3.3 Solutions
133
M V = mv
(1)
or v =
M
m
V =
10
1
V = 10 V
(2)
Condition for describing a full vertical circle is that the minimum velocity of
the test tube should be
V =
5gr =
√
5 × 980 × 5 = 156.5 cm/s
Therefore the minimum velocity of the cork which flies out ought to be
v = 10 V = 1565 cm/s = 15.65 m/s
3.43 Equating centripetal force to frictional force
mv 2
r
= μ mg
μ =
v 2
gr
=
(14) 2
9.8 × 45
=
4
9
133
M V = mv
(1)
or v =
M
m
V =
10
1
V = 10 V
(2)
Condition for describing a full vertical circle is that the minimum velocity of
the test tube should be
V =
5gr =
√
5 × 980 × 5 = 156.5 cm/s
Therefore the minimum velocity of the cork which flies out ought to be
v = 10 V = 1565 cm/s = 15.65 m/s
3.43 Equating centripetal force to frictional force
mv 2
r
= μ mg
μ =
v 2
gr
=
(14) 2
9.8 × 45
=
4
9