Question

a 100 g ball on a 70 - cm - long string is swung in a vertical circle about a point 200 cm above the floor. the string suddenly breaks when it is parallel to the ground and the ball is moving upward. the ball reaches a height 600 cm above the floor.
part a
what was the tension in the string an instant before it broke?
express your answer with the appropriate units.
t = value units
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Explanation:

Step1: Find the vertical displacement

The initial height of the ball (when the string breaks) above the floor is \(h_1 = 200\space cm\), and the final height is \(h_2=600\space cm\). The vertical displacement \(\Delta h=h_2 - h_1\).
\(\Delta h=(600 - 200)\space cm = 400\space cm=4\space m\)

Step2: Use kinematic equation to find the initial velocity

Using the kinematic equation \(v^2 = u^2- 2g\Delta h\). At the maximum height, \(v = 0\). So, \(u=\sqrt{2g\Delta h}\) (where \(g = 9.8\space m/s^2\))
\(u=\sqrt{2\times9.8\times4}=\sqrt{78.4}\approx 8.85\space m/s\)

Step3: Apply Newton's second law for circular motion

When the string is parallel to the ground, the centripetal force \(F_c=T\) (since there is no component of weight contributing to centripetal force in the radial direction at this instant). The centripetal force formula is \(F_c=\frac{mv^2}{r}\)
Given \(m = 100\space g=0.1\space kg\), \(r = 70\space cm = 0.7\space m\) and \(v = u\approx8.85\space m/s\)
\(T=\frac{mv^2}{r}=\frac{0.1\times(8.85)^2}{0.7}\)
\(T=\frac{0.1\times78.3225}{0.7}=\frac{7.83225}{0.7}\approx11.2\space N\)

Answer:

\(T = 11.2\space N\)