m kg 2000 8 cm 8 s 60 min 1 min rev 800 1 m s N 10 18 9 2 5 3 2 2 2 5 sediment

M kg 2000 8 cm 8 s 60 min 1 min rev 800 1 m s n 10 18

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m kg 2000 8 cm 0 . 8 s 60 min 1 min rev 800 1 m s N 10 1.8 9 2 5 3 2 2 2 5 sediment = × = Δ t ( b ) In Problem 72 it was shown that the rate of fall of the particles in air is 2.42 cm/s. Find the time required to fall 8.0 cm in air under the influence of gravity: s 31 . 3 cm/s 42 . 2 cm 8.0 Δ Δ air = = = v x t Find the ratio of the two times: 86 ms 38.47 s 31 . 3 Δ Δ sediment air = t t With the drag force in Problem 72 it takes about 86 times longer than it does using the centrifuge.
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Additional Applications of Newton’s Laws 461 Motion Along a Curved Path 74 A rigid rod with a 0.050-kg ball at one end rotates about the other end so the ball travels at constant speed in a vertical circle with a radius of 0.20 m. What is the maximum speed of the ball so that the force of the rod on the ball does not exceed 10 N? Picture the Problem The force of the rod on the ball is a maximum when the ball is at the bottom of the vertical circle. We can use Newton’s second law to relate the force of the rod on the ball to the mass m of the ball, the radius r of its path, and the constant speed of the ball along its circular path. The diagram to the right shows the forces acting on the ball when it is at the bottom of the vertical circle. g m F r r = g max T r r Apply radial radial ma F = to the ball: r v m mg T 2 max max = Solving for v max yields: r g m T v = max max Substitute numerical values and evaluate v max : () m/s 2 . 6 m 20 . 0 m/s 81 . 9 kg 050 . 0 N 10 2 max = = v 75 [SSM] A 95-g stone is whirled in a horizontal circle on the end of an 85-cm-long string. The stone takes 1.2 s to make one complete revolution. Determine the angle that the string makes with the horizontal.
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Chapter 5 462 Picture the Problem The only forces acting on the stone are the tension in the string and the gravitational force. The centripetal force required to maintain the circular motion is a component of the tension. We’ll solve the problem for the general case in which the angle with the horizontal is θ by applying Newton’s second law of motion to the forces acting on the stone. Apply = a F r r m to the stone: r v m ma T F x 2 c cos = = = (1) and 0 sin = = mg T F y (2) Use the right triangle in the diagram to relate r , L , and : cos L r = (3) Eliminate T and r between equations (1), (2) and (3) and solve for v 2 : cos cot 2 gL v = (4) Express the speed of the stone in terms of its period: rev 1 2 t r v π = (5) Eliminate v between equations (4) and (5) and solve for : = L gt 2 2 rev 1 1 4 sin Substitute numerical values and evaluate : ( )( ) () ° = = 25 m 0.85 4 s 1.2 m/s 9.81 sin 2 2 2 1 π 76 •• A 0.20-kg stone is whirled in a horizontal circle on the end of an 0.80- m-long string. The string makes an angle of 20º with the horizontal. Determine the speed of the stone.
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Additional Applications of Newton’s Laws 463 Picture the Problem The only forces acting on the stone as it moves in a
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