A ball is suspended on an inextensible thread of length $l$ in the field of gravity with free fall acceleration $g$. At what speed $v$ must the suspension point be moved horizontally for the ball to make a complete revolution in the vertical plane?
Decision:
From the fixed frame of reference, let us move to a frame of reference rigidly connected to the suspension point. In this system, the suspension point is stationary, and the ball is given an initial horizontal velocity and, as a result of which it undergoes rotational motion in the vertical plane. In this rotational motion, the thread tension force must be different from zero at any point on the trajectory, except perhaps at the top of the trajectory, where it can go to zero. At the top point, the force of gravity $mg$ and the thread tension force $\bar{T}$, both pointing downward, act on the ball. They produce a centripetal acceleration $\frac{v^{2}_{1}}{l}$, where $v_{1}$ is the velocity of the ball at the top point. According to Newton's second law.
$mg+T=m\frac{v^{2}_{1}}{l}$
Hence
$T=m \left ( \frac{ v^{2}_{1}}{l} - g \right )$
We can say that the body will make a complete revolution if at the upper point of the trajectory $T \geq 0$, i.e., if
$v^{2}_{1} \geq gl$ (1)
The velocity $v_{1}$ depends on the initial velocity $v$. According to the law of conservation of energy
$\frac{m v^{2}_{1}}{2} + 2mgl=\frac{m v^{2}}{2}$. (2)
From (2) and (1) we find:
$v^{2} \geq 5gl$.
Thus, for the ball to make a complete revolution, it, or its suspension point, must be given a horizontal velocity $v$ satisfying the condition $v \geq \sqrt{5gl}$.
Decision:
From the fixed frame of reference, let us move to a frame of reference rigidly connected to the suspension point. In this system, the suspension point is stationary, and the ball is given an initial horizontal velocity and, as a result of which it undergoes rotational motion in the vertical plane. In this rotational motion, the thread tension force must be different from zero at any point on the trajectory, except perhaps at the top of the trajectory, where it can go to zero. At the top point, the force of gravity $mg$ and the thread tension force $\bar{T}$, both pointing downward, act on the ball. They produce a centripetal acceleration $\frac{v^{2}_{1}}{l}$, where $v_{1}$ is the velocity of the ball at the top point. According to Newton's second law.
$mg+T=m\frac{v^{2}_{1}}{l}$
Hence
$T=m \left ( \frac{ v^{2}_{1}}{l} - g \right )$
We can say that the body will make a complete revolution if at the upper point of the trajectory $T \geq 0$, i.e., if
$v^{2}_{1} \geq gl$ (1)
The velocity $v_{1}$ depends on the initial velocity $v$. According to the law of conservation of energy
$\frac{m v^{2}_{1}}{2} + 2mgl=\frac{m v^{2}}{2}$. (2)
From (2) and (1) we find:
$v^{2} \geq 5gl$.
Thus, for the ball to make a complete revolution, it, or its suspension point, must be given a horizontal velocity $v$ satisfying the condition $v \geq \sqrt{5gl}$.