Question

A water bucket of mass $'m'$ is revolved in a vertical circle with the help of a rope of length $'r'$ . If the velocity of the bucket at the lowest point is $\sqrt {7gr} $ . Then the velocity and tension in the rope at the highest point are:
(A) $\sqrt {3gr} ,2mg$
(B) $\sqrt {2gr} ,mg$
(C) $\sqrt {gr} ,mg$
(D) zero, zero

Answer 1

Hint Use the work energy theorem and substitute the known parameters and simplify it to find the velocity of the bucket at the highest point. Use the tension formula and substitute the velocity of the bucket at the highest point there to find the tension at that point.
Useful formula:
(1) The formula of the work energy theorem is given by
${W_{net}} = \Delta K$
Where ${W_{net}}$ is the net work done and the $\Delta K$ is the change in the kinetic energy.
(2) The formula of the kinetic energy is given by
$K = \dfrac{1}{2}m{v^2}$
Where $m$ is the mass of the bucket and the $v$ is the velocity of the bucket.
(3) The tension of the bucket is given by
$T = C - W$
Where $T$ is the tension of the bucket, $C$ is the centripetal force and $W$ is the weight of the bucket.
(4) The centripetal force is given by
$C = \dfrac{{m{v^2}}}{r}$
Where $r$ is the radius of the rotation.

Complete step by step answer
It is given that the velocity of the bucket at the lowest point, $v = \sqrt {7gr} $
Let us use the formula of the work energy theorem,
${W_{net}} = \Delta K$
Expanding the formula, by substituting the change in kinetic energy in it.
$2mgr = \dfrac{1}{2}mv_b^2 - \dfrac{1}{2}mv_t^2$
$2mgr = \dfrac{1}{2}m{\left( {\sqrt {7gr} } \right)^2} - \dfrac{1}{2}mv_t^2$
By simplification of the above equation, we get
$2mgr = \dfrac{7}{2}mgr - \dfrac{1}{2}mv_t^2$
By grouping the known terms in one side and the term need to find in the other side, we get
$v_t^2 = 3gr$
By further simplification, we get
${v_t} = \sqrt {3gr} $
Hence the velocity of the bucket at the top is found as $\sqrt {3gr} $ .
Using the formula of the tension,
$T = C - W$
$T = \dfrac{{m{v^2}}}{r} - mg$
Substituting the velocity of the bucket at the top,
$T = \dfrac{{m{{\left( {\sqrt {3gr} } \right)}^2}}}{r} - mg$
By simplification,
$T = 2\,mg$
Hence the tension of the rope is $2\,mg$ .

Thus the option (A) is correct.

Note The bucket is rotated in the circular motion with its force on the centre, hence the centripetal force is substituted while calculating the tension of the rope. This tension mainly depends on the mass of the bucket used, if the bucket is more mass then its tension will also be more.