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Three quarters of a circle of track of radius R = 0.800 m is unraveled such that it lies flat, as in the diagram to the right. Two sensitive scales are placed beneath the track that are able to measure changes in the normal force on the track directly above them. Scale #1 is placed at the exact point where the track goes from being curved to being straight, while scale #2 is placed at the open end of the extended track. A ball is rolled down the circular portion (not necessarily from rest or from the top of the arc), and it rolls off the end, encountering negligible friction. The reading on scale #1 as the ball rolls over it is three times as great as the reading on scale #2 as the ball rolls over it. a. Draw force diagrams of the ball when it is above each scale. Erroneous Solution: scale #1 scale #2 The ball is in its last instant of circular motion when it hits the first scale, so the force up on it is centripetal. The force up on it at the second scale is just the normal force. In both cases the forces balance (ball is moving horizontally), so: N R above scale #1: above scale #2: mg mg b. Find the time it takes the ball to roll from scale #1 to scale #2. Erroneous Solution: We solve for v using the free body diagram (FBD) over scale #1: m v2 R = mg⇒ v = √√gR (1) This velocity is constant on the frictionless, horizontal surface, and since the distance it travels from one scale to the other is three- quarters of the circumference of the circle, we can compute the time required to traverse the distance: 3 (2R) 3πT R 3πT = √9R 2 9 2 0.800 m 9.80 m/s² = 1.35 s (2)

Three quarters of a circle of track of radius R = 0.800 m is unraveled such that it lies flat, as in the diagram to the right. Two sensitive scales are placed beneath the track that are able to measure changes in the normal force on the track directly above them. Scale #1 is placed at the exact point where the track goes from being curved to being straight, while scale #2 is placed at the open end of the extended track. A ball is rolled down the circular portion (not necessarily from rest or from the top of the arc), and it rolls off the end, encountering negligible friction. The reading on scale #1 as the ball rolls over it is three times as great as the reading on scale #2 as the ball rolls over it. a. Draw force diagrams of the ball when it is above each scale. Erroneous Solution: scale #1 scale #2 The ball is in its last instant of circular motion when it hits the first scale, so the force up on it is centripetal. The force up on it at the second scale is just the normal force. In both cases the forces balance (ball is moving horizontally), so: N R above scale #1: above scale #2: mg mg b. Find the time it takes the ball to roll from scale #1 to scale #2. Erroneous Solution: We solve for v using the free body diagram (FBD) over scale #1: m v2 R = mg⇒ v = √√gR (1) This velocity is constant on the frictionless, horizontal surface, and since the distance it travels from one scale to the other is three- quarters of the circumference of the circle, we can compute the time required to traverse the distance: 3 (2R) 3πT R 3πT = √9R 2 9 2 0.800 m 9.80 m/s² = 1.35 s (2)

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Three quarters of a circle of track of radius R = 0.800 m is unraveled such that it lies flat, as in the diagram to the right. Two sensitive scales are placed beneath the track that are able to measure changes in the normal force on the track directly above them. Scale #1 is placed at the exact point where the track goes from being curved to being straight, while scale #2 is placed at the open end of the extended track. A ball is rolled down the circular portion (not necessarily from rest or from the top of the arc), and it rolls off the end, encountering negligible friction. The reading on scale #1 as the ball rolls over it is three times as great as the reading on scale #2 as the ball rolls over it. a. Draw force diagrams of the ball when it is above each scale. Erroneous Solution: scale #1 scale #2 The ball is in its last instant of circular motion when it hits the first scale, so the force up on it is centripetal. The force up on it at the second scale is just the normal force. In both cases the forces balance (ball is moving horizontally), so: N R above scale #1: above scale #2: mg mg b. Find the time it takes the ball to roll from scale #1 to scale #2. Erroneous Solution: We solve for v using the free body diagram (FBD) over scale #1: m v2 R = mg⇒ v = √√gR (1) This velocity is constant on the frictionless, horizontal surface, and since the distance it travels from one scale to the other is three- quarters of the circumference of the circle, we can compute the time required to traverse the distance: 3 (2R) 3πT R 3πT = √9R 2 9 2 0.800 m 9.80 m/s² = 1.35 s (2)
Transcribed Image Text:Three quarters of a circle of track of radius R = 0.800 m is unraveled such that it lies flat, as in the diagram to the right. Two sensitive scales are placed beneath the track that are able to measure changes in the normal force on the track directly above them. Scale #1 is placed at the exact point where the track goes from being curved to being straight, while scale #2 is placed at the open end of the extended track. A ball is rolled down the circular portion (not necessarily from rest or from the top of the arc), and it rolls off the end, encountering negligible friction. The reading on scale #1 as the ball rolls over it is three times as great as the reading on scale #2 as the ball rolls over it. a. Draw force diagrams of the ball when it is above each scale. Erroneous Solution: scale #1 scale #2 The ball is in its last instant of circular motion when it hits the first scale, so the force up on it is centripetal. The force up on it at the second scale is just the normal force. In both cases the forces balance (ball is moving horizontally), so: N R above scale #1: above scale #2: mg mg b. Find the time it takes the ball to roll from scale #1 to scale #2. Erroneous Solution: We solve for v using the free body diagram (FBD) over scale #1: m v2 R = mg⇒ v = √√gR (1) This velocity is constant on the frictionless, horizontal surface, and since the distance it travels from one scale to the other is three- quarters of the circumference of the circle, we can compute the time required to traverse the distance: 3 (2R) 3πT R 3πT = √9R 2 9 2 0.800 m 9.80 m/s² = 1.35 s (2)
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