Rotational Motion

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3672N
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#1
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#1
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3672N
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#2
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#2
I found the moment of inertia of R, but then i don't know how to solve this further, It's a great help if someone can explain me how to solve this
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mqb2766
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#3
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#3
The diagram is a bit misleading as the step height is 1/2 the radius of the disk.

You're looking at a rotating disk having enough energy to rotate about the point of contact with the step and move up. I'd imagine the key is to consider the moment (energy) about the point of contact with the step (hence the calculation in part a) and use conservation of energy to get it to move up.
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Pangol
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(Original post by Nimantha)
I found the moment of inertia of R, but then i don't know how to solve this further, It's a great help if someone can explain me how to solve this
This is slightly beyond my knowledge in rotational problems, but I think you might have to use conservation of angular momentum.
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3672N
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(Original post by mqb2766)
The diagram is a bit misleading as the step height is 1/2 the radius of the disk.

You're looking at a rotating disk having enough energy to rotate about the point of contact with the step and move up. I'd imagine the key is to consider the moment (energy) about the point of contact with the step (hence the calculation in part a) and use conservation of energy to get it to move up.
Thankyou...
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3672N
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(Original post by Pangol)
This is slightly beyond my knowledge in rotational problems, but I think you might have to use conservation of angular momentum.
Thankyou...
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3672N
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#7
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Not sure I can use conservation of angular momentum because there is a component of the weight of R, perpendicular to the line AC
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3672N
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I feel like the angular speed omega must be a function of time rather than a constant as it reduces when it goes up..
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ghostwalker
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#9
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(Original post by Nimantha)
Not sure I can use conservation of angular momentum because there is a component of the weight of R, perpendicular to the line AC
(Original post by Nimantha)
I feel like the angular speed omega must be a function of time rather than a constant as it reduces when it goes up..
\omega is the angular speed immediately after impact. It doesn't refer to the speed throughout the subsequent motion. Conservation of angular momentum is the way to go in dealing with the impact itself.

Edited: for clarity.
Last edited by ghostwalker; 1 year ago
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3672N
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#10
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(Original post by ghostwalker)
\omega is the angular speed immediately after impact. It doesn't refer to the speed throughout the subsequent motion. Conservation of angular momentum is the way to go.
Thankyou...
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mqb2766
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#11
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(Original post by Nimantha)
I feel like the angular speed omega must be a function of time rather than a constant as it reduces when it goes up..
After the impact, think about energy. You're told omega is after the impact. It will decrease as it goes up the step.
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3672N
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#12
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(Original post by mqb2766)
After the impact, think about energy. You're told omega is after the impact.
Okay, Thankyou...
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mqb2766
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#13
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(Original post by Nimantha)
Okay, Thankyou...
That's obviously for the ke/last part.
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3672N
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Name:  IMG_20210112_150201-compressed.jpg.jpeg
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DFranklin
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#15
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#15
You seem to have calculated the linear momentum before the impact, not the angular... (Edit, or you've ignored the rotational angular momentum of the wheel before the impact, or something...)
Last edited by DFranklin; 1 year ago
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DFranklin
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#16
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(Original post by ghostwalker)
\omega is the angular speed immediately after impact. It doesn't refer to the speed throughout the subsequent motion. Conservation of angular momentum is the way to go in dealing with the impact itself.

Edited: for clarity.
I know what you suggest gives the given answer (and I'm sure is the expected approach), but are you sure angular momentum would really be conserved? (I don't think it would be unless the step is perfectly smooth).
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3672N
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#17
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(Original post by DFranklin)
You seem to have calculated the linear momentum before the impact, not the angular... (Edit, or you've ignored the rotational angular momentum of the wheel before the impact, or something...)
Thankyou for pointing out the error
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ghostwalker
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#18
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(Original post by DFranklin)
I know what you suggest gives the given answer (and I'm sure is the expected approach), but are you sure angular momentum would really be conserved? (I don't think it would be unless the step is perfectly smooth).
My reasoning is that since we are considering motion about the point A, then any impulse due to the impact would have zero moment, and so angular momentum is conserved.
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DFranklin
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#19
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(Original post by ghostwalker)
My reasoning is that since we are considering motion about the point A, then any impulse due to the impact would have zero moment, and so angular momentum is conserved.
That makes sense (I think!). Thanks + PRSOM.
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TeeEm
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#20
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#20
(Original post by Nimantha)
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there are 2 or 3 questions that have ideas similar to your problem here
https://www.madasmaths.com/archive/m...nal_motion.pdf
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