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translation and rotation |
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| May30-08, 03:56 PM | #1 |
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translation and rotation
Heres the diagram for the problem...
http://img78.imageshack.us/img78/3248/35090397eg3.jpg heres my work... http://img141.imageshack.us/img141/8352/33iw6.jpg I made my conversions.... im starting by finding the speed at D.....its not zero because of the angular velocity.....but Im trying to figure how to add the two x components... the translational x component should be .833 ft/s......but for the rotational I need to account for omega..... |
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| May30-08, 04:51 PM | #2 |
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What's the rotational speed? Is that the problem? It's r*omega. So that's the rotational speed. Now just split the velocity vector into xy components. This is basically an identical problem to you last post.
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| May30-08, 04:52 PM | #3 |
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And what is the question?
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| May30-08, 04:57 PM | #4 |
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translation and rotation |
| May31-08, 10:25 AM | #5 |
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The problem is asking for the velocity at points D and B...as well as if the object is slipping.....Im still trying to figure out point D first
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| May31-08, 10:32 AM | #6 |
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ok I found D....it was just the rotational - translational ....
I used the V=r*omega to get the rotational [tex] V_{d} = .999 - .833 = .166 ft/s = 2 in/s [/tex] but actually I think the signs should be reversed if you look at the vectors in my FBD...that makes sense because the magnitude is the same and the direction should be to the left which is negative... |
| May31-08, 10:49 AM | #7 |
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ok I found the velocity at point B as well.....im just not sure how to tell if the wheel is slipping or not...
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| May31-08, 10:56 AM | #8 |
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If the velocity at D is non-zero, then it's slipping. If it's not slipping then point D is moving at the same speed as the road and has zero velocity.
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| May31-08, 11:27 AM | #9 |
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Thank You...thats what I thought....
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