3.3 Solutions
129
3.3.3 Loop-the-Loop
3.34 If the bob of the pendulum has velocity u at B, the bottom of the vertical circle
of radius r such that
2gr < u <
5gr
(1)
then the bob would leave some point P on the are DA (Fig. 3.20). Here
2gr =
√
2 × 9.8 × 1 = 4.427 m/s and
5gr =
√
5 × 9.8 × 1 = 7 m/s
Fig. 3.20
Therefore (1) is satisfied for u = 6 m/s.
Drop a perpendicular PE on the horizontal CD. Let PE = h and PO make an
angle θ with OD. When the bob leaves the point P, the normal reaction must
vanish.
mv 2
r
− mg sin θ = 0
( 2 )
Loss in kinetic energy = gain in potential energy
1
2
m u
2
−
1
2
mv
2
= mg(h + r )
(3)
sin θ =
h
r
(4)
Eliminating v 2 between (2) and (3) and using (3), with u = 6 m/s and
r = 1.0 m,
h =
1
3
u 2
g
− 2r
= 0.558 m
129
3.3.3 Loop-the-Loop
3.34 If the bob of the pendulum has velocity u at B, the bottom of the vertical circle
of radius r such that
2gr < u <
5gr
(1)
then the bob would leave some point P on the are DA (Fig. 3.20). Here
2gr =
√
2 × 9.8 × 1 = 4.427 m/s and
5gr =
√
5 × 9.8 × 1 = 7 m/s
Fig. 3.20
Therefore (1) is satisfied for u = 6 m/s.
Drop a perpendicular PE on the horizontal CD. Let PE = h and PO make an
angle θ with OD. When the bob leaves the point P, the normal reaction must
vanish.
mv 2
r
− mg sin θ = 0
( 2 )
Loss in kinetic energy = gain in potential energy
1
2
m u
2
−
1
2
mv
2
= mg(h + r )
(3)
sin θ =
h
r
(4)
Eliminating v 2 between (2) and (3) and using (3), with u = 6 m/s and
r = 1.0 m,
h =
1
3
u 2
g
− 2r
= 0.558 m