130
3 Rotational Kinematics
3.35 Let the height of the incline be h. Then the velocity of the block at the bottom
of the vertical circle will be v =
√
2gh. Minimum height is given by the
condition that v =
√
5gr which is barely needed for the completion of the
loop.
2gh =
5gr
or h =
5
2
r =
5
2
× 12 = 30 cm
3.36 The analysis is similar to that of prob. (2.34). The velocity of the particle at
the bottom of the circular groove will be given by
v =
(2g)(2r ) =
4gr
(1)
which satisfies the condition
2gr < u <
5gr
The particle leaves the circular groove at a height h above the centre of the
circle, Fig. 3.20.
h =
1
3
u 2
g
− 2r
(2)
But u
2
= 4gr
(1)
∴ h =
2
3
r
Thus, the particle leaves the circular groove at a height of h + r =
5
3
r above
the lowest point.
3.37 Let the velocity at B be v.
Kinetic energy gained = potential energy lost
1
2
mv
2
= mg(5R − R)
∴ m
v 2
R
= 8 mg
which is the centrifugal force acting on the track horizontally. The weight acts
vertically down. Hence the resultant force
F =
(8 mg) 2 + (mg) 2 =
√
65 mg