The bob of a pendulum at rest is given a sharp hit to impat a horizontal velocity `sqrt(10 gl)` where l is the length of the pendulum. Find the tension in the string when a. the string is horizontal. B. The bob is at its highest point and c. the string makes an angle of `60^0` with teh upward vertical.

Answer

A, B, C

Explanation

a.Let the velocity at B be `v_2`
`1/2 mv_1^2=1/2mv_2^2+mgl`
`rarr 1/2m(10gl)=1/2mv_2^2+mgl`
`v_2^2=8gl`

so the tension in the string at horizontal position
`T=(mv^2)/R=m(8gl)/l`
`=8mg`
b. Let the velocity at C be `v_3`
`1/2 mv_1^2 = 1/2 mv_3^2+mg2l`
`rarr 1/2 m 10 gl = 1/2 mv_3^2+2mgl`
`rarr v_3^2=6gl`
So, the tension in the string is given by
`T_C=(mv_3^2)/l-mg=5mg`
c. Let the velocity at point D be `v_4
Again `1/2 mv_1^2=1/2 mv_4^2+mgl(1+cos60^0)`
`rarr v_4^2=7gl`
so, the tension in the string is
`T_D=(mv^2)/l-mgcos 60^0`
`=m(7gl)/l-0.5 mg`
`=7mg=0.5mg`
`6.5 mg`