
#1
Nov204, 09:37 PM

P: 322

Two lightweight rods d = 23 cm are mounted perpendicular to an axle and at 180° to each other. (Fig. 1124). At the end of each rod is a 700 g mass. The rods are spaced 40 cm apart along the axle. The axle rotates at 41 rad/s.
http://www.geocities.com/sinceury/1124alt.gif (a) What is the component of the total angular momentum along the axle? (b) What angle does the vector angular momentum make with the axle? [Hint: Remember that the vector angular momentum must be calculated about the same point for both masses.] 



#2
Nov304, 04:15 PM

P: 322

No one can help? =(




#3
Nov304, 04:45 PM

P: 4,008

Well the quantity that you will need in order to answer your questions is the rotational inertia or moment of inertia given by:
[tex]I = \Sigma_{i} m_{i}r_{i}^2 [/tex] r is distance between origin and position of the mass. Mass1 is the right hand siderod and mass2 is the other one. So we have (in the right units ofcourse) : [tex]I = 0,7 * 0,23^2 + 0,7(0.4^2 + 0.23^2)[/tex] mass = 0.7 kg d_1 = 0.23 m d_2 = sqrt(0.4²+0.23²) Then you need to find the designated formula's( as a function of I) for calculation your questions. marlon good luck 



#4
Nov304, 11:31 PM

P: 322

Equilibrium
So I calculated the angular around the center of mass:
L=I*w L=0.7*v*r, and v=d*w So L=0.7*0.23*41*sqrt(0.4²+0.23²) So the total angular momentum is 2 times the above...but I was told that it's wrong. Any ideas? 



#5
Nov404, 05:44 AM

P: 4,008

If we calculate with respect to the centre of mass(positioned at the intersection of the diagonal between the two masses and the axis.) you would get : [tex]I = 0.7*(0.2^2 + 0.23^2) + 0.7*(0.2^2 + 0.23^2) [/tex] marlon 



#6
Nov404, 09:37 AM

P: 322

If you times that by the angular speed of the rod, then you would get the answer for a, right? But I got it wrong...




#7
Nov404, 10:36 AM

P: 4,008

yes, what should you get ??? what is the answer ??? Normally it should work
marlon 



#8
Nov404, 11:18 AM

P: 322

I don't know, but the online homework submission thingy is not accpting my answer. =( Also, I'm getting different answers with the way you did it and the way I did it...




#9
Nov404, 03:35 PM

P: 4,008

Look you are gonna have to be more specific here. What did you get ???
marlon 



#10
Nov404, 03:53 PM

P: 4,008

Sorry, but i made a mistake in the distance from the two masses to the axis. In the formula for I the r represents the PERPENDICULAR distance to the axis so this is just 0,23 meters.
[tex]I = 0.7 * 0.23^2 + 0.7 * 0.23^2[/tex] try this one marlon 



#12
Nov404, 09:33 PM

P: 322

nope =( Can someone jump in and help?



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