c) the aircraft is doing a horizontal loop of radius R, at a constant speed of v= vgR. During the loop, the aircraft is in an unbanked position(zero roll) R We will also neglect dynamic effects. That is, we will consider that the plumb and the balloon are at rest relative to the aircraft and in equilibrium under the instantaneous forces Solution The mass of the plumb is mp=(2/g) kg. For the balloon, we have that the weight of the volume of air displaced is Pair VBg, and the weight of the helium inside the balloon is helium VBg, where VB is the volume of the balloon. If we neglect the weight of the balloon itself, we have -(Helium - Pair)VB=(1/g kg a)At the bottom of the vertical loop, the acceleration will be ac=v2/R=9. Therefore, a vertical Tp=mp(2g)=4 N and for the balloon Tr=-(Helium -Pair)Bl29/=24 he plumb line will be brings for the plumb and for the balloon will remain vertical. The tension in the plumb line will be b)At the top of the loop, a vertical axis accelerometer will show a zero reading(i.e. a=g). Therefore, the effective gravity will be zero, and we will have Tp=0 and TB=0 c)In this case, the centripetal acceleration, ac =v2/R=9, is in the horizontal direction The vector geff=g-a=-v2g(i+j)has a magnitude of v2g and forms an angle of 45with the ertical. Therefore, the tension in the plumb line will be Tp=mp(v2g)=2 2N,c) the aircraft is doing a horizontal loop of radius R, at a constant speed of v = √ gR. During the loop, the aircraft is in an unbanked position (zero roll). We will also neglect dynamic effects. That is, we will consider that the plumb and the balloon are at rest relative to the aircraft and in equilibrium under the instantaneous forces. Solution: The mass of the plumb is mP = (2/g) kg. For the balloon, we have that the weight of the volume of air displaced is ρairVBg, and the weight of the helium inside the balloon is ρheliumVBg, where VB is the volume of the balloon. If we neglect the weight of the balloon itself, we have −(ρhelium − ρair)VB = (1/g) kg. a) At the bottom of the vertical loop, the acceleration will be ac = v 2/R = g. Therefore, a vertical axis accelerometer will show an acceleration of 2g. The effective gravity will be gef f = −2gj. The strings for the plumb and for the balloon will remain vertical. The tension in the plumb line will be TP = mP (2g) = 4 N, and for the balloon TB = −(ρhelium − ρair)VB(2g) = 2 N. b) At the top of the loop, a vertical axis accelerometer will show a zero reading (i.e. a = g). Therefore, the effective gravity will be zero, and we will have, TP = 0 and TB = 0. c) In this case, the centripetal acceleration, ac = v 2/R = g, is in the horizontal direction. The vector gef f = g − a = − √ 2g(i + j) has a magnitude of √ 2g and forms an angle of 45o with the vertical. Therefore, the tension in the plumb line will be TP = mP ( √ 2g) = 2√ 2 N, 5