CT1: When I whirl a ball in a vertical circle attached to a rubber band, which statement is true? A. The rubber band will contract to provide an outward force on the nurf ball. B. The rubber band will contract because of the inward force on the nurf ball. C. The rubber band will not change in length. D. The rubber band will stretch because of the outward force on the nurf ball. E. The rubber band will stretch to provide an inward force on the nurf ball.
A. B. C. D. CT2
Chapter 6 Circular Motion and Other Applications of Newton’s Laws 6.1 Newton’s Second Law for a Particle in Uniform Circular Motion a r = v t 2 /r F r = ma r = mv t 2 /r taking inward as positive Note: a r = a c = centripetal acceleration a t =0, but because it is to a r, it has no bearing on the radial direction
P6.1 (p.155)
A.E.F.B.C.D. P6.55 (p.160) CT3
Chapter 6 Circular Motion and Other Applications of Newton’s Laws 6.1 Newton’s Second Law for a Particle in Uniform Circular Motion Comparison of Linear and Circular Motion
Chapter 6 Circular Motion and Other Applications of Newton’s Laws 6.2 Nonuniform Circular Motion F r = ma r = mv t 2 /r taking inward as positive F t = ma t Note: a t = tangential or acceleration Because a t is to a r, they have no bearing on each other’s motion. You can do two separate 1D problems (Ch.2).
CT4: At the top of the path when I whirl a bucket of water over my head, the water in the bucket will A. stay in the bucket because it is forced outward and stopped by the bottom of the bucket. B. stay in the bucket because gravity is temporarily suspended. C. stay in the bucket only if I whirl the bucket with enough speed so that the bottom of the bucket must supply an inward force. D. stay in the bucket at any speed that I whirl the bucket. E. not stay in the bucket.
rsin r P6.58 (p.161)
Chapter 6 Circular Motion and Other Applications of Newton’s Laws 6.4 Motion in the Presence of Resistive Forces A. Smaller objects, lower speed: R = bv B. Larger objects, higher speeds: R = DAv 2 /2 D = drag coefficient; = fluid density; A = cross-sectional area Both forces oppose motion.
P6.27 (p.157)
Fig. P6.60, p.178 Bs = Bk = Ps = Pk = 0.450