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REAL TEST 4

SAMPLE FINAL EXAM FOR AU 12

QUIZ 9 (ANSWERS)

READ ALL EXAMPLES. SOME PROBLEMS ARE JUST LIKE AN EXAMPLE.

CH. 9

SAMPLE TEST3 ; REAL TEST 3; REAL TEST 4

CH. 9 MULTIPLE CHOICE TBA - Exercise/problems Angular Velocity/Acceleration 9, constant angular acceleration
17, linear and angular variable relationships 26, 29 Kinetic Energy of Rotation and Moment of Inertia I 30, 42, 43 Rotation about moving Axis-Hint --Use Conservation of Energy 49, 50, 52 Rolling 51

TURN IN: Exercise/problems 9, 17, 26, 29, 30, 42, 43, 49, 50, 52, 51

* DISCUSSIONS PROVIDED.

SELECTED DISCUSSIONS TO exercises/problems BELOW.
ALL PROBLEMS ARE DUE EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE DESIGNED TO HELP YOU DO RELATED EXERCISES.

BELOW COUNTER-CLOCKWISE IS THE POSITIVE DIRECTION OF MOTION. Thus if rotation is CCW, the angular velocity w > 0; otherwise w < 0 (cw rotations) . WITH THESE CONVENTIONS, the angular acceleration alpha is positive in the case of speeding up in CCW rotation, etc

9. SEE TEST #1, PROBLEM 3. (a) angular velocity = w = 1900rev/min*6.28*rad/rev*(1/60 min/s) (b) angle = w*time
(c) 18 rad/s * (1/6.28) rev/rad* 60 s/min.

17. The mass and radius are not directly relevant though they give you an idea of the physical scope of typical automobile fly wheels. (a) Assume wheel does not stop: w2 = w1 -|alpha|*time where time = 30.0 seconds. This would be great if we knew the first angular velocity but we do not so let's attack this more directly using
theta = 200 rev*6.28 rad/rev = [(w1 + w2)/2]*time. Solve for second angular velocity given the first w = 500 rpm; please convert rpm rad/s !!
(b) It's time to solve for and use alpha: w2² = w1² - 2*|alpha|*theta, where theta = 200 rev before conversion to radians ; find alpha and solve for new time' if 0 = w1 - |alpha|*time' and new theta' if 0 = w1² - 2*|alpha|*theta'.

26. Note 2 types of accelerations, center seeking (centripetal) , and tangential, which add upon up vectorially to the net acceleration in a direction neither radial or tangent, but pointed inward away from the tangent because the centripetal component "pulls" the resultant toward the center.
a_R = V²/R, where V is the tangential velocity given by w*R, with R the radius & w the angular velocity. The initial angular velocity is zero and the final is given by w² = 2*(alpha)*theta , where theta -= 60 degrees (convert to radians!!) and alpha -= 0/600 m/s². Meanwhile, tangential acceleration a_t = R*alpha = constant of motion . REPEAT FOR 120 DEGREES.

29. ratio of forces = ratio of centripetal accelerations = ratio of square of speeds = (speed f)² /(speed i)² , where speedi is the initial linear velocity and speedf is the final linear velocity. In each case the linear velocity is R*w, where R and w are given (convert w to radians/sec)

30. Ibaton = Irod + 2*Icap. Note: Icap = m*(L/2)². and Irod (about the central axis perpendicular to rid) ) is given in Table 9.2 by the formula for a slender rod.

For the next two problems, see examples 9.7, 9.8 AND 9.9. READ PAGE 281 about gravitational potential energy being defined at the center of mass in vignette 'POTENTIAL ENERGY OF A RIGID BODY." USE CONSERVATION OF ENERGY ~ Import knowledge and formulas from QUIZ 7 remembering we now have rotational kinetic energy given by (1/2)*I*w². NOTE : v =w/R below.

42. USE CONSERVATION OF ENERGY :
(1/2)*mv_i² + (1/2)*I*w_i²+mgy_i = (1/2)*m*v_f² + (1/2)*I*w_f²+ mgy_f. , where I = (1/2)*M*R², and unknown M is the cylindrical pulley's mass. NOTE : w =v/R. Radius R and final linear speed of the hanging mass are all given and m = 3.00 kg. The system begins from rest so initial kinetic energies are zero but please note y_i= 2.50 m and final value of y is ZERO. Find I and M.

43. (a) The cylinder's axis is stationary, meaning it does not drop downward but the 1.50 kg weight does.
USE CONSERVATION OF ENERGY :
(1/2)*mv_i² + (1/2)*I*w_i²+ mgy_i = (1/2)*m*v_f² + (1/2)*I*w_f² + mgy_f. , where I = (1/2)*M*R², and M is the cylindrical pulley's mass. NOTE : w =v/R. Radius R, M and the final linear speed of the hanging mass are all given and m = 1.50 kg. The system begins from rest so initial kinetic energies are zero but please note y_i= h > 0 and final value of y is ZERO. Find h.
(b) Find final value of w = (final linear velocity)/R. Note | w | = (final linear speed)/R.

FOR THE NEXT 2 PROBLEMS, SEE EXAMPLE 9.10

49. USE CONSERVATION OF ENERGY :
(1/2)*mv_i² + (1/2)*I*w_i²+ mgy_i = (1/2)*m*v_f² + (1/2)*I*w_f² + mgy_f. , where I = (2/3)*m*R², R is the radius and m is the spherical shell's mass. NOTE : w =v/R. The initial value of y and final kinetic energy are zero. Note h = final y value = 5.00 m.
(a) Find the initial linear velocity and angular velocity = ( linear velocity)/R.
(b) Find (1/2)*mv_i² + (1/2)*I*w_i².

50. In the case of the marble, NOTE : w =v/R. mgH = (1/2)*m*v_f² + (1/2)*I*w_f² , where I = (2/5)*m*R².
In the block of ice case, mgH = (1/2)*m*v_f² . Compare your two results for the final linear speeds.
(c) Compare (1/2)*m*v_f² + (1/2)*I*w_f²(with w =v/R) for marble and (1/2)*m*v_f² for ice block. Which is larger? Or are these quantities the same?

52. SEE EXAMPLE 9.9:
Use w =v/R in the equation below. USE CONSERVATION OF ENERGY :
(1/2)*mv_i² + (1/2)*I*w_i²+ mgy_i = (1/2)*m*v_f² + (1/2)*I*w_f² + mgy_f. , where I = M*R².
The initial kinetic energies, translational and rotational, are zero but the initial value of y = 0.750 m. The final y value is ZERO. Find the final value of angular velocity w by using the substitution v = w*R.

51. Total kinetic energy = (1/2)*m*v_f² + (1/2)*I*w_f² = (1/2)*m*v_f² + (1/2)*I*(v_f/R)², where I = (1/2)*m*R² , (2/5)*m*R², (2/3)*m*R² or (1/2)*m*[(R/2)² + R²] . Find ratio:
1/2)*I*(v_f/R)² / [ (1/2)*m*v_f² + (1/2)*I*(v_f/R)² ] in each case. Note mass m will cancel.