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Big POPa

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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Big POPa

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  1. Big POPa Group Members: Evan Hudspeth Stuart Haase Charlie Gang Dave Dickenson

  2. Initial Step

  3. 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

  4. 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

  5. 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 .

  6. 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

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

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