Question

The Bob of a simple pendulum of length l is released from point P.What is the angle made by the net acceleration of the Bob with the string at point Q

Answer 1

Using diagram ,

We need to find the acceleration of the pendulum at any position that makes an angle θ with the vertical at the point of support.

pendulum length = L.  mass = m.  

Le the tension force in the string connected to the pendulum bob be = T.

The pendulum is released from position A.  It moves along the circular arc APG.  At P the direction of movement of pendulum bob is along PF.

The pendulum at position P does not move along  OP or PC.  Hence, the forces in this direction are balanced.  The bob moves along PF.

T = m g Cos θ

 - m g Sin θ = net force acting on the bob at P.

a_t = linear instantaneous acceleration along the tangent to the arc along PF

= Force/ mass = - g sin θ

As the pendulum is moving in a circular arc, there is a centripetal force on the bob of the pendulum.   Hence there is a centripetal acceleration resulting from it.  

a_r = v² / L,    v = tangential instantaneous velocity.

In the ideal simple pendulum, there is no loss of energy and the force that gives potential energy is the gravitational force.

 Change in P.E. of the bob when drops from A to P   =  m g BP = m g L Sin θ

  So KE of bob at P = m g L Sin θ

1/2  m v² = m g L Sin θ

v² = (2 g L Sin θ)

Radial centripetal acceleration along PO = 2 g Sin θ

The net acceleration of the bob = √(a_r² + a_t²)

a    = √[ (-g Sinθ)² + (2gSinθ)² ]

magnitude   a = √5 g Sinθ

Its direction is given by:  α with the tangential PF, say.

then according to the addition of vectors:

tan α = (2g Sinθ) / [g Sinθ + 2g Sinθ Cos 90^°)

    = 2

So α = tan⁻¹ 2 = 63.43^° with PF.

So the angle between the two accelerations is always 90^° and the angle between the resultant and each component of acceleration is always 63.43^°  or (90^° - 63.43^°).

 Thus, the resultant acceleration has magnitude √5gSinθ   and is horizontal when α = θ = 63.43^° and that is when the pendulum makes an angle 63.43^° with the vertical.

the resulting acceleration of the accelerations along PO and PF will be horizontal if their vertical components along PB and PD cancel each other.

component of tangential acceleration a_t along PD =- g sin θ * sin θ

component of radial acceleration along PB = 2 g Sin θ * Cos θ

Hence, they cancel, if SinФ = 2 Cos θ

 ie.,  Tan θ = 2   

=>  θ = 63.43 degrees with the vertical OG.

 In that case the acceleration will be along PE.