QUIZ 7  (REAL EXAM 2)   (ANSWERS) 
SEC. 5.4:    44, 45, 46, 47, 49 (NASA-AMES!), 51 (SEE 116), 53, 54, 101 (TRICKY BUT NOT TOO HARD--SEARCH THE INTERNET FOR THIS CLASSIC PROBLEM!), 116 (SEE 51), 120
FOR MANY OF THESE PROBLEMS GO TO THE LECTURE NOTES TIED TO 3/22/13. SCROLL DOWN A BIT TO SEE ACTIVPHYSICS EXERCISES,  BOOK EXAMPLE RENDITIONS (INTERPRETATIONS)
44. SEE EXAMPLE 5.21
45. SEE EXAMPLE 5.22.
46. SEE EXAMPLE 5.20. HOWEVER, TOTAL RADIUS R INCLUDES THE 3.00 m LENGTH OF THE HORIZONTAL  BAR, SIMILAR TO THE CENTRIPETAL FORCE LABORATORY. As in example 5.20, include  horizontal component of inclined string length making 30 degree angle with vertical.

(a) Using example 5.20 find speed V  given:  mV2/R = T*sin 30, where T is tension magnitude in string. R includes THE 3.00 m LENGTH OF THE HORIZONTAL  BAR.  As in example 5.20, include  horizontal component of string length.

(b) Solve for tan 30  in terms of g and  possibly other  variables following example 5.20.
47. SEE PREVIOUS PROBLEM, ONLY NOW   mV2/R = T*sin 40 + T', WHERE T AND T ' ARE  TENSIONS IN INCLINED AND  HORIZONTAL CABLES. NOTE ALSO mg = T*cos 40. Note V is given. Isolate tan 40 by dividing the two equations,  canceling out T. Solve for T'. Then find  T.
49. IF YOU ARE IN AN ARTIFICIAL GRAVITY ENVIRONMENT, THE NORMAL FORCE OF THE SPACESHIP FLOOR ON YOUR FEET IS THE CENTRIPETAL FORCE CAUSING YOUR BODY TO MOVE IN A CIRCLE THAT  NORMAL FORCE ALSO TO GIVES YOU A SENSE YOU HAVE A WEIGHT CAUSING THE FLOOR TO REACT TO YOUR "WEIGHT" WITH A PRESSURE ON YOUR FEET EQUIVALENT TO THE  NORMAL FORCE JUST AS YOU WERE STANDING ON THE GROUND AT EARTH'S SURFACE .

mV2/R = N, WHERE N = mg. FIND SPEED V.
51. (a) TOP: POSITIVE  DIRECTION IS DOWN. Sum of forces in positive direction is  mV2/R = N + mg, where V is speed at top. AT TOP, N = 0.
(b) BOTTOM: POSITIVE DIRECTION IS UP. Sum of  forces in positive direction is mV'2/R = N' - mg, WHERE V'  AND N'  ARE THE SPEED AND NORMAL FORCE AT BOTTOM.  
53. TOP: POSITIVE  DIRECTION IS DOWN. Sum of forces in positive direction is  mV2/R = N + mg, where V is speed at top. When you are at the minimum speed V,  AT TOP, N = 0 on water in  pail due to the pail's circular surface AND WATER LOSES CONTACT WITH SURFACE, SPILLING TOWARD THE GROUND. NOTE: m is the water mass, which cancels out.
54. BOTTOM: POSITIVE DIRECTION IS UP. Sum of  forces in positive direction is mV2/R = T - mg, WHERE V  AND T ARE THE SPEED AND TENSION MAGNITUDE AT BOTTOM.
(a) Find direction and magnitude of the centripetal acceleration. 
(b) Use  above equation to find T.
101. This hard problem will be discussed in lecture or lab during exam review sessions.
116. NET FORCE OF SEAT INCLUDES VERTICAL UPWARD NORMAL FORCE AND THE  HORIZONTAL STATIC FRICTION AND OTHER FORCES CAUSING PASSENGER TO MOVE IN A VERTICAL CIRCLE. IN EACH CASE, BEFORE OR AFTER THE HIGHEST POINT, THE POSITIVE DIRECTION IS HORIZONTAL  TOWARD THE CENTER OF  CIRCLE. IN HORIZONTAL DIRECTION WE HAVE SUM OF FORCES IN POSITIVE DIRECTION =  mV2/R = fs .  IN THE  VERTICAL DIRECTION WE HAVE  N = mg.  FIND THE ANGLE THETA THE NET FORCE MAKES WITH THE HORIZONTAL BY NOTING tan theta = mg/( mV2/R ) = R*g/V2  .    
120. See Problem 51.51. BUT HERE SPEED V IS CONSTANT.  IN THIS CASE,   NORMAL FORCE ALWAYS POINTS TO CIRCLE'S CENTER UNLIKE PROBLEM 116. 
(a)  BOTTOM: POSITIVE DIRECTION IS UP. Sum of  forces in positive direction is mV2/R = N' - mg, WHERE V  AND N'  ARE THE SPEED AND NORMAL FORCE AT BOTTOM.  
(b) TOP: POSITIVE  DIRECTION IS DOWN. Sum of forces in positive direction is  mV2/R = N + mg, where V AND N are the  speed and normal force at top.