# 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 | Add Yours

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)` **

**Sources:**