Question

A cylindrical bucket filled with water is whirled around in a vertical circle of radius r. What can be the minimum speedat the top of the path if water does not fall out from the bucket? If it continues with this speed, what normal contact force the bucket exerts on water at the lowest point of the path?

• A. $V_{min}=\sqrt{rg}$
  $N_{lowest}=Mg$
• B. $V_{min}=\sqrt{rg}$
  $N_{lowest}=2Mg$
• C. $V_{min}=\sqrt{\frac{rg}{2}}$
  $N_{lowest}=Mg$
• D. $V_{min}=\sqrt{rg}$
  $N_{lowest}=Mg$

Answer 1

The correct option is B

$V_{min}=\sqrt{rg}$
$N_{lowest}=2Mg$

Consider water as the system. At the top of the circle its acceleration towards the centre is vertically downward with magnitude . The forces on water are

(a) weight Mg downward and

(b) normal force by the bucket, also downward.

So, from Newton's second law
$Mg+N=M{v}^2/r$
For water not to fall out from the bucket,$N\ge 0$
Hence,$M{v}^2/r\ge Mgor,v^2\ge rg$
The minimum speed at the top must be $\sqrt{rg}$
If the bucket continues on the circle with this minimum speed $\sqrt{rg}$ are
(a) Weight Mg downward and
(b) Normal contact force N' by the bucket upward.
The acceleration is towards the centre which is vertically upward,s
or,$N^{\prime }-Mg=M{v}^2/r$
$N^{\prime }=M(g+v^2/r)=2Mg$