Saved Bookmarks
| 1. |
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» 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"` |
|