|REAL TEST 4|
|SAMPLE FINAL EXAM FOR AU 12|
|QUIZ 10 (ANSWERS)|
|READ ALL EXAMPLES. SOME PROBLEMS ARE JUST LIKE AN EXAMPLE.|
|CH. 10 ; REAL TEST 4 ; TAKE HOME TEST 3 FALL '13|
|CH. 10 MULTIPLE CHOICE 15 (Equilibrium) + more TBA - Exercise/problems Torque: 2 Newton's Second Law of Rotation: 6, 7 , 8, 11, 12 , 14 (See example 4) Power 19, Angular Momentum 22, Conservation of Angular Momentum 28, 34, 74, Equilibrium 42, 43, 46, 47|
|TURN IN: Multiple Choice 15 : exercise/problems 2 , 7 , 8, 11, 12, 19, 22, 34 (Try 74 !) 42, 43,|
|* DISCUSSIONS PROVIDED.|
|SELECTED DISCUSSIONS TO exercises/problems
ALL PROBLEMS ARE DUE EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE DESIGNED TO HELP YOU DO RELATED EXERCISES.
|2. net torque = (2 m)*sin 30 *12.0 N - (5.00 m)*(8.00 N). Note the minus sign between the terms. Torque components that cause CCW (counterclockwise) rotation are considered positive and torque components causing clockwise (CW) motions are considered negative. See figs. 10.3 and example 10.1|
|7. See example 10.2. Torque = (1.15 m)*(35.0 N) = I*alpha. Find alpha. See table 9.2 for I about an END axis for slender rod with length L = 2.50 m.|
|8. | Torque | = (0.125 m)*F =
I*|alpha|, assuming a force of magnitude F is applied at the rim
and is tangent to wheel.
Alpha can be found from: 0 = (220)*6.28*/60 rad/s - |alpha |*time, where time = 45.0 seconds. Find |alpha| and note alpha is negative. Substitute into the first equation to get |torque| and also F , the magnitude of the slowing friction force. Consult table 9.2 to get I.
|11. See example 10.3 and class notes.|
|12.You could use energy methods to solve :
Go to Quiz 9 and see a string of
problems using it in different ways. You pick the best way for this
problem; the correct choice is clear. See:
42, 43, 49, 50, 52, 51, Quiz 9 . In particular see 42. USE CONSERVATION OF ENERGY :
(1/2)*mvi2 + (1/2)*I*wi2 + mgyi = (1/2)*m*vf2 + (1/2)*I*wf2 + mgyf. , where I = (1/2)*M*R2, and I is the unknown moment of inertia of the cylindrical pulley. Note: w =v/R. Radius R and final linear speed of the hanging mass are all given and m = 15.0 kg. The system begins from rest so initial kinetic energies are zero but please note yi = 4.00 m and final value of y is ZERO. The final linear speed is 3.50 m/s.
(a) Find final w = v/R after falling distance 4.00 m from rest.
(b) Find I.
Alternative method from Ch. 10: Find alpha, using v2 = 2*a*h, where v = 3.50 m/s and h = 4.00 m. Write ma = mg - T to find tension T in the string. Then write I*alpha = R*T, where alpha = a/R. Find I.
|19. (a) I*alpha = torque =
(4.40 m)*(25.0 N). Find alpha. The find w from w = alpha*t
(b) Work = |force|*distance = |force|*R*theta, where theta = (w/2)*t, where t = 20.0 seconds and w was found in part (a). You can also get theta from (1/2)*alpha*t2.
(c) Power = torque*average w = torque*(w/2)., where w and torque are mentioned above.
|THE NEXT PROBLEMS/DISCUSSIONS, POSTED SOON, DEAL DIRECTLY WITH ANGULAR MOMENTUM, AN IDEA IN THE "BACKGROUND" UP 'TIL NOW.,|
|22. L = I*w, where I = I = (2/5)*m*R2.|
|The next deal with conservation of angular momentum|
Ii*wi = If*wf
(1/2)*MR2*wi = [(1/2)*MR2 +m*R2]*wf, where m = 70.0 kg , M = 120 kg and R = 2.00 m. Find wf.
|42. See class notes. I did this in
class !! See also example 10.13. Note in that example the cable leans
rightward instead of leftward as in figure 10.60. See also this sample
exam problem. After clicking the link scroll down to
problem 5. Problem 42 is just like the sample exam problems except
you must add the torque exerted by the weight of the load at the
right end of the beam. That torque takes the form L*Mg, where M is
the load mass. Here you are given Mg =
300 N. Thus , 0 = L*T*sin 37 - (L/2)*mg - L*Mg , where m is the mass of the beam. Solve for tension T.
|43. Check back soon but try finding examples that might look like this last turn in, including likenesses to previous Quiz 10 problems.|