QUIZ 6 (ANSWERS) |
SAMPLE TEST2 |
SAMPLE TEST3 |
READ ALL EXAMPLES, SOME OF WHICH ARE JUST LIKE ASSIGNED PROBLEMS. |
CH. 6 CIRCULAR MOTION; GRAVITY |
CH. 6 - Multiple choice THESE ARE EXTREMELY
IMPORTANT--HAND THEM ALL IN WITH THE CORRECT LETTER MARKED ONLY. On the
multiple choice , no written explanation is required though
you should understand the reasons behind your decision. 1, 3,
4, 5, 7, 8, 9, 10, 12, 14, 15 and
exercise/problems GENERAL CENTRIPETAL FORCE 2, 4, 5
(try 54 and 55)) ,
6, 9, 14(try this carefully at home--see U-tube video ~ stay dry !), 50
(try 9), 10, 11, 12, 47 (thrilling park ride), 54 (try 55),
55 GRAVITATION AND SATELLITE MOTION:22, 26, 27, 37, 38, 39, 40, 41 |
TURN IN : All multiple choice, letter only; 4, 10, 11, 12, 26, 37, 38, 47, 54 |
* DISCUSSIONS PROVIDED. |
SELECTED DISCUSSIONS TO exercises/problems
BELOW.
ALL PROBLEMS ARE DUE EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE DESIGNED TO HELP YOU DO RELATED EXERCISES. |
4. See example 6.3; the force of static friction points toward the
center of the circle and is the centripetal force: sum of forces toward center = m*v^{2}/R = pos - neg = f_{s} - 0, where R is the circle radius and v the speed. Note: v^{2}/R is the magnitude of the centripetal acceleration. |
10. Note at the bottom, the positive direction is up, toward the
circle center: sum of forces toward center = m*v^{2}/R = pos - neg =T - mg. From this simple relation you can answer all parts. Note: v^{2}/R is the magnitude of the centripetal acceleration. |
11. See example 6.5. The normal force
of the seat on person always points upward, though at circle's top that force
points sway from the center and at the bottom points toward
the center; just a heads up on geometry. (a) speed = circumference/period, where period is the time for one revolution, given. (b) At top: m*v^{2}/R = pos - neg = mg - N At bottom: m*v^{2}/R = N - mg, where in both cases v is the uniform (constant, non-changing) speed. You are given mg; find N at top and bottom. (c) That means you are changing your speed v to a value that causes N to be zero at top. N = mg - m*v^{2}/R is the equation at the top slightly rearranged. Set N = 0 and find v and then the period. (d) Re-compute N at the bottom by solving the following equation with new speed from part (c) replacing old value: m*v^{2}/R = N - mg. Find N. |
12. (a) See #38, Ch. 3, Quiz 4. At the lowest point described in problem, the airplane is traveling in a vertical, circular path. | centripetal acceleration| = a_{R} = v^{2}/R, where v is the speed. At the bottom the acceleration vector points upward vertically. Find R such that the |centripetal acceleration | is at most 4.00*g, where g is the magnitude of the acceleration of gravity . Above that level, the force of the seat of the airplane on the pilot , i.e. normal force N**, will be so large as to cause internal injuries. In Quiz 4's discussion of #38 we said we'd explore this issue further ; now's our chance. (b) At bottom: m*v^{2}/R = N - mg, where v is the given speed at that moment. Find N. ** I really meant to say magnitude of the normal force N. But if upward is positive, then magnitude N equals the positive upward component . |
26. See example 6.9 and retro-fit the solution to locations at and far above the surface of the Earth. |
37. See example 6.10 and pertinent book
discussions STart with this: mv^{2}/r = Gm*M/r^{2}. , where m is the satellite mass and M is the Earth's. MORE DISCUSSIONS LATER ! |