|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.
|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.
|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
(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) .
|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
(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 :
+ 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.
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.....|