Chapter 4: Do listed problems.  
Simulations
Quiz 4 
ANSWERS
1 8 10 11 12 13 18 21 23 29 30
32 34 40 41 42 43 50 69
Helpers!! Click this link then click the links "Practice Questions" and "Practice Problems" in the left hand column.
Since I covered  non-uniform circular motion, where the particle changes speed while moving around its orbit, all problems are due as part of Quiz 4. 
error log #1:  Problem 11 hint has been corrected by removing a reference to an ICQ from  a previous semester.

error log #2:  Problem 13  hint has been clarified.

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hints posted later; check back soon !
1. The curve is shaped like a smooth backward L. 

#On the first straight segment, which is horizontal, the acceleration points opposite the velocity, i.e. leftward, since the particle is slowing down while moving right.   

#On the second straight segment, vertical, the particle accelerates in the same direction as the velocity, i.e. upward, since it's speeding up while moving upward.  

#On the curved part, a quarter circle, the acceleration points toward the center. The velocity seems to have constant magnitude while changing direction from rightward to up. This arc  of movement behaves like uniform circular motion. 
8. 
(a) Vector-V = (vx, vy) .  At t = 2 seconds, both components are positive. Thus the vector points in the first quadrant. Find the angle with the positive x - axis =  tan-1 (| vy |/| vx |) 
(b) Vector - r = (x, y) . Here x = integral of vx from 0 to 5 seconds and y = integral of vy from 0 to 5 seconds.  Perform these integrals.   x = area under the triangle with  base  5 seconds and height  40 m/s.  y = area under the rectangle  of  base  5 seconds and height  30 m/s.  With these components,  find the magnitude r from r2 = x2 + y2. 
10. Note: vx is constant = 2 m/s throughout the motion.  Compute  vy at each time. That's easy: 
vBy  = vAy  - g(tB  - tA) , where B is a point later in time than A. We  assume the positive y-direction is up. Thus  write:   vAy  = vBy  +  g(tB  - tA) 

(a) Below answers are in  m/s. 
At t = 0: vy  =  2 + g·(1). Notice we add because 0 is a second earlier. 
At t = 2 sec: vy  =  2 - g·(1). Note we subtract since 2 is a second later.
At t = 3 sec: vy  =  2 - g·(2). Note we subtract since 3 is two seconds later.

Now,  it would be nice if you knew  g from the curve on  Planet Exidor. But  the  velocity y-component  vanishes at t = 2 seconds on the TOP.  At t = 2, vy = 0 = 2 - g·(1). Find g. 

With  g, find the other two values of vy.  

(b) See part (a). 

(c) Launch angle = tan-1 (| vy |/| vx |) at t = 0. You already know  the initial x and y- components of velocity.  
 
11.  Use (change in x)  = 50 m = 25t. Solve for t. Then find (change in y) using y-motion equations.  
12. 
(a) (change in y) = - ½gt2, assuming upward is  positive y.  Note: The initial  velocity y-component vanishes, explaining the previous equation.  Find t from given information. 
(b) (change in x ) = Vx·t. Note: (change in x ) and t are  known.  Find Vx . 
13.
(change in y) = - ½gt2, assuming upward is  positive y.  Note: The initial  velocity y-component vanishes, explaining the previous equation.  Find t from given information. (change in x ) = Vx·t. Note:  Vx is known and t can be found from the y-motion equation at the top of this hint.  Find (change in x ) . 
18. See my notes and example 4.9. If you rotate figure of ex. 4.9 by 90 degrees counter-clockwise, you have this problem exactly, except for the numbers. Replace the velocity of the air (relative to the ground) of magnitude 50 mph with the velocity of the water (relative to shore)  of magnitude 3.0 m/s.   Replace the velocity of the airplane (relative to the air) of magnitude 500 mph with the velocity of the boat (relative to the water) of  magnitude 2.0 m/s. Find x and y as shown in example 4.9: Find the time t it takes the boat to move 100 m in the y-direction across the river; use the y-component of velocity  Vy  = +2.0 m/s. With the time t, find x = (3.0 m/s)·t.  Please also draw the diagram suggested.  Thanks!
21. See table 4.1. 


wf = wi + (alpha)*(change in t).
thetaf = thetai + wi + (1/2)*(alpha)*(change in time)2  .
wf2 = wi2 + 2*(alpha)*(change in theta).

The angular acceleration is zero for each time interval. Thus, use the second equation in the left column with the angular  acceleration set to zero: (change in theta) = w·(change in t). In the first 2-second time interval,  w = 10 rad/s. In the second 2-second time interval w =  20 rad/s. Add both changes in theta to get the total change in theta. Then divide by 6.28 to get the total number of revolutions.  Note: 6.28 = 2·pi. where pi = 3.14.  

NOTE YOU CAN ALSO GET THE TOTAL ANGLE CHANGE USING THE AREA UNDER THE CURVE.

