Chapter 12: Do listed problems.  
Simulations
Quiz 12
Helpers!! Click this link then click the links "Practice Questions" and "Practice Problems" in the left hand column. 
Q12 Part 1: 6*, 14*, 19*, 22*, 25*, 30*, 32, 35, 36, 37, 39*, 
46, 47, 48, 49, 62,67, Q12 Part 2: 72, 84,86*, example 12.21, 90 (EC)
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6. Xcm  = 0.300*0.10/(0.200 + 0.100 + 0.300)   and Ycm = 0.200*0.10/(0.200 + 0.100 + 0.300) . Compute these. Assume the origin is at point A.  
14. 
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 (a) See the previous problem: Xcm  =(0.200*0.100 + 0.200*0.10)/(0.200 + 0.200 + 0.200 + 0.100)   and 
Ycm =(0.200*0.100 + 0.200*0.10)/(0.200 + 0.200 + 0.200 + 0.100) . Compute these. Assume the origin is at point A.

(b) Here are the distances of the rectangles sides: Let the distance between A and B = dAB,  between A and C = dAC , and between A and D = dAD .   The axis is at A.  I =  mB* dAB2 + mC*dAC2   + mD*dAD2    . Note that you must use the Pythagorean Theorem to  get  dAC .   
19. DONE DURING LECTURE SECTIONS.  The magnitude of the net torque = | net torque| = R(30 - 20) N*m. This net torque causes clockwise angular acceleration about the center. 
22. Let get our torques organized. 

Here  are the  forces causing counter -clockwise acceleration (and counter clockwise motion if the disk starts from rest.)
20 (N), down
30 (N), 45 degrees North West

Here is  the force causing clockwise acceleration (and clockwise motion if the disk starts from rest.)
30 (N), rightward.

Now let us find the magnitude of the net torque. Choose counter clockwise (CCW)  to be  the positive direction of rotation.  
|net torque| = pos - neg
net torque =  ( 0.05 m)*sin45*30 N  + (0.05 m)*20 N - (0.10 m)*30 N. 
Note that  NO torque is caused by the 20 N force whose line of action runs through the axis at center.  
25. I*alpha  = |net torque| . Compute | alpha | = 4.0 rad/s2  .   
30. Classic problem allowing you to compute torque using the perpendicular dropped  from the axis to the line of action of the force.  Let CCW be the positive direction of rotation. Let the axis be at the black dot shown (pivot) at rod's left end. The force's line of action runs vertically up and down  the page. Draw the horizontal line segment  between the axis and force's action line crossing the line of action at 90 degrees.  

Now let us find the magnitude of the net torque. Choose counter clockwise (CCW)  to be  the positive direction of rotation. 

 |net torque| = pos - neg
(0.60 m*sin 45)*4.0 N = I*alpha.

Now let us get I about the axis at the rod's left end : 
I = IRod  + Irocket   . 
I  = (1/3)m*L2  + M*L2, where m  and M are the  rod and rocket mass, respectively. 
Find alpha.
32. 
Equilibrium is defined as a system having NO net torque about any axis, and NO net force in any direction. 
In other  words, rotational and translation equilibrium. Let's check them both:

ROTATIONAL EQUILIBRIUM CHECK
Choose for convenience the axis of rotation at the very left and see if the applied torques cause a rotation about that axis. If I choose the left end, then I eliminate the upward force of magnitude 40 N since the line of action goes through that axis.  Having done that, we  compute the torques from the 100 N and 60 N forces about the axis at  left end. The downward 100 N force  rotates the beam clockwise, whereas the the upward 60 N force rotates it counter-clockwise. 
For the 60 N force: |Torque| = d·60 Nm, where d =  distance from the left to right beam end,   i.e.  beam length. 
For the 100 N force: |Torque| = d'·100 Nm, where d' =  distance from the left end to the point on beam where the 100 N force is applied. Look at the book diagram. 
DO these torques have the same magnitude?  I another words,  does   0 = d·60 Nm  - d'·100 Nm?  If so then we have rotational equilibrium.
TRANSLATIONAL  EQUILIBRIUM CHECK
Now you do need all the forces listed in diagram. 
The  forces are directed vertically along the y-axis.  Compute the sum of the forces in the y-direction. Is the sum zero? If so then the system is in translational equilibrium.

