QUIZ 12 - CH 10 - ROTATIONAL DYNAMICS : TORQUE, ROTATIONAL AND TRANSLATIONAL KINETIC ENERGY,  ANGULAR MOMENTUM, NEWTON'S SECOND LAW OF ROTATION,  CONSERVATION OF ANGULAR MOMENTUM  * means try out only but still on exam

REAL TEST 4.

1**, 2**, 4**, 8**,  16**, 17** (TRY 73**), 70**, 20**, 21**,68**,22**, 34*, 33**, 36**, 37**, 43**,45**,46**, 48**, 49**, 95**
WE  WISH TO GROUP THESE PROBLEMS INTO IMPORTANT SEVERAL CATEGORIES:

1. Rotating Pulley, String, and a Single Translating Object connected to it.
2. Rotating Pulley, String and Two OR MORE Translating Objects connected to it.
3. Single Rotating and Translating object connected to a string (YO YO)
4. A Single Object Rolling without slipping, on level surface or an incline or curve.

In all these problems you will generally use  

a. Conservation of Energy.
b. Newton's Laws of Translation and Rotation. 

WE WILL CLASSIFY PROBLEMS BELOW ACCORDING TO THIS SCHEME WHERE POSSIBLE. For example # 14 would be classified  as 1 a or 1 b since you can  either use Conservation of Energy or Newton's Law's  of Translation and Rotation. 

THIS IS A WORK IN PROGRESS : CHECK BACK SOON.

