Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 6th Edition
5
Circular Motion; Gravitation
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5-1 to 5-3: Uniform Circular Motion; Highway Curves
5-4: Nonuniform Circular Motion
5-6 and 5-7: Law of Universal Gravitation
5-8: Satellites; Weightlessness
5-9: Kepler's Laws

Problem 9
A
v=25m/sv=25m/s
Giancoli 6th Edition, Chapter 5, Problem 9 solution video poster
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VIDEO TRANSCRIPT

So the centripetal force for this car going around in a level curve is going to be provided by friction. Now the centripetal force formula is: ‘m’ times ‘v’ squared over ‘r’. The friction formula is: ‘µ’ times ‘FN’ the normal force and ‘FN’ in this case would be ‘m’ times ‘g’ the weight of the car and these will be equal because the friction that is providing the centripetal force, so we have ‘m’ times ‘v’ squared over ‘r’ equals ‘µ’ times ‘m’ times ‘g’. The ‘m’s cancel so this answers part of our question and indeed our answer will be independent of the masses. Mass won’t matter. So solving for ‘v’ multiply both sides by ‘r’ and square root, where ‘v’ is the square root of ‘µ’ times ‘r’ times ‘g’ that gives us the square root of zero point eight times seventy seven meters times nine point eight and we have our answer twenty five meters per second. In the 5th Edition you have seventy meters here instead of seventy seven and that’s going to give you a slightly different answer, the 5th Edition answer will be twenty three meters per second.

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