A Classical Approach of Newtonian Mechanics

7 CIRCULAR MOTION 7.7 Motion on curved surfaces

Hence,

vJ =

m v

.

M + m

Consider the case where the bob is suspended by a rigid rod. If the bob and

bullet only just manage to execute a vertical loop, then their initial kinetic energy

(1/2) (M + m) vJ^2 must only just be sufficient to lift them from the bottom to the

top of the loop—a distance 2 l. Hence, in this critical case, energy conservation

yields
1
(M + m) vJ^2 = (M + m) 2 g l,
2
which implies

or
(M + m)

√

4 g l

v =

m

vJ^2 = 4 g l,

=

1.31 ×

√

4 × 9.81 × 0.6

0.01

= 635.6 m/s.

Consider the case where the bob is suspended by a flexible rod. The velocity v JJ

of the bob and bullet at the top of the loop is obtained from energy conservation:

1

(M + m) vJJ^2 =

1

(M + m) vJ^2 − (M + m) 2 g l.

2 2

If the bob and bullet only just manage to execute a vertical loop, then the tension
in the rod is zero at the top of the loop. Hence, the acceleration due to gravity

g must account exactly for the required acceleration v JJ^2 /l towards the centre of
the loop:

It follows that, in this critical case,

vJJ^2

l

= g.

or
(M + m)

√

5 g l

v =

m

vJ^2 = 5 g l,

=

1.31 ×

√

5 × 9.81 × 0.6

0.01

= 710.7 m/s.