106
3 Rotational Kinematics
Motion in a Vertical Plane
Typical problems are of the following type:
(i) An object of mass m tied to a string is whirled in a vertical plane such that at
the top (A) of the circular path its speed is v A and at the bottom (B) it is v B .
Calculate v A and v B given the tension T B = xT A , where x is a number.
Solution: At the top the weight acts down while the centrifugal force acts up.
Therefore,
T A =
mv 2
A
r
− mg
while at the bottom both weight and centrifugal force act downwards. Therefore,
T B =
mv 2
B
r
+ mg
We get one another equation from the conservation of mechanical energy:
mg2r = ½ mv
2
B − ½ mv
2
A
Finally, by problem
T B = xT A
The four equations can be solved to permit the determination of v A and v B .
(ii) The bob of a simple pendulum of length L is drawn on one side such that it
makes an angle θ with the vertical passing through the equilibrium position.
If the bob is released from rest it passes through the equilibrium position with
velocity v. Find v.
Here we use the principle gain in kinetic energy = loss in potential energy
1
2
mv
2
= mgL(1 − cos θ)
whence v =
√
2gL(1 − cos θ)
(iii) A particle of mass m is placed at A, the highest point of a smooth sphere of
radius R with the centre at O. If it is gently pushed, it will slide down along the
arc of a great circle and leave the surface at B, at depth h below A, Fig. 3.16.
Determine the position where the particle leaves the sphere.
Here we balance the radial component of g at B with the centripetal force.
mg cos θ = mv
2
/R
Energy conservation gives another equation:
mgh = ½ mv
2