Problem 6 :
At a given instant, two beads, “A” and “B” of equal mass “m” are connected by a string. They are held on a circular grove of radius “r” as shown in the figure. Find the accelerations of beads and the tension in the string, just after they are released from their positions.

Solution : Since motion is initiated, velocity of the beads are zero. No centripetal acceleration is associated at this instant. At this instant, the beads “A” and “B” have tangential accelerations in horizontal and vertical direction respectively. As the beads are connected with a string, magnitude of accelerations of the beads are equal in magnitude. As such, magnitude of tangential accelerations are equal when beads are released.

We can find acceleration and tension by considering free body diagram of each of the beads. The initial positions of the beads are at the end of a quarter of circle. It means that triangle OAB is an isosceles triangle and the string makes an angle 45° with the radius at these positions. Let
N
1
N
1
and
N
2
N
2
be the normal force at these positions on the beads. Let also consider that the magnitude of acceleration of either bead is “a”.

The free body diagrams of two beads are shown in the figure.

Considering FBD of “A” and forces in horizontal direction, we have :

T
cos
45
0
=
m
a
T
cos
45
0
=
m
a

⇒
T
=
2
m
a
⇒
T
=
2
m
a

Considering FBD of “B” and considering forces in vertical, we have :

m
g
−
T
sin
45
0
=
m
a
m
g
−
T
sin
45
0
=
m
a

Substituting for “T”,

⇒
m
g
−
2
m
a
X
1
2
=
m
a
⇒
m
g
−
2
m
a
X
1
2
=
m
a

⇒
m
g
−
m
a
=
m
a
⇒
m
g
−
m
a
=
m
a

⇒
a
=
g
2
⇒
a
=
g
2

We find tension in the string by putting value of acceleration in its expression as :

T
=
2
m
g
2
=
m
g
2
T
=
2
m
g
2
=
m
g
2