Question

A block of mass m is kept on a horizontal ruler. The friction coefficient between the ruler and the block is µ. The ruler is fixed at one end and the block is at a distance L from the fixed end. The ruler is rotated about the fixed end in the horizontal plane through the fixed end. (a) What can the maximum angular speed be for which the block does not slip? (b) If the angular speed of the ruler is uniformly increased from zero at an angular acceleration α, at what angular speed will the block slip ?

Answer 1

(a) Let the Maximum Angular Speed by which the ruler is rotated be ω.

At this angular speed the block will experience an outward force which will be balanced by the friction.  

F= mω²L.

∴ mω²L=µmg

ω²=µg/L  

ω=√(µg/L)

(b) When the speed is increased with an angular acceleration, the circular motion becomes non-uniform. So the block will have both radial and tangential accelerations.  

∴ The radial acceleration will be same as above =ω²L.  

The tangential acceleration = dv/dt

=d(ωL)/dt

= Ldω/dt  [Since, I is Constant]

=Lα    

So resultant acceleration will be                  

=√{(ω²L)²+(Lα)²}  

= L√(ω4+α²)    

So the force on it will be ,

mL√(ω4+α²) = µmg  

ω4+α²  =µ²g²/L²    

ω =  [(µ²g²/L²-α²)] ¹⁾⁴.

Hope it helps.