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Angluar momentum question please help!

  • Thread starter Epiphone
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  • #1
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Homework Statement



Two identical spheres, each of mass M and negligible radius, are fastened to opposite ends of a rod of negligible mass and length 2l. This system is initially at rest with the rod horizontal and is free to rotate about a frictionless, horizontal axis through the center of the rod and perpendicular to the plane of the page. A bug of mass 3M, lands gently on the sphere on the left. Assume that the size od the bug is small compared to the length of the rod. Express your answers to all parts of the question in terms of M, l, and physical constants.

a.) Determine the torque about the axis immediately after the bug lands on the sphere.

b.) Determine the angular acceleration of the rod-spheres-bug system immediately after the bugs lands.

At the instant that the rod is vertical (pointing up and down meaning it has rotated 90 degrees counterclockwise) determine each of the following:

c.) The angular speed of the bug.

d.) The angular momentum of the system

e.) The MAGNITUDE and DIRECTION of the force that must be exerted on the bug by the wphere to keep the bug from being thrown off the sphere.



Homework Equations



T = r x F
T = Ialpha
L = r x p
L = Iw

The Attempt at a Solution



I realize this question has been asked on the forum before, but i have not found the other posts very helpful because they needed help with different parts

here are my solutions so far:
a) T = 3Mgl

b) alpha = 9g/16Ml
(solved by setting T = r x F = Ialpha)

c) w = sqrt(9(pi)g/16Ml)
(solved using w^2 = w^2 + 2alpha...)

d) L = 4Ml^2(sqrt(pi)g/Ml)
(solved by L = Iw)

e) I am stuck here! i dont know what concept to use!

Thanks in advance
 

Answers and Replies

  • #2
Doc Al
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Hint: What's the bug's acceleration?
 
  • #3
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hm, since alpha = a/l
alpha = 9g/16Ml
so a = 9g/16M?

and f = ma, so f = (3M*9g)/16M = 27g/16M?

i feel like i am making up rules here :/
 
  • #4
Doc Al
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hm, since alpha = a/l
alpha = 9g/16Ml
so a = 9g/16M?
Two problems here:
(1) alpha will just give you angular acceleration (and tangential); what about radial acceleration?
(2) that's alpha at the initial position, not when the system is vertical.

Another hint: What kind of motion is the bug exeriencing at the lowest point?
 
  • #5
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i think radial is V^2/r

im a little confused though. why would it be at the "lowest point" if the rotational axis is perpendicular to the page? isnt the height always the same?

its instantaneous velocity is tangent to the circular path and perpendicular to the radius, but im not sure if that is the type of motion you were referring to in your hint
 
  • #6
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and does this problem make my answer to part c wrong? because i used the initial alpha in my calculation of w^2 = w^2 + 2alpha(delta(theta))
 
  • #7
Doc Al
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i think radial is V^2/r
Exactly. What's the bug's speed?

im a little confused though. why would it be at the "lowest point" if the rotational axis is perpendicular to the page? isnt the height always the same?
The axis remains at the same height, but not the masses. Certainly not the bug.

its instantaneous velocity is tangent to the circular path and perpendicular to the radius, but im not sure if that is the type of motion you were referring to in your hint
Good. And the motion (and acceleration) of the bug is particularly simple to analyze when the rod is vertical.
 
  • #8
Doc Al
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and does this problem make my answer to part c wrong? because i used the initial alpha in my calculation of w^2 = w^2 + 2alpha(delta(theta))
That's certainly not true. (Sorry, I didn't even check your other answers.) Hint for this part: Use conservation of energy.
 
  • #9
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The axis remains at the same height, but not the masses. Certainly not the bug.
right right, for some reason i thought i was looking at the problem from above. that clears up some things!

and for the conservation of energy, initially i think there is only grav potential, and when it is at its lowest point, there is rotational kinetic
so that would mean mgh = .5Iw^2 + mgh
and then i got w = sqrt(9g/8l)
 
Last edited:
  • #10
Doc Al
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and for the conservation of energy, initially i think there is only grav potential, and when it is at its lowest point, there is rotational kinetic
Exactly.
so that would mean mgh = .5Iw^2 + mgh
and then i got w = sqrt(9g/8l)
I don't understand how you got this answer. What's the change in gravitational PE? What's the rotational inertia of the system?

(You'd better redo part b while you're at it.)
 

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