Question

A 3 kg pendulum bob on a 2 m long string is released with a velocity of 2 m/s when the support string makes an angle of 44 degrees with the vertical.  Calculate the tension in the string at the highest position of the bob.

Answer 1

The height of fall is given as

          h1 = L ( 1- cos θ )

              = 2 ( 1- cos 44 )

             = 0.56132 m

According to conservation of energy of the pendulum

       m g h1 + (1/2) m V1^2 = (1/2) mV2^2

       V2^2 = V1^2 + 2 g h1

                = (2)^2 + 2 (9.8) (0.56132 m)

                = 15.0019

Applying conseration of energy when the pendulum goes to the extreme point on the right is

          m g h2 = (1/2) m V2^2

                h2 = V2^2 / 2 g

                    = (15.0019) / 2 (9.8)

                   = 0.7654 m

The angle can be calcualted from

          h2 = L ( 1- cos φ )

       0.7654 = 2 ( 1- cos φ )

            φ = 51.88

The tension at the extreme point of the pendulum is

           T = m g cos φ = (3)(9.8) cos 51.88

                                 = 18.14 N