Question

A plank is rotating in a vertical plane about one of its ends with a constant angular velocity `omega=sqrt(2)rad//s`. A block of mass `m=2kg` is placed at a distance `l=1m` from its end `A` (see figure) which is hinged. The block starts sliding down when the plank makes an angle `theta=30^(@)` with the horizontal. If coefficient of friction between the plank and the block is `mu` and given that `mu^(2)=k//25`. Find the value of `k`.

Answer 1

`(1)/(v)-(1)/(u)=(1)/(f)`
`v-u=x`,
`(1)/(u+x)-(1)/(u)=(1)/(f)`
`rArr (1)/(u+x)=(1)/(u)+(1)/(f)`
`u+x=(uf)/(u+f)`
`x=(uf)/(u+f)-u=u[(f)/(u+f)-1]`
`x=(-u^(2))/(u+f)`
differentiating w.r.t. `x`,
`:. deltax=[(-2u)/(u+f)+(u^(2))/((u+f)^(2))]deltau`
`=(-2u(u+f)+u^(2))/((u+f)^(2))deltau`
`=-(u^(2)+2uf)/((u+f)^(2))deltau`
`deltax=((u+f)^(2)+f^(2))/((u+f)^(2))=[((f)/(u+f))-1]deltau`
`deltau=a lt lt |u|=|3f|`
`deltax=[((f)/(-3f+f))^(2)-1]a=-(3)/(4)a`
Amplitude `=|deltax|=(3)/(4)a`