|REAL TEST 4|
|SAMPLE FINAL EXAM FOR AU 12|
|QUIZ 9 (ANSWERS)|
|READ ALL EXAMPLES. SOME PROBLEMS ARE JUST LIKE AN EXAMPLE.|
|SAMPLE TEST3 ; REAL TEST 3; REAL TEST 4|
|CH. 9 MULTIPLE CHOICE TBA - Exercise/problems
Angular Velocity/Acceleration 9, constant angular
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
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.
aR = 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 at = 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)*mvi2 + (1/2)*I*wi2 + mgyi = (1/2)*m*vf2 + (1/2)*I*wf2 + mgyf. , 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 yi = 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
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 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 yi = 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)*mvi2 + (1/2)*I*wi2 + mgyi = (1/2)*m*vf2 + (1/2)*I*wf2 + mgyf. , 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)*mvi2 + (1/2)*I*wi2 .
|50. In the case of the marble, NOTE :
w =v/R. mgH
= (1/2)*m*vf2 + (1/2)*I*wf2
, where I = (2/5)*m*R2.
In the block of ice case, mgH = (1/2)*m*vf2 . Compare your two results for the final linear speeds.
(c) Compare (1/2)*m*vf2 + (1/2)*I*wf2 (with w =v/R) for marble and (1/2)*m*vf2 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)*mvi2 + (1/2)*I*wi2 + mgyi = (1/2)*m*vf2 + (1/2)*I*wf2 + mgyf. , 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*vf2
= (1/2)*m*vf2 + (1/2)*I*(vf
/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*(vf /R)2 / [ (1/2)*m*vf2 + (1/2)*I*(vf /R)2 ] in each case. Note mass m will cancel.