Rotational inertial and energy.

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J-dizzal
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Homework Statement


The figure shows a rigid assembly of a thin hoop (of mass m = 0.25 kg and radius R = 0.13 m) and a thin radial rod (of length L = 2R and also of mass m = 0.25 kg). The assembly is upright, but we nudge it so that it rotates around a horizontal axis in the plane of the rod and hoop, through the lower end of the rod. Assuming that the energy given to the assembly in the nudge is negligible, what is the assembly's angular speed about the rotation axis when it passes through the upside-down (inverted) orientation?

Diagram of a thin ring labelled Hoop, attached at its bottom to a thin vertical Rod, with the lower end of the rod on a dashed line labelled Rotation axis.

Homework Equations


ΔU=mgΔy
I=1/3 ML2 (rod)[/B]

The Attempt at a Solution


Handwritten physics working on yellowed, stained paper: a small hoop with a radial rod drawn above, and calculations of moment of inertia and energy with labels like m = 0.25 kg and R = 0.13 m.


Im having trouble finding the rotational inertial of the rod and the hoop.
for the rod I am using 1/3 ML2 because its I of the rod rotating about its end.
for the hoop i tried using its com and applying that to the formula I=MR2, where M=.25kg and R=.39m
Thanks
edit. complete[/B]
 
Last edited:
to relate rotational inertia to angular velocity use the kinetic energy-work formula. ΔK=½ Iωf2 - ½ Iωi2.
where ΔK = change in gravitational potential engery, because there is no other forces acting on the system.
I=the sum of the rod and hoop, and the hoop you need to use the parallel axis theorem
ωi=0
 
J-dizzal said:
to relate rotational inertia to angular velocity use the kinetic energy-work formula. ΔK=½ Iωf2 - ½ Iωi2.
where ΔK = change in gravitational potential engery, because there is no other forces acting on the system.
I=the sum of the rod and hoop, and the hoop you need to use the parallel axis theorem
ωi=0
The rod and hoop are rotating about the end of the rod opposite of the hoop, so don't worry too much about calculating the MOI about the c.g.of the combo.

Be careful about which formula you use to calculate the MOI of the hoop about its c.g. If you are using a table of formulas, inspect very carefully which formula goes with which axis of rotation. There's two different formulas for the MOI of a hoop.

The image of your calculations looks like you spilled something all over the paper. Try to show a little pride in your work. This stained paper suggests to your instructor that you are trying not to spend a lot of time on your work.
 
SteamKing said:
The rod and hoop are rotating about the end of the rod opposite of the hoop, so don't worry too much about calculating the MOI about the c.g.of the combo.

Be careful about which formula you use to calculate the MOI of the hoop about its c.g. If you are using a table of formulas, inspect very carefully which formula goes with which axis of rotation. There's two different formulas for the MOI of a hoop.

The image of your calculations looks like you spilled something all over the paper. Try to show a little pride in your work. This stained paper suggests to your instructor that you are trying not to spend a lot of time on your work.
My instructor will not see this, I just have to enter the correct answer into online homework website.
 
J-dizzal said:
My instructor will not see this, I just have to enter the correct answer into online homework website.
So far it doesn't look like you're getting the correct answer, are you ?

By the way: What's the purpose of having you do homework?
 
SammyS said:
So far it doesn't look like you're getting the correct answer, are you ?

By the way: What's the purpose of having you do homework?
Yes, i have completed this problem. in post#2 I tried to explain how i found the answer.
The purpose of doing homework is to learn by practice. And its worth points toward my grade.
 
SteamKing said:
don't worry too much about calculating the MOI about the c.g.of the combo.
I don't think J-dizzal did that.
 
haruspex said:
I don't think J-dizzal did that.
Quite frankly, it's hard to tell exactly what he did. It's also not clear that the snap shot of his calculations shows his latest work, either.