## 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?

1. 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?

2. anonymous

I do know that the velocity vector is tangent to the wheel. Here's the image attached.

3. IrishBoy123

|dw:1442571304048:dw|

4. anonymous

So the angle for velocity vector should be 30 degrees?

5. Abhisar

You can find the velocity vector using conservation of mechanical energy.

6. anonymous

And how do I do that? Could you elaborate a bit?

7. 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)

8. 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

9. Abhisar

one min..

10. anonymous

m is not given in the question.

11. Abhisar

No that will not work anyway...My assumptions are wrong. m is not required because they will cancel out

12. anonymous

Oh okay. So $80 \pi \cos 30$ is not the correct horizontal velocity component?

13. anonymous

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

14. Abhisar

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

15. Abhisar

That's what i think..

16. Abhisar

@IrishBoy123 , what do you think?

17. 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

18. IrishBoy123

@emcrazy14 does that make sense to you? does it need to be broken down?

19. 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?

20. IrishBoy123

let me do it ignoring gravity as i think the numbers make no sense just a sec

21. anonymous

Okay, sure.

22. 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?

23. 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?

24. anonymous

25. IrishBoy123

of course - because gravity has virtually no time to act, which is why the question looks so odd

26. anonymous

Okay. Thank you so much. The question is indeed very odd. Really appreciate your help throughout. :)

27. Abhisar

WoW!!

28. Abhisar

Thanks irishboy c: