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Constants I Periodic Table Submit Previous Answers (Figure 1)The car shown in the figure has mass m (this includes the mass of the wheels). The wheels have radius r, mass mw, and moment of inertia I = kmwr2. Assume that the axles apply the same torque T to all four wheels. For simplicity, also assume that the weight is distributed uniformly so that all the wheels experience the same normal reaction from the ground, and so the same frictional force. Correct With this expression for the normal force on each tire we can now examine an interesting side-note. This isn't crucial for the remainder of this question, but it is interesting. The coefficient of friction u between rubber and cement is about 0.9. Also, the maximum frictional force is fm = uN Substituting for f and N in the scalar expression of Newton's second law Fnet car would go from 0 to 60 mph in only about 3 seconds! Of course, air resistance becomes significant at higher speeds, so this simple model does not give accurate results. However, many race car models can accelerate from 0 to 60 mph in about 3.3 seconds. max = ma gives (a)max = /ug^ 8.8 m/s2. With this acceleration, the Part D Now assume that the frictional force f is not at its maximum value. What is the relation between the torque T applied to each wheel by the axles and the acceleration a of the car? Once you have the exact expression for the acceleration, make the approximation that the wheels are much lighter than the car as a whole. Figure 1 of 1 Express your answer in terms of some or all of the variables m, r, T and the magnitude of the acceleration due to gravity g. View Available Hint(s) ν ΑΣφ ? Submit Provide Feedback Next

Constants I Periodic Table Submit Previous Answers (Figure 1)The car shown in the figure has mass m (this includes the mass of the wheels). The wheels have radius r, mass mw, and moment of inertia I = kmwr2. Assume that the axles apply the same torque T to all four wheels. For simplicity, also assume that the weight is distributed uniformly so that all the wheels experience the same normal reaction from the ground, and so the same frictional force. Correct With this expression for the normal force on each tire we can now examine an interesting side-note. This isn't crucial for the remainder of this question, but it is interesting. The coefficient of friction u between rubber and cement is about 0.9. Also, the maximum frictional force is fm = uN Substituting for f and N in the scalar expression of Newton's second law Fnet car would go from 0 to 60 mph in only about 3 seconds! Of course, air resistance becomes significant at higher speeds, so this simple model does not give accurate results. However, many race car models can accelerate from 0 to 60 mph in about 3.3 seconds. max = ma gives (a)max = /ug^ 8.8 m/s2. With this acceleration, the Part D Now assume that the frictional force f is not at its maximum value. What is the relation between the torque T applied to each wheel by the axles and the acceleration a of the car? Once you have the exact expression for the acceleration, make the approximation that the wheels are much lighter than the car as a whole. Figure 1 of 1 Express your answer in terms of some or all of the variables m, r, T and the magnitude of the acceleration due to gravity g. View Available Hint(s) ν ΑΣφ ? Submit Provide Feedback Next

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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Part D

Constants I Periodic Table Submit Previous Answers (Figure 1)The car shown in the figure has mass m (this includes the mass of the wheels). The wheels have radius r, mass mw, and moment of inertia I = kmwr2. Assume that the axles apply the same torque T to all four wheels. For simplicity, also assume that the weight is distributed uniformly so that all the wheels experience the same normal reaction from the ground, and so the same frictional force. Correct With this expression for the normal force on each tire we can now examine an interesting side-note. This isn't crucial for the remainder of this question, but it is interesting. The coefficient of friction u between rubber and cement is about 0.9. Also, the maximum frictional force is fm = uN Substituting for f and N in the scalar expression of Newton's second law Fnet car would go from 0 to 60 mph in only about 3 seconds! Of course, air resistance becomes significant at higher speeds, so this simple model does not give accurate results. However, many race car models can accelerate from 0 to 60 mph in about 3.3 seconds. max = ma gives (a)max = /ug^ 8.8 m/s2. With this acceleration, the Part D Now assume that the frictional force f is not at its maximum value. What is the relation between the torque T applied to each wheel by the axles and the acceleration a of the car? Once you have the exact expression for the acceleration, make the approximation that the wheels are much lighter than the car as a whole. Figure 1 of 1 Express your answer in terms of some or all of the variables m, r, T and the magnitude of the acceleration due to gravity g. View Available Hint(s) ν ΑΣφ ? Submit Provide Feedback Next
Transcribed Image Text:Constants I Periodic Table Submit Previous Answers (Figure 1)The car shown in the figure has mass m (this includes the mass of the wheels). The wheels have radius r, mass mw, and moment of inertia I = kmwr2. Assume that the axles apply the same torque T to all four wheels. For simplicity, also assume that the weight is distributed uniformly so that all the wheels experience the same normal reaction from the ground, and so the same frictional force. Correct With this expression for the normal force on each tire we can now examine an interesting side-note. This isn't crucial for the remainder of this question, but it is interesting. The coefficient of friction u between rubber and cement is about 0.9. Also, the maximum frictional force is fm = uN Substituting for f and N in the scalar expression of Newton's second law Fnet car would go from 0 to 60 mph in only about 3 seconds! Of course, air resistance becomes significant at higher speeds, so this simple model does not give accurate results. However, many race car models can accelerate from 0 to 60 mph in about 3.3 seconds. max = ma gives (a)max = /ug^ 8.8 m/s2. With this acceleration, the Part D Now assume that the frictional force f is not at its maximum value. What is the relation between the torque T applied to each wheel by the axles and the acceleration a of the car? Once you have the exact expression for the acceleration, make the approximation that the wheels are much lighter than the car as a whole. Figure 1 of 1 Express your answer in terms of some or all of the variables m, r, T and the magnitude of the acceleration due to gravity g. View Available Hint(s) ν ΑΣφ ? Submit Provide Feedback Next
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