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What is the component of the total angular momentum along the axle?

by physicsss
Tags: equilibrium
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physicsss
#1
Nov2-04, 09:37 PM
P: 319
Two lightweight rods d = 23 cm are mounted perpendicular to an axle and at 180 to each other. (Fig. 11-24). 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/11-24alt.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.]
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physicsss
#2
Nov3-04, 04:15 PM
P: 319
No one can help? =(
marlon
#3
Nov3-04, 04:45 PM
marlon's Avatar
P: 4,006
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 side-rod 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(0.4+0.23)

Then you need to find the designated formula's( as a function of I) for calculation your questions.

marlon

good luck

physicsss
#4
Nov3-04, 11:31 PM
P: 319
Unhappy What is the component of the total angular momentum along the axle?

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(0.4+0.23)

So the total angular momentum is 2 times the above...but I was told that it's wrong. Any ideas?
marlon
#5
Nov4-04, 05:44 AM
marlon's Avatar
P: 4,006
Quote Quote by physicsss
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(0.4+0.23)

So the total angular momentum is 2 times the above...but I was told that it's wrong. Any ideas?
Well that is because the I that i calculated is not with respect to the center of mass but with respect to the origin.

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
physicsss
#6
Nov4-04, 09:37 AM
P: 319
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...
marlon
#7
Nov4-04, 10:36 AM
marlon's Avatar
P: 4,006
yes, what should you get ??? what is the answer ??? Normally it should work

marlon
physicsss
#8
Nov4-04, 11:18 AM
P: 319
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...
marlon
#9
Nov4-04, 03:35 PM
marlon's Avatar
P: 4,006
Look you are gonna have to be more specific here. What did you get ???

marlon
marlon
#10
Nov4-04, 03:53 PM
marlon's Avatar
P: 4,006
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
marlon
#11
Nov4-04, 05:15 PM
marlon's Avatar
P: 4,006
Am i right now ???

maybe someone else can help us out here...

marlon
physicsss
#12
Nov4-04, 09:33 PM
P: 319
nope =( Can someone jump in and help?
marlon
#13
Nov5-04, 04:53 AM
marlon's Avatar
P: 4,006
Quote Quote by physicsss
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(0.4+0.23)

So the total angular momentum is 2 times the above...but I was told that it's wrong. Any ideas?
But it is just L = I * w and w = 41 rad/sec. Just multiply the two...

marlon


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