7 CIRCULAR MOTION 7.7 Motion on curved surfaces
R
experiences an acceleration v^2 /r towards the centre of the loop, Newton’s second
law of motion yields
v^2
m = m g − R.
r
Now, the reaction R is equivalent to the apparent weight of the pilot. In partic-
ular, if the pilot is “weightless” then he/she exerts no force on the plane, and,
therefore, the plane exerts no reaction force on the pilot. Hence, if the pilot is
weightless at the top of the loop then R = 0 , giving
r = v^2 =
g
(500 × 1000/3600)^2
9.81
= 1.97 km.
Worked example 7.5: Ballistic pendulum
Question: A bullet of mass m = 10 g strikes a pendulum bob of mass M = 1.3 kg
horizontally with speed v, and then becomes embedded in the bob. The bob is
initially at rest, and is suspended by a stiff rod of length l = 0.6 m and negligible
mass. The bob is free to rotate in the vertical direction. What is the minimum
value of v which causes the bob to execute a complete vertical circle? How does
the answer change if the bob is suspended from a light flexible rod (of the same
length), instead of a stiff rod?
Answer: When the bullet strikes the bob, and then sticks to it, the bullet and bob
move off with a velocity vJ which is given by momentum conservation:
m v = (M + m) vJ.
mg
r