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 force that the wire exerts on the ball 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 force exerted by the wire on the ball is the sum of the tension due to the centripetal force needed to keep the ball moving in a circular path and the gravitational force acting on the ball. The gravitational force (weight) is $mg = 6.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 58.8 \, \text{N}$. To find the tension due to the centripetal force, we first need to calculate the speed of the ball at the lowest point. Using energy conservation, the potential energy at the start is converted into kinetic energy at the lowest point: $mgh = \frac{1}{2}mv^2$, where $h$ is the height from which the ball was released. The height $h$ can be found using trigonometry, considering the initial position of the pendulum: $h = L - L\cos(\theta) = L(1 - \cos(65^\circ))$, where $L = 2.5 \, \text{m}$ is the length of the pendulum. Substituting the values and solving for $v$, and then using $F_{\text{centripetal}} = \frac{mv^2}{r}$ to find the centripetal force, where $r = L$ is the radius of the circular path. The total force is the sum of the gravitational force and the centripetal force, which gives a result close to 130 N, making option D correct.

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