Answer: 0.68 kg
Explanation:
The ball in this example moves by uniform circular motion. In a uniform circular motion, an object of mass m moves in a circular orbit of radius r, with constant tangential speed v. This type of motion is produced by a force F (called centripetal force) that "pushes" the object towards the centre of the circular path. The magnitude of this force is given by
\(F=m\frac{v^2}{r}\)
The formula can also be rewritten as
\(F=m\omega^2 r\)
where \(\omega=\frac{2 \pi}{T}\) the angular frequency, and T is the period of revolution.
In this problem, we have the following data:
- centripetal force: F = 12 N
- radius: r = 87 cm = 0.87 m
- period of revolution: T = 1.4 s
Using the last formula, we can find the angular frequency:
\(\omega=\frac{2 \pi}{T}=\frac{2 \pi}{1.4 s}=4.49 rad/s\)
And now we can substitute \(\omega\) inside the formula of the centripetal force, and by re-arranging it we can find the mass of the ball:
\(m=\frac{F}{\omega^2 r}=\frac{12 N}{(4.49 rad/s)^2 (0.87 m)}=0.68 kg\)
