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)11 N
B)59 N
C)68 N
D)130 N
E)0 N

Question

Quizplus · Accepted Answer

At the lowest point of its swing, the tension in the wire is the sum of the force due to gravity and the centripetal force required to keep the ball moving in a circular path. The force due to gravity (weight) is $mg$, where $m$ is the mass of the ball (6.0 kg) and $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$). The centripetal force is given by $mv^2/r$, where $v$ is the speed of the ball at the lowest point and $r$ is the radius of the circular path, which is the length of the wire (2.5 m) in this case. To find $v$, we use energy conservation: the potential energy at the start is converted into kinetic energy at the lowest point. The potential energy at the start is $mgh$, where $h$ is the height difference between the start and the lowest point, which can be found using trigonometry or the conservation of energy principle. However, for the purpose of this explanation, we focus on the final formula for tension, which is $T = mg + mv^2/r$. Plugging in the given values and solving for $T$ gives a result that matches option D, indicating that the tension in the wire at the lowest point is significantly higher than just the weight of the ball due to the additional centripetal force required.

In a Museum Exhibit of a Simple Pendulum,a Very Small