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 = (w_{f} - w_{i})/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) w_{2} = w_{1} +alpha*(t_{2}
- t_{1}), 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
displacement. |
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) . |
21. 4*pi = (1/2)*alpha*t^{2}. Find t. Note: 4*pi = (1/2)*(w_{f} + w_{i})*t, where the initial angular velocity is zero and pi = 3.14... Find the final angular velocity. Then find a_{RAD} 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) a_{RAD } = R*w^{2}, where w is the final angular velocity and a_{t} = R*alpha. These acceleration components are perpendicular to each other; find the magnitude of the net acceleration using the Pythagorean Theorem. |
28. (a) a_{t} = R*alpha ;
note a_{t }< 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*w^{2} = 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 :
KE_{1}
+ U_{1} + W_{P} = KE_{2} + U_{2},
where W_{P} = P*D, D = 5.00 m, KE_{2} = (1/2)*I*w^{2}
, w = v/R and v = 6.00 m/s. I = (1/2)*M*R^{2}. , where mass M can be found from the weight given. The potential energies U can be set equal to zero. |
47. KE_{1} + U_{1} = KE_{2} + U_{2 . }The initial (1) kinetic energy is zero, U_{1} = mgh, U_{2} = 0 and _{ }KE_{2 }=_{ }(1/2)*Iw^{2} + (1/2)*mv^{2}. You can find w and v using the condition: (1/2)*I*w^{2} = 4.50 J and v = R*w. Note: I = (1/2)*M*R^{2}. FIND h. |
MORE TO COME..... |