The bob of a pendulum at rest is given a sharp hit to impart 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 the upward vertical.

Explanation

Let at lowest point we have given `v_1=sqrt(10gl)` at A and when the string is horizontal the velocity at B is `v_2`.
Using conservation of mechanical energy, `DeltaK+DeltaU=0`
`(1/2mv_2^2-1/2mv_1^2)+(mgl)=0`
`1/2m(10gl)=1/2mv_2^2+mglimpliesv_2^2=8gl`
a. Now using force equation, the tension in the string at horizontal position,
`T=(mv^2)/(R)=m(8gl)/(l)=8mg`

b. Similarly, using conservation of mechanical en energy between A and C,
`DeltaK+DeltaU=0`
Let the velocity at C at `v_3`,

`[1/2mv_3^2-1/2mv_1^2]+(mg2l)=0`
`implies1/2m10gl=1/2mv_3^2=2mgl`
`impliesv_3^2=6gl`
So, the tension in the string is at C given by
`TC=(mv_3^2)/(l)-mg=5mg`
c. Let the velocity at point D be `v_4`.
Again using conservation of mechanical energy between A and D, `DeltaK+DeltaU=0`
`(1/2mv_4^2-1/2mv_1^2)+[mgl(1+cos60^@)]=0`
`impliesv_4^2=7gl`

So, the tension in the string is
`T_D=(mv^2)/(l)-mgcos60^@`
`=m((7gl))/(l)-0.5mg`
`=7mg-0.5mg=6.5mg`
The bob of a stationary pendulum is given a sharp hit to impart it a horizontal speed of `sqrt(3gl)`.