Answer to Question #155365 in Physics for Andu

Question

The amplifier shows the force F = 60 N. Find the velocity with which the stone with mass = 400 g turns in the circle with radius R = 30 cm.

Answer 1

Since the stone turns in the circle, the net force acting on it is a centripetal force. According to the second Newton's law, it is equal to:

$$F = ma$$

where $m = 400g = 0.4kg$ is the mass of the stone, and $a$ is the centripetal acceleration. $a$ is given as follows:

$$a = \dfrac{v^2}{R}$$

where $v$ is the speed of the stone, and $R = 30cm = 0.3m$ is the radius of the circle. Subsituting $a$ into the expression for $F$, and expressing $v$, obtain:

$$F = m\dfrac{v^2}{R}\\ v = \sqrt{\dfrac{RF}{m}}\\ v = \sqrt{\dfrac{0.3m\cdot 60N}{0.4kg}} \approx 6.7m/s$$

Answer. 6.7 m/s.