You swing a bucket full of water in a vertical circle at the end of a rope. The mass of the bucket plus the water is $\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{\text{kg}}$. The center of mass of the bucket plus the water moves in a circle of radius $\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{\text{m}}$. At the instant that the bucket is at the top of the circle, the speed of the bucket is 4 m/s. What is the tension in the rope at this instant?

The mass of the bucket plus water is$m=3.5\text{kg}$.

The center of mass of the bucket plus water moves in a circle of radius$R=1.3\text{m}$

The force applied to an item in curved motion that is pointed toward the axis of rotation or the centre of curvature is known as the centripetal force.

The net force on an object is equal to the rate of change of momentum and can be written as the sum of two components.

The parallel rate of change of momentum${\frac{d\overrightarrow{p}}{dt}}_{\left|\right|}$and the perpendicular rate of change of momentum${\frac{d\overrightarrow{p}}{dt}}_{\perp }$are the two elements that we are concerned with.

$$\begin{gathered} F_{net}=\frac{d\overrightarrow{p}}{dt} \\ ={\frac{d\overrightarrow{p}}{dt}}_{\left|\right|}+{\frac{d\overrightarrow{p}}{dt}}_{\perp } \end{gathered}$$

In this case, the force acting on the object is in $+yor-y$direction, therefore, there is no parallel force and equals zero.

${\frac{d\overrightarrow{p}}{dt}}_{\left|\right|}=0$

As a result, the rate of change is the direction change owing to the perpendicular rate of change.

The net force exerted on the object equals the rate change of the momentum and the magnitude of the perpendicular rate change at speeds much less than the speed of light given by

$$\begin{gathered} F_{net}={\frac{d\overrightarrow{p}}{dt}}_{\perp } \\ =\frac{m{v}^2}{R} \end{gathered}$$

Also, the object is under two forces, the tension force$F_T$of the string, and its weight mg, so the net force of both forces is:

$F_{net}=F_T+mg$

Therefore,

$$\begin{gathered} F_T+mg=\frac{m{v}^2}{R} \\ {F}_T=\frac{m{v}^2}{R}-mg \end{gathered}$$

Now, put the values for $m,v,gandR$to get the tension force

$$\begin{gathered} F_T=\frac{m{v}^2}{R}-mg \\ =\frac{{\displaystyle (3.5\text{kg}){\left(4m/s\right)}^2}}{{\displaystyle 1.3\text{m}}}-(3.5\text{kg})\left(9.81m/s^2\right) \\ =8.74\text{N} \end{gathered}$$

Thus, the tension in the rope at this instant is $8.74N$.