Q. The spool shown in figure is placed on rough horizontal surface and has the inner radius `r` and outer radius `R.` The angle `theta` between the applied force and the horizontal can be...
Q. The spool shown in figure is placed on rough horizontal surface and has the inner radius `r` and outer radius `R.` The angle `theta` between the applied force and the horizontal can be varied.The critical angle`(theta)` for which the spool does not roll and remains stationary is given by:-
A) `theta = cos^-1(r/R)`
B) `theta=cos^-1(2r)/R`
C) `theta=cos^-1sqrt(r/R)`
D) `theta=sin^-1(r/R)`
1 Answer
The figure is below. The outer circle of radius `R` rolls on the rough surface while the force is applied to the inner circle making an angle `theta` with the horizontal.
The condition for static equilibrium is equality of moments and equality of forces on both horizontal and vertical axis.
The external applied force has the components
`Fh =F*cos(theta)`
`F_v =F*sin(theta)`
On the horizontal and vertical axis the equilibrium condition are
`F*cos(theta) =Ff` (1)
`F*sin(theta)=N`
(The reaction `N` from the surface is not shown in the figure.)
The momentum equality is
`F*r = Ff*R`
which combined with (1) gives
`F*r = F*R*cos(theta)`
or equivalent
`cos(theta) = r/R`
Thus the angle for which the spool does not roll is
`theta =arccos(r/R)`
The correct answer is A) `theta =cos^(-1)(r/R)`
