One end of a 1.0-m string is fixed; the other end is attached to a 2.0-kg stone.The stone swings in a vertical circle, passing the top point at 4.0 m/s.The tension force of the string at this point is about:
A)0 N
B)12 N
C)20 N
D)32 N
E)52 N

Question

Quizplus · Accepted Answer

At the top of the swing, the tension in the string must provide the centripetal force required to keep the stone moving in a circle, in addition to counteracting the force of gravity. The centripetal force ($F_c$) is given by $F_c = \frac{mv^2}{r}$, where $m$ is the mass, $v$ is the velocity, and $r$ is the radius of the circle. Here, $m = 2.0 \, \text{kg}$, $v = 4.0 \, \text{m/s}$, and $r = 1.0 \, \text{m}$. Thus, $F_c = \frac{(2.0)(4.0)^2}{1.0} = 32 \, \text{N}$. The force of gravity acting on the stone is $F_g = mg = (2.0)(9.8) = 19.6 \, \text{N}$. At the top of the circle, the tension in the string must support the weight of the stone and provide the centripetal force, but since the stone is also being pulled down by gravity, the net force acting towards the center is the centripetal force minus the gravitational force. Therefore, the tension in the string is actually $F_c - F_g = 32 - 19.6 = 12.4 \, \text{N}$, which is approximately 12 N.

One End of a 1