Is the beam in equilibrium?  
35. See class notes. All velocities are computed relative to ground (G). Let rightward be the positive direction.   Assume the center of mass (cm) moves rightward.
(a) See class notes and book (section 12.9)  to get the magnitude of the angular velocity w of the rolling tire about an axis at the center. 
(b) For this part, the linear velocity of the point P at the top of the  wheel rim is directed horizontally right.
AT TOP, VPG = VPcm + VcmG   = |wR| + |wR| = 2|wR|, where w and R are the angular velocity and radius of the wheel.  VcmG  = 25.0 m/s = |wR|.
(c) AT BOTTOM, VPG = VPcm + VcmG   = - |wR| + |wR| = ?  Again, note VcmG  = 25.0 m/s = |wR|.
36. KE = KErot + KEcm  .  See equation 12.39
37. See my notes and the discussion in book surrounding Fig. 12.47.  I did an example like this using conservation of energy
39.  
(a) See pages 368-370. Magnitude =  A*B* sin 60, where A = 4.0 and B =  6.0.  The  cross product's direction is either into or out of the page.   With your right fingers, wrap the vector-A into the vector - B and your thumb points out, the cross product's direction. Use  identical methods in part (b).  
46. See equation 12. 48. Here's how you do this following my "easy way"  covered in class. Let's deal with this in two parts

DIRECTION:
The position vector of magnitude r is drawn from the origin (tail of vector) to the point where the bird is flying at that moment (tip of vector). As you can see the position vector is "rotating" clockwise at that moment. If you  wrap your right fingers clockwise, you thumb points IN (in the negative z-direction). Thus IN is the direction of the angular  momentum of magnitude L.   
MAGNITUDE:
Drop a perpendicular from the axis at the origin to the line of action of the linear momentum  of magnitude p = mv. That perpendicular will make an angle of 45 degrees with the positive x-axis. That's because the linear momentum and line of action make an angle of -45 degrees with the  positive x-axis. Now find the angle Ø that the position  vector makes  with the positive  x axis.  Note:  tanØ = 2/1. Find Ø in degrees.  Compute  Ø' = Ø - 45.  This is the angle between the position vector and the perpendicular.  It can be shown that the length  of the perpendicular segment between the origin (axis) and the line of action is d = rcosØ' .  You must compute r from the fact that 
r2 = 12 + 22. The angular momentum  has magnitude L = d·p = d·mv.  

The other way of getting the magnitude and direction of the angular momentum  is computing the cross product as shown in section 12.10. 
47. See equation 12. 48. Again,  here's how you do this following my "easy way"  covered in class. We  deal with this in two parts. 

DIRECTION:
The position vector of magnitude r is drawn from the origin (tail of vector) to the point where the bird is flying at that moment (tip of vector). As you can see the position vector is "rotating" counter-clockwise at that moment. If you  wrap your right fingers counter-clockwise, your thumb points OUT (in the positive z-direction). Thus OUT is the direction of the angular  momentum of magnitude L.   
MAGNITUDE:
Drop a perpendicular from the axis at the origin to the line of action of the linear momentum  of magnitude p = mv. That perpendicular is along the positive  x-axis. That's because the linear momentum and line of action are pointed upward along the positive y-axis. It should be obvious that the length  of the perpendicular segment between the origin (axis) and the line of action is d = 3.0 m.   The angular momentum  has magnitude L = d·p = d·mv.  

The other way of getting the magnitude and direction of the angular momentum  is computing the cross product as shown in section 12.10. 
48. 
DIRECTION: Wrap your right fingers counter-clockwise in the direction of rotation. Thus your thumb points OUT. That is the direction of the angular momentum of magnitude L. 
MAGNITUDE: L = Iw. In this case w is positive because  the rotation is counter clockwise. So I do not have to take the absolute value. (Otherwise I would.)  The moment of inertia about the center is  I = ML2/12 from Table 12.2. 

