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Example 8-12 A Simple Pulley I A pulley is a solid uniform cylinder of mass Mpulley and radius R that is free to rotate around an axis through its center. A lightweight rope is wound around the pulley. You exert a constant force of magnitude F on the rope, which makes the rope unwind and rotates the pulley (Figure 8-25). The rope does not stretch and does not slip on the pulley. What is the angular acceleration of the pulley about its axle? Pulley Lightweight rope Figure 8-25 Exerting a torque on a pulley What is the angular acceleration of the pulley as the F force F makes the rope unwind? Set Up We begin by drawing a free-body diagram for the pulley, taking care to draw each force at the point where it acts. We are told that the rope is lightweight (that is, it has much less mass than the pulley), so the force that you exert on the free end of the rope has the same magnitude F as the force that the rope exerts on the pulley. This force exerts a torque on the pulley and causes an angular acceleration. We’ll use the rotational form of Newton’s second law to determine this angular acceleration. Solve The free-body diagram shows that only the tension force F exerts a torque on the pulley causing it to rotate around its axle. (The support force Fsupport and the weight of the pulley Wpulley both act at its rotation axis, so the lever arm is zero for both of these forces.) The lever arm for the tension force is R, so the tension torque is R multiplied by F. From Table 8-1 in Section 8-3, the moment of inertia of a solid cylinder around its central axis is I = MR2 >2. Insert this into Newton’s second law for rotational motion and solve for apulley, z. Reflect Newton’s second law for rotational motion: a tz = Ipulley apulley, z (8-20) Fsupport wpulley Torque due to tension force: tz = r›F = RF F lever arm R (This torque makes the pulley rotate in the clockwise direction, so we take clockwise to be the positive rotation direction.) This is the only torque acting on the pulley, so a tz = RF a tz = Ipulley apulley,z rotation axis F Substitute values of a tz and Ipulley = MpulleyR2 >2: 1 RF = Mpulley R2apulley,z 2 2RF 2F Solve for angular acceleration az: apulley,z = = Mpulley R Mpulley R2 Our result for the angular acceleration apulley,z depends on the pulling force F, the pulley mass Mpulley, and the pulley radius R. It makes sense that our result is proportional to the ratio F>Mpulley: A greater force F means a stronger pull and a greater angular acceleration, while a greater mass Mpulley means the pulley is more difficult to rotate and gives a smaller angular acceleration. Our result also shows that the larger the pulley radius R, the smaller the angular acceleration apulley,z. This may seem backwards, since a larger radius means that the force F causes a larger torque t = RF. However, increasing the radius increases the moment of inertia I = Mpulley R2 >2 by a larger factor than it increases the torque. (Doubling the radius doubles the torque but quadruples the moment of inertia.) So the moment of inertia plays a more important role, which is why apulley,z decreases with increasing pulley radius.