DISCUSSIONS
1**, 2**, 4**, 8**,  16**, 17** (TRY 73**), 70**, 20**, 21**,68**,22**, 34*, 33**, 36**, 37**, 43**,45**,46**, 48**, 49**, 95**
1. ** DO These exercises on at a time, an connect the dots mentally. Note the torque in (f)  is ZERO since the  force goes through the axis . THIS IS IMPORTANT AND WILL REPEATEDLY BE ENCOUNTERED when you selectively choose which force to use in  CHAPTER    11, STATICS. 
2.** I will make an interesting comment only . If I had chosen the axis a the right end of the beam, the force of magnitude F1 would exert NO TORQUE.  
4.** We mentioned the force of magnitude 11.9 N exerts no torque about the axis through the center. The 14.6 N-force twists disk clockwise---R*sin 40 is the moment arm  for that force)---and the 8.50 N-force rotates counter clockwise. 
8.** CLOCKWISE ROTATION, TORQUE POINTS OUT. Right fingers rotate counter clockwise and your right  thumb points OUT. Moment arm = L*sin 37. 
16.**   GROUP 1a or 1b
(a)  Use I*alpha = R*T or I*(a/R2) = T and ma = mg - T.  Solve these simultaneously for T and a.
(b) Use v2 = 2*a*h from Ch. 2.
You can also use conservation of energy:
mgh = (1/2)*m*v2 + (1/2)*I*w2 where w = v/R. 
(c)  Use Ch. 2 methods.
(d) This is critical for ENGINEERING 36 "Engineering Mechanics -Statics" . On the Pulley the vertical forces sum to  ZERO: 0 = N - M*g - T , where N is the magnitude of the normal force on the pulley from the axle.
Note:  Rotational Inertia I restricts the translational acceleration downward  such that a < g. 
17.** GROUP 2b 
Labels: A, block on table; B, falling mass  and P, pulley.
(I) MA*a = TA,  (II) I*(a/R2) = TB - TA, (III) MB*a = MB*g  - TB  . 
(a) and (b) : Solve these three equations simultaneously for a and  two tensions. TIP: Add the equations to cancel the tensions, go back and substitute.
(c)  ENGINEERING 36 "Engineering Mechanics -Statics" : Vertically: 0 = Fv - Mg -  TB .  Horizontally: 0 = Fh - TA.  
70.** GROUP  1b
M*a = Mg*sin 36.9 - T -  uk*Mgcos 36.9, where uk is the coefficient  of kinetic friction. 
 I*(a/R2) = T
Note: Rotational Inertia I restricts the translational acceleration down the incline such that a < g*sin theta,  the result for a block sliding down an inclined plane without being connected  to a string wrapped around a pulley and without friction.
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20.** GROUP  3a
mgh = (1/2)*m*v2 + (1/2)*I*w2 where w = v/R and I = m*R2, useful for simplifying purposes.   
21.**  GROUP  4a
KEtotal  = (1/2)*m*v2 + (1/2)*I*w2 , where w = v/R and I is given by formulae for several objects. 
Find [(1/2)*I*w2 ]/KEtotal   in each case. Any guesses on  which is the largest ? smallest? What happens ion the limit  I  goes to infinity?
SEE EXAMPLE 10.5 FOR AN APPLICATION OF CONSERVATION OF ENERGY TO  AN INCLINE 
68**. GROUP  1b
PULLEY:  I*(a/R) = r*F -RT, where F is the magnitude of  applied force and T is  rope  tension magnitude. Note tension acts at distance R  = 0.25 m from axis whereas the applied force acts   0.12 m away. 
Also m*a = T - mg, the equation for the rising mass attached to  rope.
22.** GROUP 4b
m*a = mg*sin theta - fs.  
 I*(a/R2) =  fs
Solve simultaneously for a and  fs .
 fsmax
= us*N, where us is the coefficient of static friction. Evaluate this from what you know from chapter 5. 
Check results for doubled mass.
34.** Mathematically, Group 1and b---Rotating Pulley (wheel), Applied Torque portion of principle.
I*alpha = L*F allows you to get alpha, from the entire range of qu4eesiotn are answered. . 
We skipped lecture discussion of power. Let us go over this now.
differential work = dW = (torque)*d theta
Integrating both sides we get
W = (1/2)*I*w2 = rotational KE , using torque = T*(alpha) and a couple other calculus tricks; consult the textbook
Note: W = Torque*total theta
Instantaneous Power = dW/dt = (1/2)*2*I*w*dw/dt = I*alpha*w = torque*w.
(a) and (b) Use kinematics to get alpha, which you know from above discussion and w
(c) W = (1/2)*I*w2 . Note:  W = Torque*total theta
(d) Average power = (change in W)/time =   (1/2)*I*w2/time
(e) Instantaneous Power = torque*w.
33.** Mathematically, Group 1and b---Rotating Pulley (Propeller), Applied Torque portion of principle.
I*alpha = L*F allows you to get alpha, from the entire range of qu4eesiotn are answered. . 
We skipped lecture discussion of power. Let us go over this now.
differential work = dW = (torque)*d theta
Integrating both sides we get
W = (1/2)*I*w2 = rotational KE, using torque = T*(alpha) and a couple other calculus tricks; consult the textbook.
W = Torque*total theta
Instantaneous Power = dW/dt = (1/2)*2*I*w*dw/dt = I*alpha*w = torque*w.
(a) Find alpha kinematically (Ch. 9) then use it to complete the question
(b) Pure kinematics Ch. 9.  
(c) W = Torque*total theta
(d) (1/2)*I*w2 = rotational KE; compare with part (c). 
1**, 2**, 4**, 8**,  16**, 17** (TRY 73**), 70**, 20**, 21**,68**,22**, 34*, 33**, 36**, 37**, 43**,45**,46**, 48**, 49**, 95**
ANGULAR MOMENTUM AND ITS CONSERVATION
36.** Reviewed in class, but still as important as ever. For a single particle I = mr2, where r is the perpendicular distance from the axis of rotation to the particle that is rotating about it.  In general Lz = I*w. 
If the point particle was rotating by itself, then it has Lz = mr2*w 
If the point particle was rigidly attached to a rotating solid cylinder about an axis through the center and perpendicular to the cylinder's face, then Itotal = Icylinder + mr2  and Lz = = Itotal*w
37.** (a) (R*sin 36.9)*p = | Lz |
(b) (Rcos36.9)*mg =| torque| = |dLz/dt | 
43.** More complex than you think: I includes a central cylinder (hands  and arms wrapped) and a long, slender cylindrical  rod (hands and arms outstretched).  Ii = 0.40 kg*m2 + Irod and  If =  0.40 kg*m2 +  Icylinder  .  See table 9.2.
Liz = Lfz .Ii*wi = If*wf
45.**  
Liz = Lfz .
Ii*wi = If*wf
Ii = Icylinder .   
If  = Iperson  +  Icylinder  .  

