**(a)** i) Consider a simple pendulum of mass ‘m’ and length ‘L’.

L = l + r,

where, l = length of string

r = radius of bob

ii) Let OA be the initial position of pendulum and OB, its instantaneous position when the string makes an angle θ with the vertical.

In displaced position, two forces are acting on the bob:

Gravitational force (weight) ‘mg’ in downward direction Tension T′ in the string.

iii) Weight ‘mg’ can be resolved into two rectangular components:

Radial component mg cos θ along OB and Tangential component mg sin θ perpendicular to OB and directed towards mean position.

iv) mg cos θ is balanced by tension T′ in the string, while mg sin θ provides restoring force

∴ F = − mg sin θ

where, negative sign shows that force and angular displacement are oppositely directed.

Hence, restoring force is proportional to sin θ instead of θ. So, the resulting motion is not S.H.M.

v) If θ is very small then,

sin θ ~ θ =

∴ F = - mg

∴

= - g

∴

= - g

∴ a = -

x ... (i)

∴ a α - x ... [Since g/L = constant]

vi) In S.H.M,

a = −

ω2 x ….(ii)

Comparing equations (i) and (ii), we get,

ω2 =

But, ω =

∴

()2 =

∴

=

√ ∴ T = 2π

√ ... (iii)

Equation (iii) represents time period of simple pendulum.

vii) Thus period of simple pendulum depends on the length of the pendulum and acceleration due to gravity.

(b)

ω1 = 2.5 π rad/s =

2πn1 ∴

I2 =

I1,n2 (r. p. m) = ?

By conservation of angular momentum,

I2ω2 =

I1ω1 or

I2(2πn2) =

I1(2πn1) ∴

n2 =

=

=

∴

n2 =

r.p.s.

=

× 60 = 00 rpm

The new speed of rotation in r.p.m is 100 r.p.m