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anonymous

  • one year ago

A lump of wet putty moves in uniform circular motion as it rides at a radius of 20.0 cm on the rim of a wheel rotating counterclockwise with a period of 5.00 ms. The lump then happens to fly off the rim at the 5 o’clock position (as if on a clock face). It leaves the rim at a height of h = 1.20 m from the floor and at a distance d = 2.50 m from a wall. At what height on the wall does the lump hit?

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  1. anonymous
    • one year ago
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    This is what I've done so far. Displacement vectors: \[x=0.2\cos 60 \] \[y= -0.2\sin 60\] Velocity vectors: \[x=80 \pi \cos \theta\] \[y= 80 \pi \sin \theta\] I cannot work out the angle for this. Help?

  2. anonymous
    • one year ago
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    I do know that the velocity vector is tangent to the wheel. Here's the image attached.

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  3. IrishBoy123
    • one year ago
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    |dw:1442571304048:dw|

  4. anonymous
    • one year ago
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    So the angle for velocity vector should be 30 degrees?

  5. Abhisar
    • one year ago
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    You can find the velocity vector using conservation of mechanical energy.

  6. anonymous
    • one year ago
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    And how do I do that? Could you elaborate a bit?

  7. Abhisar
    • one year ago
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    When the putty is at the bottom position it will have kinetic energy = mg2R. At 50 clock position it will have KE = mg2R - mg(1.20)

  8. Abhisar
    • one year ago
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    KE = 1/2mv^2 , equate the quantities to get the value of v. You know the angles as stated by irishboy. So you can find the height

  9. Abhisar
    • one year ago
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    one min..

  10. anonymous
    • one year ago
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    m is not given in the question.

  11. Abhisar
    • one year ago
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    No that will not work anyway...My assumptions are wrong. m is not required because they will cancel out

  12. anonymous
    • one year ago
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    Oh okay. So \[80 \pi \cos 30\] is not the correct horizontal velocity component?

  13. anonymous
    • one year ago
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    I worked out 80*pi by determining the distance covered by a point on the wheel in one revolution. So, \[v= \frac{ 2 \pi (0.2) }{ 5\times 10^{-3} }\]

  14. Abhisar
    • one year ago
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    K.E = mg(2R+h)-mgh=mg2r=1/2mv^2 \(\sf V^2=4gr=8\\ \Rightarrow V=2\sqrt{2}\) Horizontal component = Vcos\(\sf 30=\sqrt{6}\)

  15. Abhisar
    • one year ago
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    That's what i think..

  16. Abhisar
    • one year ago
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    @IrishBoy123 , what do you think?

  17. IrishBoy123
    • one year ago
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    hi abhisar i am still thinking about the fact the gum is travelling at 560mph when it leaves the wheel!! so it will travel pretty much in a straight line along the tangent to the wall :-)) maybe there something wrong with this question.... but yes, get \(\large v = \omega \, r= \frac{2 \pi r}{T} = \frac{2 \pi \, \, 0.2}{0.005}\) which is that stupidly big number it will hit the wall at \(t = \frac{2.5}{v \, \cos \theta} = \frac {2.5 \times 0.005}{2 \pi \, \, 0.2 \ \cos(\pi /6)}\) at which time t it will have increased ite height by \(v \sin(\pi /6) t - \frac{1}{2}gt^2\) latexing as i go so pls expect a typo or 2

  18. IrishBoy123
    • one year ago
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    @emcrazy14 does that make sense to you? does it need to be broken down?

  19. anonymous
    • one year ago
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    Yes, I just matched it with my calculations. I did it in the same way. I'm getting the height= 3.129m from the floor. Do you think that's a reasonable answer?

  20. IrishBoy123
    • one year ago
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    let me do it ignoring gravity as i think the numbers make no sense just a sec

  21. anonymous
    • one year ago
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    Okay, sure.

  22. IrishBoy123
    • one year ago
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    |dw:1442574162890:dw| the highest it could go [assuming no gravity] as shown by the blue lines is 2.64m do you see that?

  23. anonymous
    • one year ago
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    Okay. I redid the calculations by the equations you wrote. I got 1.44m to which I added the height 1.20m alreday given. So the final answer I got is 2.64m. This agrees with the answer you got assuming no gravity. Is it okay now?

  24. anonymous
    • one year ago
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    *already

  25. IrishBoy123
    • one year ago
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    of course - because gravity has virtually no time to act, which is why the question looks so odd

  26. anonymous
    • one year ago
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    Okay. Thank you so much. The question is indeed very odd. Really appreciate your help throughout. :)

  27. Abhisar
    • one year ago
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    WoW!!

  28. Abhisar
    • one year ago
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    Thanks irishboy c:

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