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lecture notes-Chapter8-posted

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

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1 Chapter 8: Torque and Angular Momentum Rotational Kinetic Energy Rotational Inertia Torque Work Done by a Torque Equilibrium (revisited) Rotational Form of Newton’s 2 nd Law Rolling Objects Angular Momentum
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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 = ϖ 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 ϖ ϖ ϖ I r m r m K n i i i n i i i = = = = =
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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.
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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. ϖ 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.
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5 2 2 2 2 2 1 1 2 1 2 m kg 67 . 0 = + = = = 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. ϖ r 1 r 2 m 1 m 2
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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 . 0 m 2 . 0 kg 0 . 49 2 1 2 1 = = = MR I
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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.
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8 Torque method 1: = rF τ Hinge end F F || F θ 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 θ ). Top view of door
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9 The units of torque are Nm (not joules!). By convention: When the applied force causes the object to rotate counterclockwise (CCW) then τ is positive. When the applied force causes the object to rotate clockwise (CW) then τ is negative.
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10 Torque method 2: F r = τ 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.
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11 Line of action of the force θ Hinge end F θ r Top view of door Lever arm The torque is: θ θ sin sin r r r r = = θ τ sin rF F r = = Same as before
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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. What magnitude torque does the cord apply to the drum?
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