Understanding Motion in a Vertical Circle
When a mass $m$ is rotated in a vertical circle of radius $r$, the forces acting on it change depending on its position. The two primary forces are gravity ($mg$) and the tension ($T$) in the string.
The Physics of the Situation
At any point, the net force towards the center of the circle provides the necessary centripetal force, $F_c = \frac{mv^2}{r}$.
- At the Top: Both tension ($T_{top}$) and gravity ($mg$) point towards the center. So, $T_{top} + mg = \frac{mv^2}{r}$.
- At the Bottom: Tension ($T_{bot}$) points towards the center, while gravity points away. So, $T_{bot} - mg = \frac{mv^2}{r}$.
Because the speed $v$ is constant, the centripetal force is constant. Therefore, the minimum tension occurs at the top, and the maximum tension occurs at the bottom.
Step-by-Step Problem Solving
Given:
- Mass $m = 0.2 \text{ kg}$
- Radius $r = 1 \text{ m}$
- Minimum Tension $T_{min} = 3 \text{ N}$
- $g \approx 9.8 \text{ m/s}^2$
1. Calculate the speed ($v$): At the top, $T_{min} = \frac{mv^2}{r} - mg$. $3 = \frac{0.2 v^2}{1} - (0.2 \times 9.8)$ $3 = 0.2v^2 - 1.96$ $0.2v^2 = 4.96$ $v^2 = 24.8$ $v = \sqrt{24.8} \approx 4.98 \text{ m/s}$
2. Calculate the maximum tension ($T_{max}$): At the bottom, $T_{max} = \frac{mv^2}{r} + mg$. $T_{max} = 0.2(24.8) + (0.2 \times 9.8)$ $T_{max} = 4.96 + 1.96 = 6.92 \text{ N}$