QUIZ 11 - CH 9 - ROTATIONAL KINEMATICS AMD ENERGY; MOMENT OF INERTIA
 (SAMPLE TEST 4 ; REAL TEST 3)
 * means try out only but still on exam
8, 16, 18, 20, 21, 24, 28, 44, 47, 54, 56, 68, 83, 84, 87,  89; TRY 30  for the basic definition  of I about various  axes.)
THEMES: KINEMATICS,  ENERGY CONSIDERATIONS; PARALLEL AXIS THEOREM; GENERAL PROBLEMS.
WE  WISH TO GROUP THESE PROBLEMS INTO IMPORTANT SEVERAL CATEGORIES:

1. Rotating Pulley, String, and a Single Translating Object connected to it.
2. Rotating Pulley, String and Two OR MORE Translating Objects connected to it.
3. Single Rotating and Translating object connected to a string (YO YO)
4. A Single Object Rolling without slipping, on level surface or an incline or curve.

In all these problems you will generally use  Conservation of Energy. 


THIS IS A WORK IN PROGRESS : CHECK BACK SOON.
 MORE DISCUSSIONS SOON
  8. SEE TABLE 9.1.

 (a) alpha = (wf - wi)/change in time, where change in time = 7.00 seconds. Note the final angular velocity is positive and the initial angular velocity is negative.

(b) w2 = w1 +alpha*(t2 - t1), in general.      The SPEED is decreasing between t = 0 and when the angular velocity is zero,  Given alpha , find the time when the angular velocity is zero by setting w2 = 0 and w1 = -6.00 rad/s. For time greater than this time, the angular speed is increasing.  

(c) SEE TABLE 9.1. USE  angular displacement = (1/2)*( (wf + wi)*(7 s)
16.  SEE TABLE 9.1.
Between  t =  0  and 2.00 s, the wheel has a constant non-zero, positive alpha.   For t > 2.00 s, alpha is constant and negative.  
(a) To get total angle, find the angular  displacement between 0 and 2.00 seconds and add this value to 432 rad.

Between 0 and 2.00 seconds use equation 9.11 to find angular displacement.
(b) Find the time it takes the grinder to come to rest after circuit is tripped.  Use the angular displacement of 432 rad during this time period:  432 = (1/2)*(wf + wi)*time, where wf  = 0. Find time. Note:  wi= 24 rad/s + (30.0 rad/s^2)*( 2.00 seconds) from part (a). Solve for time and add to 2.00 seconds. 

(c) alpha = (wf -  wi)*time,  Plug in values for the initial angular velocity and the time from part (b). The final angular velocity is zero.   
18.  
(a) v  = 25.0 cm/s = R*w. Find w (rad/s) and convert to RPM ; R is given.
(b) a = g/8 = R*alpha 
(c)  h = s = arc length swept out by pulley =  3.25 m = R*theta, where theta = angular displacement. 
20. The linear speed v is constant for all tracks; that means the angular velocity must change as we scan from track to track. 

(a) v = R*w, where R = radius of  track. Find w for the inner most (smallest R) and outmost track (largest R) .
(b) Compute v*time; convert minutes to seconds.
(c) alpha = (change in w)/time, where time = 74 min ( convert to seconds) and (change in w) is the difference between the two values of w you found in part (b).  Note: Your answer will be negative. 
21.  4*pi = (1/2)*alpha*t2.  Find t.   Note: 4*pi =  (1/2)*(wf + wi)*t, where the initial angular velocity is zero and pi = 3.14... Find the final angular velocity.   Then find aRAD  corresponding to the final angular velocity. Note you can also do this problem using equations 9.12 to get the final angular velocity given angular displacement = 4*pi, where pi = 3.14.... (do not round off until the end !)
24. (a) Use equation  9.7. (b) Use equation 9.10 (c)  v = R*w, where w is the final v angular velocity.  

(d)   aRAD  = R*w2, where w is the final angular velocity and   at  = R*alpha.  These acceleration components are perpendicular to each other; find the magnitude of the net acceleration  using the Pythagorean Theorem. 
28.  (a) at = R*alpha  ; note  at < 0 . (b)  At 3.00 s, v = R*w(3), where w(3) is the angular velocity at t = 3.00 s. 
 Note: alpha = [w(3) - w(0)]/( 3.00 s).  Find w(0).   

(c) Use equation 9.10.
(d) R*w2 = g.  Find w, then compute w = w(3) + alpha*time. Compute time and add to 3.00 seconds
44.. SEE EXAMPLES 9.7, 9.8  : KE1 + U1 + WP = KE2 + U2, where WP = P*D,  D = 5.00 m,   KE2 = (1/2)*I*w2 ,  w = v/R  and v = 6.00 m/s. 
 I = (1/2)*M*R2. , where mass M can be found from the weight given.  The potential energies U can be set equal to zero. 
47.   SEE EXAMPLE 9.8:  HERE YOU ARE FINDING  h.  Let m = mass of falling stone and M = pulley mass.
  KE1 + U1 = KE2 + U2  . The initial (1)  kinetic  energy is zero,  U1 = mgh,  U2 = 0 and    KE2  =   (1/2)*Iw2  + (1/2)*mv2.  
You can find w and v using the  condition: (1/2)*I*w2  = 4.50 J and v = R*w.  Note:  I = (1/2)*M*R2. FIND h.
MORE TO COME.....