Work however are quite different quantities and must

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work, however, are quite different quantities and must not be confused. Work is often expressed in joules (1 J 1 N m),but torque never is. In the next chapter we shall discuss torque in a general way as being a vector quantity. Here, however, because we consider only rotation around a single axis, we do not need vector notation. Instead, a torque has either a positive or negative value depending on the direction of rotation it would give a body initially at rest: If the body would rotate counterclockwise, the torque is positive. If the object would rotate clockwise, the torque is negative. (The phrase “clocks are negative” from Section 10-2 still works.) Torques obey the superposition principle that we discussed in Chapter 5 for forces:When several torques act on a body, the net torque (or resultant torque ) is the sum of the individual torques.The symbol for net torque is t net . F : r : F : r F : F : r ( r sin )( F ) r F , F : r : F : r : F : F : CHECKPOINT 6 The figure shows an overhead view of a meter stick that can pivot about the dot at the position marked 20 (for 20 cm).All five forces on the stick are horizontal and have the same magnitude. Rank the forces according to the magnitude of the torque they pro- duce,greatest first. 0 20 40 P ivot point 100 F 1 F 2 F 3 F 4 F 5 Fig. 10-16 ( a ) A force acts on a rigid body, with a rotation axis perpendicular to the page.The torque can be found with ( a ) angle f , ( b ) tangential force compo- nent F t , or ( c ) moment arm . r F : O P φ ( a ) O P φ ( b ) F r F t O P φ Rotation axis ( c ) φ Line of action of F r Moment arm of F Rotation axis Rotation axis F F F r r r The torque due to this force causes rotation around this axis (which extends out toward you). You calculate the same torque by using this moment arm distance and the full force magnitude. But actually only the tangential component of the force causes the rotation.
260 CHAPTER 10 ROTATION 10-9 Newton’s Second Law for Rotation A torque can cause rotation of a rigid body, as when you use a torque to rotate a door. Here we want to relate the net torque t net on a rigid body to the angular acceleration a that torque causes about a rotation axis.We do so by analogy with Newton’s second law ( F net ma ) for the acceleration a of a body of mass m due to a net force F net along a coordinate axis.We replace F net with t net , m with I , and a with a in radian measure, writing t net I a (Newton’s second law for rotation). (10-42) Proof of Equation 10-42 We prove Eq. 10-42 by first considering the simple situation shown in Fig. 10-17. The rigid body there consists of a particle of mass m on one end of a massless rod of length r .The rod can move only by rotating about its other end, around a rota- tion axis (an axle) that is perpendicular to the plane of the page.Thus, the particle can move only in a circular path that has the rotation axis at its center. A force acts on the particle.However,because the particle can move only along the circular path,only the tangential component F t of the force (the component that is tangent to the circular path) can accelerate the particle along the path.We can relate F t to the particle’s tangential acceleration a t

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