Question

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 1

(a) (i) At the lowest positin:
`T_(1)-mg=mu^(2)/(L)=(m)/(L)xx10gLimpliesT_(1)=11mg`
(ii) At the highest position:
`V^(2)=u^(2)-2gh=u^(2)-2gxx2L`
`=10gL-4gL=6gL`
`T_(2)+mg=(mv^(2))/(L)=(m)/(L)xx6gL=6mg` `T_(2)=5mg`
(b) (i) `v^(2)=u^(2)-2gh=u^(2)-2gL(1-cos 60^(@)`
`=10gL-2gL(1-(1)/(2))=9gL`
`T-mg cos 60^(@)=(mv^(2))/(L)=(m)/(L)xx9gL=9mg`
`T-(mg)/(2)=9mgimpliesT=(19)/(2)mg`
(ii) `v^(2)=u^(2)-2gh=10gL-2gL(1+cos60^(@))=7gL`
`T+mg cos 60^(@)=(mv^(2))/(L)`
`T+(mg)/(2)=(m)/(L)xx7gL=7mGimpliesT=(13mg)/(2)`