anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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?
anonymous
  • anonymous
I do know that the velocity vector is tangent to the wheel. Here's the image attached.
1 Attachment
IrishBoy123
  • IrishBoy123
|dw:1442571304048:dw|

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anonymous
  • anonymous
So the angle for velocity vector should be 30 degrees?
Abhisar
  • Abhisar
You can find the velocity vector using conservation of mechanical energy.
anonymous
  • anonymous
And how do I do that? Could you elaborate a bit?
Abhisar
  • Abhisar
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)
Abhisar
  • Abhisar
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
Abhisar
  • Abhisar
one min..
anonymous
  • anonymous
m is not given in the question.
Abhisar
  • Abhisar
No that will not work anyway...My assumptions are wrong. m is not required because they will cancel out
anonymous
  • anonymous
Oh okay. So \[80 \pi \cos 30\] is not the correct horizontal velocity component?
anonymous
  • anonymous
I worked out 80*pi by determining the distance covered by a point on the wheel in one revolution. So, \[v= \frac{ 2 \pi (0.2) }{ 5\times 10^{-3} }\]
Abhisar
  • Abhisar
K.E = mg(2R+h)-mgh=mg2r=1/2mv^2 \(\sf V^2=4gr=8\\ \Rightarrow V=2\sqrt{2}\) Horizontal component = Vcos\(\sf 30=\sqrt{6}\)
Abhisar
  • Abhisar
That's what i think..
Abhisar
  • Abhisar
@IrishBoy123 , what do you think?
IrishBoy123
  • IrishBoy123
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
IrishBoy123
  • IrishBoy123
@emcrazy14 does that make sense to you? does it need to be broken down?
anonymous
  • anonymous
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?
IrishBoy123
  • IrishBoy123
let me do it ignoring gravity as i think the numbers make no sense just a sec
anonymous
  • anonymous
Okay, sure.
IrishBoy123
  • IrishBoy123
|dw:1442574162890:dw| the highest it could go [assuming no gravity] as shown by the blue lines is 2.64m do you see that?
anonymous
  • anonymous
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?
anonymous
  • anonymous
*already
IrishBoy123
  • IrishBoy123
of course - because gravity has virtually no time to act, which is why the question looks so odd
anonymous
  • anonymous
Okay. Thank you so much. The question is indeed very odd. Really appreciate your help throughout. :)
Abhisar
  • Abhisar
WoW!!
Abhisar
  • Abhisar
Thanks irishboy c:

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