Effective length of a pendulum is 50 cm and the mass of the bob is 4g . The bob is drawn to one side until the string is horizontal , and is then released. When the string makes an angle `60^@` with the vertical , what is the velocity and the kinetic energy of the bob ?

Explanation

OB is the horizontal position of the string [Fig.1.20] .

At B, total energy of the bob
= potential energy = mgh
=`(4xx980xx50)` erg
At C, total energy of the bob = kinetic energy + potential energy
= kinetic energy `+4 xx980 xxAD`
[` cos 60^@ =(OD)/(OC) or, OD =(50)/2 =25cm therefore AD =OA -OD =50 -25 =25 cm]`
According to the law of conservation of energy ,
energy of the bob at C = energy of the bob at B
or, K.E. of bob at C + P.E . of the bob at C
= K.E. of the bob at B + P.E. of the bob at B
or, `1/2 mv^2 +4xx980 xx25 =0 +4xx980xx50`
[v= velocity of bob at C ]
or, `1/2 mv^2 =4xx980 xx25=98000`
`therefore v^2 =(98000xx2)/(4) or, v=sqrt(49000) =222.36 cm *s^(-1)`.