Question

1. a 0.2 kg ball is spun by a string in a vertical circle with a radius of 0.3 m and a speed of 4 m/s with nearly uniform circular motion. determine the tension force at the top and bottom of the circle.

Explanation:

Step1: Analyze forces at the top

At the top, tension \(T_{top}\) and weight \(mg\) act downward. Net force provides centripetal force \(F_c=\frac{mv^{2}}{r}\). So \(T_{top}+mg=\frac{mv^{2}}{r}\).
Substitute \(m = 0.2\space kg\), \(v = 4\space m/s\), \(r=0.3\space m\), \(g = 9.8\space m/s^{2}\):
\(T_{top}=\frac{mv^{2}}{r}-mg=\frac{0.2\times4^{2}}{0.3}-0.2\times9.8\)
\(=\frac{0.2\times16}{0.3}-1.96=\frac{3.2}{0.3}-1.96\approx10.67 - 1.96=8.71\space N\)

Step2: Analyze forces at the bottom

At the bottom, tension \(T_{bottom}\) acts upward and weight \(mg\) acts downward. Net force provides centripetal force. So \(T_{bottom}-mg=\frac{mv^{2}}{r}\).
Substitute values:
\(T_{bottom}=\frac{mv^{2}}{r}+mg=\frac{0.2\times4^{2}}{0.3}+0.2\times9.8\)
\(=\frac{3.2}{0.3}+1.96\approx10.67+1.96 = 12.63\space N\)

Answer:

Tension at the top: \(8.71\space N\), Tension at the bottom: \(12.63\space N\)