Question

One type of slingshot can be made from a length of rope and a leather pocket for holding the stone. The stone can be thrown by whirling it rapidly in a horizontal circle and releasing it at the right moment. Such a slingshot is used to throw a stone from the edge of a cliff, the point of release being 23.2 m above the base of the cliff. The stone lands on the ground below the cliff at a point X. The horizontal distance of point X from the base of the cliff (directly beneath the point of release) is 29.0 times the radius of the circle on which the stone is whirled. Determine the angular speed of the stone at the moment of release.

Answer 1

Answer:

The angular velocity is 13.57rad/s

Explanation:

To solve the problem, it is necessary to consult the concepts related to the description of displacement, velocity and acceleration given by the kinematic equations of motion.

For consideration given the distance traveled by the stone in the horizontal plane is 29 times the radius, that is

$x= 29r$

The relationship between the radius, the tangential velocity and the angular velocity is given by

$\omega = \frac{V_0}{r}$

While the distance is given by

$x=v_0 t$

From the kinematic equations the displacement is described as

$y=v_0+\frac{1}{2}at^2$

Re-arrange for time,

$t= \sqrt{\frac{2y}{a}}$

Here a is equal to the gravity acceleration, then

$t = \sqrt{\frac{2(23.2)}{9.8}}$

$t = 2.17s$

From the previous equation written above we have to

$\omega = \frac{V_0}{r}$

$\omega = \frac{\frac{x}{t}}{\frac{x}{29}}$

$\omega = \frac{29}{t}$

$\omega = \frac{29}{2.17}$

$\omega = 13.36rad/s$

Therefore the angular velocity of the stone at the moment of release is 13.57rad/s