Loop the loop example
A ball of mass m goes around a frictionless track in the shape of a circle of radius
r. What should be its minimum velocity at the bottom of the track so that it
never loses contact with the track as it goes around the circle?
We have started working on this problem in class. Here is a summary of what
we have done.
As the ball goes around the circle, it is located by the angle q as defined in the
figure below.
vtop
q
N BT
v
W BE
vbot
The forces acting on the ball are the normal on the ball by the track, NBT, and its
weight, W BE.
According to Newton’s second law,
NBT + W BE = m a
We can define a tangential direction (along v) and a normal direction (toward the
center of the circle).
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loopTheLoop.nb
Along these two axes, we get
tangential axis: -m g sinq = m
„v
„t
normal axis: NBT - m g cosq = m
v2
r
The ball never loses contact with the track if NBT ¥ 0. From the normal
2
component equation, we get m g cosq + m vr ¥ 0.
The inequality is always satisfied if - p2 § q §
p
,
2
that is the ball always stays on
the track when it is on the lower part of the circle.
However, if
p
2
§q §
3p
,
2
cosq § 0 and we must have v ¥
-r g cosq if the
ball is to stay on the track. The most stringent condition happens at the top of
the track where v is minimum and cos(p) = -1.
rg.
So the ball will always stay on the track if vtop ¥
It remains to relate the velocity at the top vtop to the velocity at the bottom vbot .
To do so, start from the tangential component of Newton’s second law:
-m g sinq = m „v
„t
If we can integrate the equation between q = 0 where v = vbot and q = p where
v = vtop , we will have the relationship we are looking for.
To perform the integration and find the minimum value of v at the bottom,
follow the steps listed below:
.
a) multiply the two sides of the equation by „q
„t
b) on the right hand side of the equation, replace
„q
„t
terms of v using the relationship between v and w.
c) do the integration.
d) use that vtop ¥
g r to find the condition on vbot.
Check your work with me at the end of the lab.
= w by its expression in
loopTheLoop.nb
Solution:
a) -m g sinq
b) -m g sinq
„q
„t
„q
„t
=m
=m
„v
„t
„v
„t
„q
„t
v
since „q
„t
r
= w = vr .
c) If t = 0 at the bottom and t = ttop at the top, we have
q(0) = 0, qttop = p, v(0) = vbot , and vttop = vtop .
By integrating between t = 0 and t = ttop, we get
ttop
ttop
„q
„v v
 0 -m g sinq „t  „ t =  0 m „t r „t
that is
g cosqt 
or
-2 g =
1
2r
ttop
0
=
1
2r
vt2
ttop
0
vtop2 - vbot2
which yields
vbot2 = vtop2 + 4 r g
d) And since vtop2 ¥ g r , we have vbot 2 ¥ 5 g r.
To conclude, the ball stays in contact with the track if vbot ¥
5gr .
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