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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?
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?

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This 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
 one year ago
Best ResponseYou've already chosen the best response.0I do know that the velocity vector is tangent to the wheel. Here's the image attached.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4dw:1442571304048:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the angle for velocity vector should be 30 degrees?

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1You can find the velocity vector using conservation of mechanical energy.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And how do I do that? Could you elaborate a bit?

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1When 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
 one year ago
Best ResponseYou've already chosen the best response.1KE = 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0m is not given in the question.

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.1No that will not work anyway...My assumptions are wrong. m is not required because they will cancel out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay. So \[80 \pi \cos 30\] is not the correct horizontal velocity component?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.1K.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
 one year ago
Best ResponseYou've already chosen the best response.1@IrishBoy123 , what do you think?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4hi 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
 one year ago
Best ResponseYou've already chosen the best response.4@emcrazy14 does that make sense to you? does it need to be broken down?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.4let me do it ignoring gravity as i think the numbers make no sense just a sec

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4dw:1442574162890:dw the highest it could go [assuming no gravity] as shown by the blue lines is 2.64m do you see that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay. 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?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.4of course  because gravity has virtually no time to act, which is why the question looks so odd

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay. Thank you so much. The question is indeed very odd. Really appreciate your help throughout. :)
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