Remember to convert the angular velocity from  from rpm to rad/s. 
49. 
DIRECTION: Wrap your right fingers clockwise in the direction of rotation of the disk.  Thus your thumb points in the negative x direction.  That is the direction of the angular momentum of magnitude L. 
MAGNITUDE: L = | Iw | .  The moment of inertia about the center is  I = MR2/2 from Table 12.2. 

Remember to convert the angular velocity from  from rpm to rad/s. 
62. This equilibrium problem is in a slightly different category from  problems 31, 63, and 66. (The latter three take similar forms mathematically and are similar to example 12.15. ) 

This problem is similar to example 12.16 and problem 64 below. 

Equilibrium is defined as a system having NO net torque about any axis, and NO net force in any direction. 
In other  words, rotational and translation equilibrium. Let's check them both.

ROTATIONAL EQUILIBRIUM CHECK to find  N2
Let the axis of rotation be the beam's  left end.  If I choose the left end, then I eliminate the upward normal force N1 from support 1.  See example 12.16 for useful pictures. There are three torques that tend to rotate the beam about the left end.

For the downward weight of the beam.  |Torque| = d·mg , where d =  distance from the left to the center,   i.e.  half the beam length. This torque tends to rotate the beam clockwise. (m = beam mass. )

For the downward force due to the student: |Torque| = d'·Mg, where d' =  distance from the left end to the point on beam where he stands as shown in the diagram. M = student's mass. This torque tends to rotate the beam clockwise. 

For the upward normal force of magnitude N2 at support 2.  |Torque| = d''·N2, where d'' =  distance from the left to right beam end,   i.e.  beam length. This torque tends to rotate the beam counter-clockwise. 

We can easily find N2  by requiring that the torques sum to zero. We usually say that the torques causing counter- clockwise rotation are positive and those causing clockwise rotation are negative.  Thus,  
0 =  d''·N2  -  d·mg - d'·Mg.  Find  N2  . 

TRANSLATIONAL  EQUILIBRIUM CHECK to find  N1 . 
The net force in the y-direction is zero, so 

0 =  N1 +  N2  - mg    - Mg.  Find N1 . 
67. Hint posted later, but part (a) is similar to  example 12.14.  I =MR2 (Hollow cylinder) .  See Table 12.2. Part (b) can easily be done using conservation of energy:   mgh = ½Iw2 + ½mv2 where m =  hanging block mass. I = MR2 .  Note w = v/R. From this find v after the hanging block drops the given  distance h from rest. 
72. Done in class. See example 12.12.  Use I*alpha = net torque, equation 12.31. Find alpha from Table 12.1 equations, page 342.  Then find the torque.  Then set torque magnitude = R*f, where f = friction force. Find f. 
84.  See example 12.23. We will give the hint with much less discussion: 
Use conservation of angular momentum immediately  before and after the collision at door edge:
R·mVB =  Iwf , where VB  is the bullet  velocity just before collision. R is the door length.
Solve for wf  .  Note: I = (1/3)MR2 + mR2 taking into account the door and bullet moment of inertias about the axis at edge.  
86. See example 12.22. Lf = (IB   + IT)*Wf =IT*Wi where IB = 2*MB*R2 and IT is found in table 12.2. R = Turn Table (T) radius. B stands for s block whose mass is 0.500 kg = MB. Note that you need not covert from rpm to rad/s  because you are not calculating anything like energy measured in Joules. 
90. More hints soon ! See example 12.21  as a conceptual guide.  
Li = Lf. 
Iiwi = Ifwf . 
Ii = ½MiRi2 + (2)(1/3)(md2 + (1/12)mL2), Mi = given initial effective mass = 40 kg, Ri = given radius of the initial cylindrical torso, L = given arm length, d = distance between the center of the arm and the center of the torso; d =  Ri + L/2, which comes from the  parallel axis theorem.
If  =  ½MfRf2   where the final torso radius and effective mass are given.  
Find wf  . 
Read and understand example  12.21. The the line of action of the gas forces  pushing the masses outward go through the axis; thus these forces  exert no torque and angular momentum is conserved.    
More hints  will be posted soon !