A small ball of mass m suspended on an inextensible thread in the field of gravity rotates in the vertical plane. At the upper point of the trajectory, the tension of the thread is zero. Find the tension of the thread at the lower point of the trajectory.
Decision:
Since at the upper point of the ball's trajectory the thread tension is zero, then according to Newton's law II
$\frac{mv^{2}}{R}=mg$ (1)
here $v$ is the velocity of the ball at the upper point, $R$ is the length of the thread. For the lower point of the trajectory, the equation has the form
$\frac{mv^{2}_{1}}{R}=T-mg$ (2)
Here, $v_{1}$ is the velocity of the ball at the bottom point of the trajectory and $T$ is the tension of the thread at this point. From the law of conservation of energy it follows
$\frac{mv^{2}_{1}}{2}=\frac{mv^{2}}{2}+2mgR$. (3)
Solving the system of equations (1) through (3) with respect to $T$, we find:
$T=6mg$
Decision:
Since at the upper point of the ball's trajectory the thread tension is zero, then according to Newton's law II
$\frac{mv^{2}}{R}=mg$ (1)
here $v$ is the velocity of the ball at the upper point, $R$ is the length of the thread. For the lower point of the trajectory, the equation has the form
$\frac{mv^{2}_{1}}{R}=T-mg$ (2)
Here, $v_{1}$ is the velocity of the ball at the bottom point of the trajectory and $T$ is the tension of the thread at this point. From the law of conservation of energy it follows
$\frac{mv^{2}_{1}}{2}=\frac{mv^{2}}{2}+2mgR$. (3)
Solving the system of equations (1) through (3) with respect to $T$, we find:
$T=6mg$