A block of mass m is kept on a horizontal ruler. The frilction coefficient between the ruler and the block is `mu`. The ruler is fied at one end and the block is at a distance L from the fied end. The ruler is rotated about the fixed end in the horizontal plane through the fixed end. a. What can the maximum angular speed of the ruler isuniformly increased from zero t an angular acceleration `alpha` , at what angular speed will the block slip?

Answer

A, B, D

Explanation

The frictional force provides the necessary centripetal force.
:. `mLomega^(2)=mumg`
or `omega=sqrt((mug)/(L))`
(b) Net force of circular motion will be provided by the friction
`omega=alphat` ....(i)
`F_(n et)=msqrt(a_(t)^(2)+a_(n)^(2))`
:. `mumg=msqrt((Lalpha)^(2)+(Lomega^(2))^(2))` ...(ii)
Here , `a_(t)=Lalpha` and `a_(n)=Lomega^(22)`
Substituting `omega=alphat` in Eq. `(ii)` we have,
:. `mug=sqrtr(L^(2)alpha^(2)+L^(2)alpha^(4)t^(4))`
:. `t=((mu^(2)g^(2)-L^(2)alpha^(2))/(L^(2)alpha^(4)))^(1)/(4)`
Substituting the value of `t` in Eq. `(i)` we have,
o`omega=[((mug)/(L))^(2)-alpha^(2)]^(1)/(4)`