6 - Circular Motion and Other Applications of Newton's Laws

# Ball is following the dotted circular path shown in

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Unformatted text preview: re 6.3 under the inﬂuence of a force. At a certain instant of time, the force on the ball changes abruptly to a new force, and the ball follows the paths indicated by the solid line with an arrowhead in each of the four parts of the ﬁgure. For each part of the ﬁgure, describe the magnitude and direction of the force required to make the ball move in the solid path. If the dotted line represents the path of a ball being whirled on the end of a string, which path does the ball follow if the string breaks? Let us consider some examples of uniform circular motion. In each case, be sure to recognize the external force (or forces) that causes the body to move in its circular path. EXAMPLE 6.2 How Fast Can It Spin? A ball of mass 0.500 kg is attached to the end of a cord 1.50 m long. The ball is whirled in a horizontal circle as was shown in Figure 6.1. If the cord can withstand a maximum tension of 50.0 N, what is the maximum speed the ball can attain before the cord breaks? Assume that the string remains horizontal during the motion. Solution It is difﬁcult to know what might be a reasonable value for the answer. Nonetheless, we know that it cannot be too large, say 100 m/s, because a person cannot make a ball move so quickly. It makes sense that the stronger the cord, the faster the ball can twirl before the cord breaks. Also, we expect a more massive ball to break the cord at a lower speed. (Imagine whirling a bowling ball!) Because the force causing the centripetal acceleration in this case is the force T exerted by the cord on the ball, Equation 6.1 yields for Fr mar v2 Tm r EXAMPLE 6.3 Solving for v, we have v √ Tr m This shows that v increases with T and decreases with larger m, as we expect to see — for a given v, a large mass requires a large tension and a small mass needs only a small tension. The maximum speed the ball can have corresponds to the maximum tension. Hence, we ﬁnd vmax √ Tmaxr m √ (50.0 N)(1.50 m) 0.500 kg 12.2 m/s Exercise Calculate the tension in the cord if the speed of the ball is 5.00 m/s. Answer 8.33 N. The Conical Pendulum A small object of mass m is suspended from a string of length L . The object revolves with constant speed v in a horizontal circle of radius r, as shown in Figure 6.4. (Because the string sweeps out the surface of a cone, the system is known as a conical pendulum.) Find an expression for v. Solution Let us choose to represent the angle between string and vertical. In the free-body diagram shown in Figure 6.4, the force T exerted by the string is resolved into a vertical component T cos and a horizontal component T sin acting toward the center of revolution. Because the object does 6.1 not accelerate in the vertical direction, Fy may 0, and the upward vertical component of T must balance the downward force of gravity. Therefore, (1) T cos 155 Newton’s Second Law Applied to Uniform Circular Motion Because the force providing the centripetal acceleration in this example is the component T si...
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