Question

a boy is whirling a stone around his head by means of a string. the string makes one complete revolution every second; and the magnitude of the tension in the string is f. the boy then speeds up the stone, keeping the radius of the circle unchanged, so that the string makes two complete revolutions every second. what happens to the tension in the sting?
a. the magnitude of the tension increases to four times its original value, 4f.
b. the magnitude of the tension is unchanged.
c. the magnitude of the tension increases to twice its original value, 2f.
d. the magnitude of the tension reduces to half of its original value, f/2.
e. the magnitude of the tension reduces to one - fourth of its original value, f/4.

Explanation:

Step1: Recall centripetal - force formula

The centripetal force $F_c = m\omega^{2}r$, where $m$ is the mass of the stone, $\omega$ is the angular velocity, and $r$ is the radius of the circular path. The tension in the string provides the centripetal force, so $F = m\omega_{1}^{2}r$.

Step2: Find the new angular velocity

The initial angular velocity $\omega_{1}=2\pi n_{1}$, where $n_{1} = 1$ rev/s. The new angular velocity $\omega_{2}=2\pi n_{2}$, and $n_{2}=2$ rev/s. So $\omega_{2} = 2\omega_{1}$.

Step3: Calculate the new tension

The new tension $F'=m\omega_{2}^{2}r$. Substitute $\omega_{2} = 2\omega_{1}$ into the formula: $F'=m(2\omega_{1})^{2}r=4m\omega_{1}^{2}r$. Since $F = m\omega_{1}^{2}r$, then $F' = 4F$.

Answer:

A. The magnitude of the tension increases to four times its original value, 4F.