Chapter 6 -- Conservation of Angular Momentum

 

 

 

Updated ---  4 October 2006

 

 

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Hint

Answer

Update Date

6.1

Statics problem; nothing tricky.

a) T = 2000N (along cable)

b) Rcx = 1600 N (right), Rcy = 1680 N (up) 

 

6.2

The key here is to not be mislead by the shape of the chute.(What a poor drawing!!!) In reality, it is curved and the line of action of the linear momentum carried out by the grain at B does not pass through G or C. Also you will probably need to assume that the system is steady state? Is this reasonable? Why?

Cx = 110.2 lbf →
Cy = 306.8 lbf  ↑
By = 423.0 lbf  ↑

03Oct05

6.3

Assume: (1)  There is no friction between the horizontal bar and the inclined surface.  (2) The two chains are parallel to the inclined surface.

After picking your system, the next most important decision is selecting the point about which you will calculate the angular momentum. Any point will work but some are "easier." Because there is no friction at between the bar and the inclined surface, the only reaction force that the surface can exert on the bar is a single force that is perpendicular to the inclined surface. (Without the reaction force on the inclined surface you cannot get this system to remain stationary.)

F1 = 106.7 lbf;
F2 = 116.7 lbf
R = 17.32 lbf acting on the bar and normal to the inclined surface.

03Oct05

6.4

Note that the monorail moves because of the friction force between the wheel and the track. Assuming the wheels don't slip, then the maximum friction force is governed by the maximum static friction force on each wheel. Think, what's going to happen to the normal forces between the wire and the monorail wheel? If you could increase the acceleration (or friction force) without limit what would happen to normal forces on the wheels? Ever heard of a monorail "wheelie?" Now the acceleration is limited by the size of the friction force that can be generated by the wheels on the wire.

dVG/dt = 0.3396 g ←

03Oct05

6.5

(a) Straight forward application of LM and AM. Be sure and show why the dP/dt and dL/dt terms equal zero. Note that the reaction of the ground on the wheel is directly below the trailer center of mass.


(b) Problem is to find vertical hitch force on trailer to keep trailer translating to the right with an acceleration of 4.5 m/s2. Complete solution has three unknowns; thus requires use of LM and AM equations. Although any point will work suggest you calculate AM about either point G, A, or B.

(a) Hitch forces = 0 and wheel supports all of the weight.

(b) Vertical force of ground on wheel is 10,179 N up. Vertical force of hitch on the trailer is 1350 N down. The hitch has to pull down to keep the trailer translating. Horizontal force on trailer is 4050 N to the right.

03Oct05

6.6

 

 

 

6.7

 

 

 

6.8

 

 

 

6.9

This is an open, steady-state system with water flowing in at the bottom and flowing out at the horizontal discharge. Note that the flowing water carries both linear and angular momentum across the system boundary. Also note that the pressure around the hydrant is atmospheric.

(a)  Application of conservation of LM in the x direction. (Recall that the pressure at the discharge from the hydrant is equal to the atmospheric pressure.)
(b)  Same as (a) only this time in the vertical direction. The pressure at the inlet to the hydrant is not equal to atmospheric pressure.
(c)  Application of conservation of AM around the center of the base of the hydrant. Why the base? Because the moment of Rx and Ry about the base is zero and you can solve for Mbase directly.  Would this be true if you calculated the angular momentum about the center of the hydrant outlet? No.

Rx = 216.0 lbf  →

Ry = 11,810 lbf  ↓

Mbase = 324.0 lbf-ft  CW

All of the reactions are applied by the ground on the base of the hydrant.

 

04Oct07

6.10

Application of cons. of AM and LM to a closed SS (or non-moving) system. Since you want to know information about the reactions at E, your system must cut the frame where the vertical support goes into the ground at E What kind of reactions are possible at E? If you consult the notes you will see that this type of connection can have a moment and a resultant force with components in the x and y directions.  (Study the reaction types in the notes as you must be able to correctly recognize what reactions any given support can provide.)

Ex = 90 kN ←
Ey = 200 kN ↑
ME = 180 kN-m CCW

03Oct05

6.11

 

 

 

6.12

Consider a system that includes the massless rod and the ball. What forces and moments act on this closed system during the motion of the ball from position A to position B?

Final speed is 6 in/s

03Oct05

6.13

Application of mass, LM, and AM for an open, steady-state system. Your system should an open system consisting of just the conveyor belt device and the coal on the belt. [Also you must first calculate the velocity components of the coal as it is hitting the belt.( Vx = 3 m/s; Vy = 3.27 m/s ↓) If you make the inlet to your overall system the coal leaving the horizontal conveyor belt then you also need to consider the weight of the falling coal stream in your system.] Depending upon the point you use to calculate the AM for the system, you may need to solve some equations simultaneously

Cx = 89.9 N →;  Cy = 2360 N ↑

Dx = 0; Dy = 2900 N ↑

A useful check on these answers is to just find the reactions when the conveyor belt system is shut down.

