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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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
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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?
I do know that the velocity vector is tangent to the wheel. Here's the image attached.
1 Attachment
|dw:1442571304048:dw|

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

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