 75....
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This note was uploaded on 12/23/2011 for the course PHY 2130 taught by Professor Rehse during the Fall '08 term at Wayne State University.

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lecture notes-Chapter8-posted - 1 Chapter 8: Torque and...

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