Homework Help is Here – Start Your Trial Now!
Science
Physics
Tutorial Exercise A bead with a hole through it slides on a wire track. The wire is threaded through the hole in the bead, and the bead slides without friction around a loop-the-loop (see figure below). The bead is released from rest at a height h = 3.05R. R (a) What is its speed at point Ⓐ? (b) How large is the normal force on the bead at point A if its mass is 5.95 g? Part 1 of 4 - Conceptualize is Since the bead is released above the top of the loop, it starts with enough potential energy to reach point Ⓐ and still have excess kinetic energy. The energy of the bead at point proportional to h and g. If it is moving relatively slowly, the track will exert an upward force on the bead, but if it is whipping around quickly, the normal force will push it toward the center of the loop. Part 2 of 4 - Categorize The speed at the top of the loop can be found from the conservation of energy for the bead-track-Earth system and the normal force can be found from Newton's second law. Part 3 of 4 - Analyze (a) We define the bottom of the loop as the zero level for the gravitational potential energy. For the total energy of the system at point Ⓐ, we have U¡ + K¡ = Uƒ + Kf. Now we know that U; mgh and Ki = 0 because the bead starts from rest at height h. At point Ⓐ, the bead is at height 3.05 = Your response differs from the correct answer by more than 10%. Double check your calculations.R with speed v, so we have mgh + 0 = mg 3.05 Your response differs from the correct answer by more than 10%. Double check your calculations.R) + Substituting the given expression for h in terms of R, gives mg 3.05 R) = mg ( 3.05 × 를까? S.R) + 를까? Your response differs from the correct answer by more than 10%. Double check your calculations.R) + Now we can solve for v in terms of R and g. v2 3.05 = Your response differs from the correct answer by more than 10%. Double check your calculations. 3.05 V = × Your response differs from the correct answer by more than 10%. Double check your calculations.gR

Tutorial Exercise A bead with a hole through it slides on a wire track. The wire is threaded through the hole in the bead, and the bead slides without friction around a loop-the-loop (see figure below). The bead is released from rest at a height h = 3.05R. R (a) What is its speed at point Ⓐ? (b) How large is the normal force on the bead at point A if its mass is 5.95 g? Part 1 of 4 - Conceptualize is Since the bead is released above the top of the loop, it starts with enough potential energy to reach point Ⓐ and still have excess kinetic energy. The energy of the bead at point proportional to h and g. If it is moving relatively slowly, the track will exert an upward force on the bead, but if it is whipping around quickly, the normal force will push it toward the center of the loop. Part 2 of 4 - Categorize The speed at the top of the loop can be found from the conservation of energy for the bead-track-Earth system and the normal force can be found from Newton's second law. Part 3 of 4 - Analyze (a) We define the bottom of the loop as the zero level for the gravitational potential energy. For the total energy of the system at point Ⓐ, we have U¡ + K¡ = Uƒ + Kf. Now we know that U; mgh and Ki = 0 because the bead starts from rest at height h. At point Ⓐ, the bead is at height 3.05 = Your response differs from the correct answer by more than 10%. Double check your calculations.R with speed v, so we have mgh + 0 = mg 3.05 Your response differs from the correct answer by more than 10%. Double check your calculations.R) + Substituting the given expression for h in terms of R, gives mg 3.05 R) = mg ( 3.05 × 를까? S.R) + 를까? Your response differs from the correct answer by more than 10%. Double check your calculations.R) + Now we can solve for v in terms of R and g. v2 3.05 = Your response differs from the correct answer by more than 10%. Double check your calculations. 3.05 V = × Your response differs from the correct answer by more than 10%. Double check your calculations.gR

Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter7: Conservation Of Energy
Section: Chapter Questions
Problem 3P: Review. A bead slides without friction around a loop-the-loop (Fig. P7.3). The bead is released from...
See similar textbooks
icon
Related questions
Question
Tutorial Exercise A bead with a hole through it slides on a wire track. The wire is threaded through the hole in the bead, and the bead slides without friction around a loop-the-loop (see figure below). The bead is released from rest at a height h = 3.05R. R (a) What is its speed at point Ⓐ? (b) How large is the normal force on the bead at point A if its mass is 5.95 g? Part 1 of 4 - Conceptualize is Since the bead is released above the top of the loop, it starts with enough potential energy to reach point Ⓐ and still have excess kinetic energy. The energy of the bead at point proportional to h and g. If it is moving relatively slowly, the track will exert an upward force on the bead, but if it is whipping around quickly, the normal force will push it toward the center of the loop. Part 2 of 4 - Categorize The speed at the top of the loop can be found from the conservation of energy for the bead-track-Earth system and the normal force can be found from Newton's second law.
Transcribed Image Text:Tutorial Exercise A bead with a hole through it slides on a wire track. The wire is threaded through the hole in the bead, and the bead slides without friction around a loop-the-loop (see figure below). The bead is released from rest at a height h = 3.05R. R (a) What is its speed at point Ⓐ? (b) How large is the normal force on the bead at point A if its mass is 5.95 g? Part 1 of 4 - Conceptualize is Since the bead is released above the top of the loop, it starts with enough potential energy to reach point Ⓐ and still have excess kinetic energy. The energy of the bead at point proportional to h and g. If it is moving relatively slowly, the track will exert an upward force on the bead, but if it is whipping around quickly, the normal force will push it toward the center of the loop. Part 2 of 4 - Categorize The speed at the top of the loop can be found from the conservation of energy for the bead-track-Earth system and the normal force can be found from Newton's second law.
Part 3 of 4 - Analyze (a) We define the bottom of the loop as the zero level for the gravitational potential energy. For the total energy of the system at point Ⓐ, we have U¡ + K¡ = Uƒ + Kf. Now we know that U; mgh and Ki = 0 because the bead starts from rest at height h. At point Ⓐ, the bead is at height 3.05 = Your response differs from the correct answer by more than 10%. Double check your calculations.R with speed v, so we have mgh + 0 = mg 3.05 Your response differs from the correct answer by more than 10%. Double check your calculations.R) + Substituting the given expression for h in terms of R, gives mg 3.05 R) = mg ( 3.05 × 를까? S.R) + 를까? Your response differs from the correct answer by more than 10%. Double check your calculations.R) + Now we can solve for v in terms of R and g. v2 3.05 = Your response differs from the correct answer by more than 10%. Double check your calculations. 3.05 V = × Your response differs from the correct answer by more than 10%. Double check your calculations.gR
Transcribed Image Text:Part 3 of 4 - Analyze (a) We define the bottom of the loop as the zero level for the gravitational potential energy. For the total energy of the system at point Ⓐ, we have U¡ + K¡ = Uƒ + Kf. Now we know that U; mgh and Ki = 0 because the bead starts from rest at height h. At point Ⓐ, the bead is at height 3.05 = Your response differs from the correct answer by more than 10%. Double check your calculations.R with speed v, so we have mgh + 0 = mg 3.05 Your response differs from the correct answer by more than 10%. Double check your calculations.R) + Substituting the given expression for h in terms of R, gives mg 3.05 R) = mg ( 3.05 × 를까? S.R) + 를까? Your response differs from the correct answer by more than 10%. Double check your calculations.R) + Now we can solve for v in terms of R and g. v2 3.05 = Your response differs from the correct answer by more than 10%. Double check your calculations. 3.05 V = × Your response differs from the correct answer by more than 10%. Double check your calculations.gR
Expert Solution
steps

Step by step

Solved in 2 steps with 5 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
Physics
ISBN:
9781133104261
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
University Physics Volume 1
University Physics Volume 1
Physics
ISBN:
9781938168277
Author:
William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:
OpenStax - Rice University
Classical Dynamics of Particles and Systems
Classical Dynamics of Particles and Systems
Physics
ISBN:
9780534408961
Author:
Stephen T. Thornton, Jerry B. Marion
Publisher:
Cengage Learning
Physics for Scientists and Engineers: Foundations…
Physics for Scientists and Engineers: Foundations…
Physics
ISBN:
9781133939146
Author:
Katz, Debora M.
Publisher:
Cengage Learning
College Physics
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
College Physics
College Physics
Physics
ISBN:
9781285737027
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS