Question

A ball of mass 1.25 kg is attached by a 1.4 meter long massless rope to the top of a vertical pole. The ball swings in a horizontal circle at a constant rate of ? rad/s with the rope making an angle of 29 degree with the vertical. What is the angular velocity of the ball? (include units with answer)

A ball of mass 1.25 kg is attached by a 1.4 meter long massless rope to the top of a vertical pole. The ball swings in a horizontal circle at a constant rate of ? rad/s with the rope making an angle of 29

Answer 1

The free body diagram is shown in the figure:

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From the 2nd Newton law we can obtain the next equations for the components of the forces:

$\sum_{i} F_x_i = -T\sin\theta = -ma_c$,           (1)

$\sum_{i} F_y_i = -T\cos\theta -mg= 0$,         (2)

$a_c$ is the centripetal acceleration and can be written as:

$a_c = \omega^2 r = \omega^2 l\sin\theta$.     (3)

Substituting eq (3) in (1):

$T\sin\theta = m\omega^2 l\sin\theta$

or

$T = m\omega^2 l$        (4)

Getting T from (2) and substituting in (4):

$\frac{mg}{\cos\theta} = m\omega^2 l$.

And solving for $\omega$ ge get:

$\omega = \sqrt{\frac{g}{l\cos\theta}} = \sqrt{\frac{9.8~\mathrm{m/s^2}}{(1.4 ~\mathrm{m})\cos(29^\circ)}}= 2.83~\mathrm{rad/s}$.