Answer to Question #274521 in Physics for Kiki

Question

A 2.5 kg object is whirled in a vertical circle whose radius is 0.89 m. If the time of one revolution is 0.94 s, the magnitude of the tension in the string (assuming uniform speed) when it is at the bottom is ___x10²N. (Give your answer with 2 sig digs, and do not include units)

Answer 1

Let's apply the Newton's Second Law of Motion:

$$F_T-mg=\dfrac{mv^2}{r},$$ $$F_T=\dfrac{mv^2}{r}+mg,$$ $$F_T=\dfrac{m(\omega r)^2}{r}+mg=m(\omega^2r+g).$$

We can find the angular velocity as follows:

$$\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{0.94\ s}=6.68\ \dfrac{rad}{s}.$$

Finally, we can find the magnitude of the tension in the string:

$$F_T=2.5\ kg\times((6.68\ \dfrac{rad}{s})^2\times0.89\ m+9.8\ \dfrac{m}{s^2})=1.2\times10^2\ N.$$