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82 Rotational Dynamics

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ow do you start the rotation of an object That is, how do you change its angular velocity Suppose you have a soup can that you want to spin If you wrap a string around it and pull hard, you could make the can spin rapidly Later in this chapter, you will learn why gravity, the force of Earth s mass on the can, acts on the center of the can The force of the string, on the other hand, is exerted at the outer edge of the can, and at right angles to the line from the center of the can, to the point where the string leaves the can s surface You have learned that a force changes the velocity of a point object In the case of a soup can, a force that is exerted in a very specific way changes the angular velocity of an extended object, which is an object that has a definite shape and size Consider how you open a door: you exert a force How can you exert the force to open the door most easily To get the most effect from the least force, you exert the force as far from the axis of rotation as possible, as shown in Figure 8-3 In this case, the axis of rotation is an imaginary vertical line through the hinges The doorknob is near the outer edge of the door You exert the force on the doorknob at right angles to the door, away from the hinges Thus, the magnitude of the force, the distance from the axis to the point where the force is exerted, and the direction of the force determine the change in angular velocity

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Describe torque and the factors that determine it Calculate net torque Calculate the moment of inertia

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lever arm torque moment of inertia Newton s second law for rotational motion

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Lever arm For a given applied force, the change in angular velocity depends on the lever arm, which is the perpendicular distance from the axis of rotation to the point where the force is exerted If the force is Figure 8-3 When opening a perpendicular to the radius of rotation, as it was with the soup can, then door that is free to rotate about the lever arm is the distance from the axis, r For the door, it is the distance its hinges, the greatest torque from the hinges to the point where you exert the force, as illustrated is produced when the force is in Figure 8-4a, on the next page If the force is not perpendicular, the applied farthest from the hinges perpendicular component of the force must be found (a), at an angle perpendicular The force exerted by the string around to the door (b) the can is perpendicular to the radius a No effect Little effect Maximum effect If a force is not exerted perpendicular to the radius, however, the lever arm is reduced To find the lever arm, extend the line of the force until it forms a right angle with a line from the center of rotation The distance between the intersection and the axis is the lever arm Thus, using trigonometry, the lever b arm, L, can be calculated by the equation L r sin , as shown in Figure 8-4b In this equation, r is the distance from the axis of rotation to the point where the force is exerted, and is the angle between the force and the radius from the axis of rotation to the point where No effect Some effect Maximum effect the force is applied

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Section 82 Rotational Dynamics

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Figure 8-4 The lever arm is along the width of the door, from the hinge to the point where the force is exerted (a) The lever arm is equal to r sin , when the angle, , between the force and the radius of rotation is not equal to 90 (b)

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