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Big POPa. Group Members: Evan Hudspeth Stuart Haase Charlie Gang Dave Dickenson. Initial Step. Spin It!. τ= F·r F= m·a m= 0.032 slugs r= 0.167 ft The acceleration was found by constant acceleration equation: V2 2 =V1 2 +2ad d is the distance the ball travels which is 1.417 ft.

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Presentation Transcript
slide1
Big POPa

Group Members:

Evan Hudspeth

Stuart Haase

Charlie Gang

Dave Dickenson

slide3
Spin It!
  • τ=F·r F=m·a
  • m= 0.032 slugs
  • r= 0.167 ft
  • The acceleration was found by constant acceleration equation:
  • V2 2=V12+2ad
  • d is the distance the ball travels which is 1.417 ft.
  • V2= 4.79 ft/s V1= 0 ft/s
  • (4.79 ft/s)2= (0 ft/s)2 + 2a(1.417 ft)
  • a= 8.13 ft/s2
  • F= (0.032 slugs) (8.13 ft/s2) = 0.026 lb
  • τ = (0.167 ft)(0.026 lb)= 0.0043 ft-lb
slide4
The POP
  • mgh = ½mV2
  • mass cancels out in the calculation
  • g= 32.2 ft/s2
  • h= 0.875 ft
  • (32.2 ft/s2)(0.0875 ft)= ½V2
  • V= 7.51 ft/s
slide5
Ball Drop
  • mpd+mbd/mp+mb
  • (.00014slugs)(1.29ft)+(.0013slugs)(.75ft)/(.00014slugs)+(.0013slugs)
  • Center of mass=0.88ft from tied side
  • Y: V22=V12 + 2ad
  • V1= 0 ft/s
  • a= 32.2 ft/s2
  • d=1.29 ft
  • v22=(0 ft/s) + 2(32.2 ft/s2)(1.29 ft)
  • V2= 9.11 ft/s
  • X: V1 (found with conservation of energy)= 1.036 ft/s
  • V2= 0 ft/s
  • d= 0.083 ft
  • a= -6.47 ft/s2

.

slide6
Final Step

Perfectly inelastic collision

mball (Vball) + mcup(Vcup)= (mball+mcup)Vfinal

mball=0.0015 slugs

Vball=9.11 ft/s

mcup=9.01*10-4

Vcup=0 ft/s

(0.0015slugs)(9.11 ft/s)+0=(0.0024 slugs) Vfinal

Vfinal= 5.69 ft/s

issues
Issues
  • Initial Ramp
  • Hitting the cup reliably
  • Popping the balloon (placement of needle)
  • Working out minor kinks
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