Question

A stone of mass 1 kg is whirled in horizontal circle attached at the end of a 1 m long string. If the string makes an angle of `30^(@)` with vertical, calculate the centripetal force acting on the stone. `(g=9.8" m"//"s"^(2))`.

Answer 1

Given : `m=1" kg",r=1" m",theta=30^(@)`
Now, Centripetal force `=(mv^(2))/( r )`
`:.` Centripetal force `=(1xxv^(2))/( r )`
and Velocity of body, `v=sqrt("rg "tan theta)`
`=sqrt(1xx9.8xxtan 30^(@))=2.37" m"//"s"`
Again, Centripetal force `=(mv^(2))/( r )`
`=(1xx(2.37)^(2))/(1)`
`=5.61" N"`