03Oct05

6.14

Treat the rod as the system and apply AM and LM to the rod. Also recognize that the rod is translating to the left at a known velocity. (Hint: Fc should be less than the value for when there is no acceleration.)

Force of panel on rod at C is 3.428 N and is perpendicular to the rod.

At B:  Bx = 6.971 N ← and
By = 23.36 N ↑

03Oct05

6.15

 

 

 

6.16

 

 

 

6.17

 

 

 

6.18

Straight forward application of basic definition of the moment of a force about a point and the moment of a force couple about a point: M = r x F. Part (b) is greatly simplified if you break the forces into components and then find net moment as the sum of each component, e.g. break force A into an x and y component.

(a.i)Sum about Point O:
5600 lbf-ft CW;  
(a.ii) Sum about Point P:
 2600 lbf-ft CW;
(b) Sum about Point O: 
 335.7 N-m CW;

03Oct05

6.19

For Part (a), use the definition of moment of a force to find the moment about point P produced by each weight and then sum the values.

For Part (b) Application of some basic definitions ........
(i)  Use your knowledge of rotational velocity:  V = rω. (Don't forget to show the magnitude and direction of the velocity, a vector quantity.)
(ii) Application of the definition of linear momentum of a particle,  P = mV. (Show magnitude and direction.)
(iii) Application of the definition of angular momentum of a particle about point P,  L = m(r ´ V). Remember that the direction (sense) of the angular momentum is in the direction that the velocity vector V is trying to "turn" the position vector r.  (Show magnitude and direction.)

(a) Net moment = 0

(b i)  Velocity
VA= 0.250 m/s ↑;
VB= 0.500 m/s ↓
(b ii) Linear momentum
PA = 2.5 kg-m/s ↑
PB = 2.5 kg-m/s ↓
(b iii) Angular momentum
 LP,A =  0.3125 (kg-m2/s) CW; 
 LP,B =  0.625 (kg-m2/s) CW

02Oct06

6.20

 

 

 

6.21

 

 

 

6.22

Carefully pick your system to use as much of the available information as possible. Consider the location and magnitude of the net forces due to pressures acting on the surface boundaries. Carefully determine all transports of linear and angular momentum and use them in the appropriate conservation equation to solve for the reactions at 1.

Rx =175.5 kN ←
Ry = 39.2 kN ↑
Mo = 33.6kN-m  CW

04Oct05

6.23

Even though the bike does not begin to flip over, it will still have a rate of change of angular momentum about points A and B. In other words, do not cross out the left hand side of COAM.

a) Ff = 471.6 N (right)

b) ax = 0.62g

c) muk = 0.62

 

6.24

(a) Why should you treat the bus as the system? Yes, there must be friction between the tires and the pad or the test would be of no value.  Try both angular and linear momentum equations.
(b)  For this part, you have a couple of good choices for systems -- the bus and the pad combined, or the bus and then the pad, or the pad and then the bus. Will you need to consider both types of momentum?
(c)  What system will allow you to investigate the friction force between the bus and the pad?
PLEASE be careful drawing you momentum-interaction (free-body) diagrams. Take care to correctly identify the direction of all forces acting on a system, especially the friction forces.

(a)  dVx/dt = (d/h)g ←
(b)  F = 4500 lbf ←
(c)  μs = d/h = 0.75

04Oct05

6.25

 

 

 

6.26

 

 

 

6.27

 

 

 

6.28

 

 

 

6.29

 

 

 

6.30

Probably most straight-forward if you choose an open, stationary system. Remember that the velocity direction is important in COLM and COAM.

a) tan(theta) = (4mgL1)/(rho pi d2 Vjet2 L2)

b) Ax = rho (pi d2/4) Vjet2 (cos(theta) - 1)

 

Ay = mg - rho (pi d2/4) Vjet2 sin(theta)

 

6.31

(a)  Application of conservation of linear and angular momentum to a stationary, closed system.
(b)  Application of conservation of linear and angular momentum to a translating, closed system. Note that the only force causing motion is the force of the ground acting on the rear tires. Although the truck is moving, the force of the ground is transmitted to the rear wheel through static friction. What would happen to Ay and  By if the truck was in danger of tipping?

(a)  Ay = 3000 lbf (up)

(b)  Ay = 4412 lbf (up), By = 588 lbf (up), dv/dt = 22.7 ft/s2 (right)

04Oct06