Question

a ball of mass 0.1 g is suspended by a string of length 30 cm , keeping it always taut , it describes a circle of radius 15 cm . find the angular speed of the ball.

Answer 1

A ball of mass $0.1\mathrm{g}$ is suspended by a string of length $30~\mathrm{cm}$ , keeping it always taut, it describes a circle of radius $15~\mathrm{cm}$ . find the angular speed of the ball.

Solution:

Situation of the problem is shown on the figure:

$$\sin \theta = \frac {1 5 c m}{3 0 c m} = \frac {1}{2} o r \theta = 3 0 ^ {o}$$

Right triangle ABC:

$$T _ {x} = T \sin \theta ; T _ {x} = T \cos \theta ;$$

Newton's second law for the ball:

$$\vec {T} + \overline {{m}} \vec {g} = m \vec {a}$$

The projection on the X-axis:

$$x: T \sin \theta = m a$$

centripetal acceleration of the ball (circular movement)

$$a = \frac {V ^ {2}}{r} = \omega^ {2} r$$

(2)in (1): $T\sin \theta = m\omega^2 r$ (3)

The projection on the Y-axis:

$$y: T \cos \theta - m g = 0$$ $$T \cos \theta = m g$$ $$(3) \div (4): \frac {T \sin \theta}{T \cos \theta} = \frac {m \omega^ {2} r}{m g}$$ $$\tan \theta = \frac {\omega^ {2} r}{g}$$ $$\omega^ {2} = \frac {g \cdot \tan \theta}{r}$$ $$\omega = \sqrt {\frac {g \cdot \tan \theta}{r}} = \sqrt {\frac {9 . 8 \frac {m}{s ^ {2}} \cdot \tan 3 0 ^ {o}}{0 . 1 5 m}} = 6. 1 4 \frac {r a d}{s}$$

Answer: angular speed of the ball is $\omega = 6.14\frac{rad}{s}$