 

23. 
(a) (45 rev/min)·(6.28 rad/rev)·(1/60)(min/sec) = angular velocity in rad/s.  
(b) T = period = time for one revolution = circumference/ (linear speed) , where linear speed = v =  w·r . 
See section 4.6. The circumference  = 6.28·r. Thus, T =  6.28/w . 
29. T = 1/3 s. Note the frequency = f = 1/T = 3 s-1 .  linear speed = v = circumference/T . Finally, the magnitude of the centripetal acceleration = ar = v2/r. 
30. angular acceleration = slope of w. Find the slope between 0 and 2 s, 2 s and 4 s, and 4 s and 7 s. Note that in the first interval,  the slope is negative; in the second interval,  the slope is zero; and in the third, the slope is positive. Use 
rise/ run to compute slope in each case.
32. Find area under the curve just like you might have done in problem 11, Chapter 2.
34. Exactly "half way around the curve" means the line from the center of the quarter circle to the car makes an angle of 45 degrees from the mathematical horizontal (East).  The car's acceleration vector makes an angle of 20 degrees with the mathematical horizontal. The angle between the acceleration vector and the radial line is 45 - 20 = 25 degrees.
Now,
ar = a·cos25, where a is the magnitude of the acceleration vector .
|at| = a·cos55, since the angle between the  acceleration vector and the tangential component  of acceleration at is
 90 - 25 = 55 degrees.  Note:  |at| = a·sin25, from the properties of complimentary angles.
Note also: the magnitude a is given.
PROJECTILE MOTION RESOURCES: SEC 4.3. See equations (4.17) and my gracious notes.  
40. The range equation R = Vo2sin2Ų/g   is  easily derived from example 4.5. PLEASE READ THAT EXAMPLE IN CONJUNCTION WITH MY NOTES! ALSO SEE EXAMPLE 4.6. The maximum range occurs when the launch angle is 45 degrees.   Set:

Vo2sin2Ų/g = Vo2/(2g)   = RMAX/2.  

This reduces to sin2Ų  = 1/2. Solve for all values of the angle making this equation true. Hint: There are two solutions. 
41. See examples 4.5 and 4.6 + equations (4/17) + my notes. 
(a) Use y-motion equation Vy = VosinŲ - gt. Set Vy  = 0,  solve for t, and plug into h = y - yo = (VosinŲ)·t - ½gt2 . You can also use Vy2 = (VosinŲ)2 - 2g(y - yo). Set Vy  = 0,  solve for h =  y - yo .
(b) Observe h's dependence on the angle by substituting the given values. 
42. 
(a) Evaluate y - yo = (VosinŲ)·t - ½gt2  for t = 7.5 s.  
(b) Vy2 = (VosinŲ)2 - 2g(y - yo). Set Vy  = 0,  solve for h =  y - yo .
(c) Find Vx  = constant. Vy = VosinŲ - gt for t = 7.5 s. Compute the speed V from V2 = Vx2 + Vy2. Finally, from the signs of both velocity components, find the quadrant  and related angle with either the positive or negative x-axis. 
43. The shot is fired  from an elevated position and a non zero initial angle, which means the problem is different from examples I've done in class.  I would say this problem is similar to example 4.6---it's like example 4.6 "in reverse."  Here's is a hint:
(a) Use: y - yo = -1.8  m = (VosinŲ)·t - ½gt2  to find the  time t.. In other words solve this quadratic equation. Once you get t, then find the change in x - xo = (VocosŲ)·t
(b) Repeat computations for the new angles given. 
50.  See problem 43; but only now the angle is below the horizontal axis.
Use: y - yo = -3.0 m  = -(Vosin15)·t - ½gt2  to find the  time t.. In other words solve this quadratic equation. Once you get t, then find the change in x - xo = (Vocos15)·t
69. This is an easy problem if you have had time to review my notes; here's  a hint:

You need the following equations in table 4.1 for each segment of the motion:

wf = wi + (alpha)*(change in t).
thetaf = thetai + wi + (1/2)*(alpha)*(change in time)2  .
wf2 = wi2 + 2*(alpha)*(change in theta).

alpha = angular acceleration,  which is constant in each time segment.

Between 0 and 1 seconds, the angular velocity w is constant since alpha = 0. Recall that alpha = dw/dt.

Between 1 and 2 seconds, the angular velocity is given by, 

w = w(1) + (alpha)*( t - 1) , where w(1) is the angular velocity at t = 1 second. Note w(1) = 60 rpm.
Note also that alpha = 4 rad/s2 from the graph. The problem is that you must convert w to rad/s to get the right answer. When you are done you can convert back to rpm. Here is how you make the conversion.

60 rpm = 60 rev/min = (60 rev/min)*(6.28 rad/rev)*(1/60 min/s). After making that conversion write:

w(2) = w(1) + (alpha)*(2 - 1)

or

w(2) = w(1)  + (alpha)*( 1s).

That is how you get w(2) = angular velocity at 2 seconds. Since alpha is zero after t = 2 seconds, w(3) = w(2).

Note that you can also get the answer using the area under the curve:

w(3) = w(0) + (area under the curve between 0 and 3 seconds). Here w(0)  = 60 rpm, and the area under the curve is the rectangle area between 1 and 2 seconds.