The normal force ( directed downwards ) provided by the track at the top of the loop is equal to one-half the weight of the car.
What is the speed of the coaster at this point ?
2007-11-25 03:33:34 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Centripetal Force =MV^2/r
Top of loop
Reaction + weight = centripetal force
Radius = 7.
1/2(m(9.8))+(9.8m)=MV^2/7
14.7=V^2/7
V^2=102.9
V=10.14396372 ms^-1
approx. 10.1 ms^-1
2007-11-25 03:45:13 · answer #1 · answered by Murtaza 6 · 0⤊ 0⤋
Well, take your pick! 10.1 or 8.3 m/s. Only one will get you the mark. OR perhaps ... they are both wrong!
Isn't it fun getting help online. You never know until the homework comes back.
2007-11-25 11:44:25 · answer #2 · answered by za 7 · 0⤊ 1⤋
mg + (1/2) mg = m V^2/r ---> V^2 = (3/4) g d = 102.9 (m/s)^2
V = 10.1 m/s
2007-11-25 11:39:39 · answer #3 · answered by Luigi 74 7 · 1⤊ 1⤋
no idea
2007-11-25 11:48:36 · answer #4 · answered by i like summer 1 · 0⤊ 1⤋