In a museum exhibit of a simple pendulum, a very small but dense 6.0-kg ball swings from a very light 2.5-m wire. The ball is released from rest when the pendulum wire makes a 65° angle with
The vertical, and it swings in a circular arc with no appreciable friction or air resistance. What is the
Tension in the wire just as the ball swings through its lowest position?
A) 68 N
B) 0 N
C) 59 N
D) 130 N
E) 11 N

Question

Quizplus · Accepted Answer

To find the tension in the wire at the lowest position, we use the formula for the centripetal force required to keep the ball moving in a circular path, which is $F_c = \frac{mv^2}{r}$, where $m$ is the mass of the ball, $v$ is its velocity at the lowest point, and $r$ is the radius of the circular path (length of the wire). Additionally, at the lowest point, the tension in the wire must also counteract the weight of the ball ($mg$), so the total tension is $T = F_c + mg$.First, we find the velocity of the ball at the lowest point using energy conservation. The potential energy at the release point is converted into kinetic energy at the lowest point. The change in height ($h$) can be found using trigonometry, considering the initial angle and the length of the pendulum: $h = r - r\cos(\theta) = r(1 - \cos(\theta))$, where $\theta$ is the angle from the vertical (65°) and $r$ is the length of the wire (2.5 m).The potential energy at the top is $PE = mgh$, and the kinetic energy at the bottom is $KE = \frac{1}{2}mv^2$. Setting $PE = KE$ gives $mgh = \frac{1}{2}mv^2$, which simplifies to $v = \sqrt{2gh}$.Substituting $g = 9.8 \, \text{m/s}^2$ and $h = 2.5(1 - \cos(65°))$ into the equation for $v$, and then calculating $F_c$ and adding $mg$ to find the tension gives a value that matches the option D, 130 N. This accounts for both the centripetal force needed to keep the ball moving in a circle and the force needed to counteract gravity at the lowest point.

In a Museum Exhibit of a Simple Pendulum, a Very