ANSWERS
Quiz 5
Simulations Projectile Motion 
Ch. 5:    1, 4, 7, 9, 10, 12, 13, 14, 15, 16, 18, 28, 30, 34, 35, 39, 47, 50, 51, 53, 66, 82
Turn ins: 7 (Redo for radius r = 0.70 m and speed 8.00 m/s--Hint: See #16),  12 (Hint: See #10),  18, 34 (Hint: See #28), 50,  51( redo with mass 82.0 kg) ,  53( Redo with mass 63.0 kg), 66, 82 (Hint: see #14)
)10/17/2010 02:45:42 AM  Note correction to #51 below.
10/17/2010 02:45:42 AM  Note correction to #30: masses not charges !
10/17/2010 02:45:42 AM  Note a new comment on problem #51 at the bottom of the problem. 
IN PROBLEMS BELOW INVOLVING CIRCULAR MOTION, THE POSITIVE DIRECTION IS TOWARD THE CENTER OF THE CIRCLE. 
1. 
(a) V2/R
(b)
FR = pos - neg  = fs - 0 . (friction force  acts in the the pos direction; there is no neg) 
 mV2/R  = fs ,  You are to evaluate the centripetal force using the information on V and R. 
4. FR = pos - neg  =F - 0 . (force of magnitude F  acts in the the pos direction; there is no neg) 
 mV2/R  = F.  You are to evaluate V using the information on F and R. 
7. See figure 5.8.
(a) 
FR = pos - neg = T1 + mg - 0  (there is no neg) 
mV2/R =  T1 + mg
(b) 
FR = pos - neg = T2 - mg 
mV2/R = T2 - mg 
9. FR = pos - neg = fsMAX   - 0 (there is no neg) 
mV2/R =  fsMAX  . 
Now,  fsMAX  = µN, where the magnitude N of the normal force EQUALS the weight of the car on the horizontal surface( no banking !) 
10.  
FR = pos - neg = fsMAX   - 0 (there is no neg) 
mV2/R =  fsMAX  . 
Now,  fsMAX  = µN, where the magnitude N of the normal force is the weight of the car on the horizontal surface( no banking !) Convert 90 km/h to m/s by multiplying by 5/18. Find  µ. 
12. FR = pos - neg = fsMAX   - 0 (there is no neg) 
mV2/R =  fsMAX  . 
Now,  fsMAX  = µN, where the magnitude N of the normal force equals  the weight of the coin on the horizontal surface.
Find  µ. 
Note 36 RPM (revolutions/minute) gives you the speed V in the following way: 
V= 36 Rev/Minute·(6.28 rad/rev)·(1/60 min/sec)·(0.11 m). This is easy to de-cipher.  Basically,  I got the distance in one revolution =circumference= (6.28)·(radius) , where radius = 0.11m. I then divided by the time for one revolution = 60/36 seconds.   
13. The picture for this problem is different than fig.  5.9 . In this  problem, a roller coaster,  the normal force given by N  always points to the center of the circle, the positive direction.  (A ferris wheel is different. The normal force only points up in the bottom position. ) 
AT top: 
FR = pos - neg = Ntop + mg - 0  (there is no neg) 
mV2/R =  Ntop + mg.

Set N = zero and find the speed V. 

Note the mathematical structure of this problem is like problem 7. See figure 5.8. At bottom you would have:
mV2/R = Nbott - mg , where bott is short for bottom. 
14. The picture for this problem is the same as  fig.  5.9 when  the ferris wheel is at the top. In this  problem, a sports car on a large bump in the road, the normal force given by N  is up, like fig. 5.9's rendition of the ferris wheel at  the highest point in the orbit.. 
Thus:
FR = pos - neg =  mg - N   
mV2/R = mg - N. Note that down is positive, toward the center. 

(a) Find N where m is the mass of the car. 

(b) Find N where m is now the mass of the person and N is the magnitude of the seat on the person's bottom.   

(c)   mV2/R = mg - N.
Set N = 0 and find the speed V, where m is for the driver. Note: m cancels out.  
15. FR = pos - neg =  mg - N   
mV2/R = mg - N. Note that down is positive, toward the center. 
The passenger is weightless when N = 0. Solve for V in that case. Then convert to RPM's by "undoing" the conversion I presented in problem 12: Multiply your answer by (1/6.28 rev/rad)·(60 sec/min)·(1/radius), where radius = 0.15 m.  
16. See figure 5.8.  See problem 7, and the structure of problem 13. 
(a) At bottom: 
FR = pos - neg = T2 - mg 
mV2/R = T2 - mg . Find V given the tension. . 
(b) At top:
FR = pos - neg = T1 + mg - 0  (there is no neg) 
mV2/R =  T1 + mg
Set the tension = 0 and find V. 
18. The normal force is the centripetal force. 
 FR = pos - neg  = N - 0 . (The normal force   acts in the the pos direction; there is no neg) 
 mV2/R  =N.   You are to evaluate the centripetal force using the information on the rotational frequency and  R. 
Get V with  this conversion: V =  (0.50 rev/sec)(6.26 rad/rev)(radius) , where radius = 4.6 m.  

