Saved Bookmarks
| 1. |
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» Solution :`(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` 
|
|