Note: For a single particle, representing the person, I = mr2, where r is the perpendicular distance from the axis of rotation to the particle that is rotating about it.  In general Lz = I*w. 
If the point particle was rotating by itself, then it has Lz = mr2*w 
If the point particle was rigidly attached to a rotating solid cylinder about an axis through the center and perpendicular to the cylinder's face, then Itotal = Icylinder + mr2  and Lz = = Itotal*w
(b) Use (1/2)*Iw2 to get the total KE before and after;  compare.  Kinetic energy was lost to heat during the impact as the person and disc (turn table)came to relative rest and rubbed against each other.. 

THE FOLLOWING PROBLEMS CAN BE GROUPED TOGETHER---45***( in hindsight), 46**, 48**, 49**, 95**---VIA THE IDEA   OF CONSERVATION OF ANGULAR MOMENTUM INVOLVING A POINT PARTICLE COLLIDING WITH AN  EXTENDED RIGID BODY. THE AXIS OF ROTATION RUNS THROUGH OR ALONG THE  RIGID BODY. 
46.*** See example 10-12** and  #95.** 
Liz = Lfz .
Lmudi + Ldoori =  Lmudf + Ldoorf
D*pi  +  Idoor*wdoori  = D*pf + Idoor*wdoorf
D = width of door/2  = Width/2 =  perpendicular distance from the axis to the line of action of the mud's linear momentum; pi = mvi, where vi= 12.0 m/s and m is the mud mass.  
p=   mvf, where vf  = D*wdoorf ,  where wdoorf  is the final angular speed of the door.  
Note: D*pf + Idoor*wdoorf   =  mD2*wdoorf   +  Idoor*wdoorf   = (Imud+ Idoor)*wdoorf
See  Table 9.2***  (d) for Idoor  in terms of Width.
Note: wdoori = 0. 
48.***  Let R = Earth radius,  m = mass of asteroid, M = Earth mas.
Liz = Lfz .
Lasteroid_i + Learth_i =  Lasteroid_f + Learth_f   .
0 + I earth *wi = R*pf + I earth *wf .
 I earth *wi = R*mvf + I earth *wf .
I earth *wi = R*m*R*wf + I earth *wf .
I earth *wi = m*R2*wf + I earth *wf .
I earth *wi =Iasteroid *wf + I earth *wf .
I earth *wi = (Iasteroid  + I earth )*wf .
Prove if the day is 25 % longer then wf = (4/5)w i  .  Solve for Iasteroid    and then m in terms of M, the Earth's mass.
See table 9.2 for Isphere = Iearth.   = (2/5)*MR2.
49.**  See example 10-12*** and  #95.***
Liz = Lfz .
Lball_i + Lbar_i =  Lball_f  + Lbar_f.
D*pi  +  Ibar*wbar_i  = D*pf + Ibar*wbar_f
D =1.50 m = perpendicular distance from the axis to the line of action of the ball's linear momentum;
pi = mvi, where vi= 10.0 m/s and m is the ball's mass. 
p=   mvf, where vf  = - 6.00 m/s since it rebounds in the negative direction, whether left of right in your diagram. . 
Note: wbar_i = 0.
Find wbar_f.
(b) *  Linear momentum is not conserved if you consider the bar and ball  to be the system. Some of the linear momentum gets transferred to the Earth through the pivot which exerts an extra NORMAL  force on the Earth during the collision. See  #28***, Ch. 8 .  That NORMAL force has  the same direction as the initial momentum of the ball and is applied at the pivot  during impact. Thus linear momentum is NOT conserved within the  ball-bar system. That system is   "OPEN" regarding  linear translations during the small time period just before and after collision.
*  On the other hand, even though the pivot may be  rigidly attached to the Earth,  the rotation process at the pivot is considered frictionless; there are no friction forces  between  the axle's  circular  outer  surface  and the same shaped inner pivot surface rubbing against it  angularly as bar swings. Thus  no  angular momentum get transferred to Earth since the   pivot experiences NO torque and the pivot's outer surface exerts  NO torque on Earth at the Earth-pivot interface.  Thus angular momentum is conserved within the  ball-bar system which is "CLOSED"  during the small time period just before and after  collision.