Now in the vertical direction, the person's weight acts DOWN and the force of STATIC friction acts UP  if the person does not slide down.  
In the y direction :
may = pos - neg
0 = fs - mg.
Now, fs = µN = µmV2/R . 
Solve for  µ. 
28.
F =  GMm/(RE + h)2 where h = height above the earth. Known: h . Find F. 
30. On Moon,

g = GM/R2, where M and R are moon mass and radius, available inside book jacket.  
34. g =  GM/(RE + h)2 where h = height above the earth,  M and R are Earth mass and radius, available inside book.  Known: two values of h . Find g twice. 
35.  0.98 =  GM/r2 where unknown r = distance from earth's center.  M is the Earth's mass.. 
39. Each force will point to the square center  along  diagonal. Now for the magnitude. I will use the Pythagorean Theorem repeatedly. Let the square side length be represented by  letter a. The diagonal length is 1.41a, where 1.41 = square root of 2 to two significant figures.  

Pick any mass . The force  from each of the two nearest mass  on the other end of a  side has magnitude : F = Gm2/a2  .  These two forces are perpendicular to each other, so their resultant is 1.41F and points toward the center along the   diagonal. 

You must now include the mass  on the other end of  the diagonal,  which has magnitude F' =  Gm2/2a2  and points toward the center along the  diagonal. Evaluate the net force along the diagonal:  1.41F + F'. 
47. Use mv2/(R + h) = GMm/(R + h)2 ,   or 
mv2  = GMm/(R + h) 

Solve for v given the Earth radius R and the mountain height h.  
50. See fig. 5.9. v = 6.28r/T, where T and r are  the time for one revolution and  the radius given.   

At top: 
FR = pos - neg =  mg - N   
mV2/R = mg - N. Note that down is positive, toward the center. 
At bottom:
FR = pos - neg =  N - mg   
mV2/R = N - mg. Note that up is positive, toward the center. 

In both above cases, evaluate N/mg. This will cancel out the mass m, which you do not know.   
51. HERE IS  CORRECTION TO PART (A):

(a) I misread the problem in assuming the spaceship is moving around the earth in a circular orbit at constant speed. The problem states the ship  moves at constant velocity, which means it's  moving in a straight  line when it is at a distance r = 4.2x10 6 m from the center of the Moon. In that case, the normal force points away from  Moon, and the gravitational  force  points toward  Moon. Let the positive direction be toward the Moon's center.

ma = pos - neg =  GmM/r2  - N, where a = magnitude of the acceleration. But since the velocity = constant,  a = 0, so the normal force N =    GmM/r2   .  

Still,  the hint  below assuming a circular orbit IS  important . In at case, when you follow my hint and  plug in all the numbers,  you should get N = 0. So yes, check out the alternative hint to (a) below. Note: the hint to (b) is correct.
 
Alternative (a):  Assuming a circular orbit around Moon at constant speed V, let the pos direction point toward the center of the Moon center.  The astronaut of mass m has two oppositely directed forces on her: In the negative direction away from Moon is the normal force of magnitude N. In the pos direction is the gravitational force. 

FR = pos - neg =  GmM/r2  - N   
mV2/r =  GmM/r2  - N. Note that down is positive, toward the center. M = moon mass. 

Note that  r is given. Solve for N. Note that you need to get V. That's easy: 

The equation for the spaceship is 
m'V2/r =  Gm'M/r2  , where m' = ship mass. Thus:

V2/r =  GM/r2  or,

V2 =  GM/r    .

Solve for V.  Then plug into the first equation to find N.  You should get N = 0.
(b) ma = pos - neg =  GmM/r2  - N   
ma = GmM/r2  - N.  m = astro-person's  mass.  a = 2.9 m/s2.     Find N. 

Note that you will find that N is negative, which is not physical. That is because in this situation, the acceleration toward Moon is so large that you would be "pinned" to the roof of the spaceship. Thus the more proper way of doing the problem is to assume you are "walking" on the roof, in which case the normal  force acts toward the Moon as you move toward it.   Thus the equation would become:
ma = pos - neg  = GmM/r2  + N - 0 
ma = pos - neg  = GmM/r2  + N.
Now solve for N.  
 
53.
Below let  N be force magnitude of scale on woman's feet.  Find N in each case.
(a) Let up be the pos direction:
ma =  pos - neg
ma = N - mg
But constant velocity  means a = 0.
(b)
Let down be the pos direction:
ma =  pos - neg
ma = mg - N
But constant velocity  means a = 0.
(c)
Let up be the pos direction:
ma =  pos - neg
ma = N - mg
(d)
Let down be the pos direction:
ma =  pos - neg
ma = mg - N
(e)
Let down be the pos direction:
ma =  pos - neg
ma = mg - N
But a = g. What must be  N?
66. See figure 5.8 for the bottom location.

FR = pos - neg = T - mg 
mV2/R = T - mg .
Find V given T = 1400 (N),  R = 5.5 (m) and m = 80 kg.  
82. SEE PROBLEM 14 and  50 . 

At the top of the hill, at point A, we have:   
FR = pos - neg =  mg - N   
mV2/R = mg - N. Note that down is positive, toward the center. 
At point B, The acceleration is zero; thus: 0 = mg - N
At point C, bottom of valley: 
FR = pos - neg =  N - mg   
mV2/R = N - mg .  Note that  up is positive, toward the center. 
(a) Find N at each point and compare.
(b) Heavy means a large N; light means a small N. 
(c)  At  point A, we have:   
FR = pos - neg =  mg - N   
mV2/R = mg - N. Set N = 0 and find V.  
more hints later.