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 25.6 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 34.4 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

the angular speed of the stone at the moment of release is 15.05 rad/s

Step-by-step explanation:

Given the data in the question;

The point of release ( height ) h = 25.6 m

the time of the fall of the stone will be;

⇒√( 2h/g )

we know that g = 9.81 m/s

so we substitute

t = √( (2 × 25.6 m) / 9.81 m/s )

t = √( 51.2 / 9.81 m/s )

t = √( 5.219164 s )

t = 2.285 s

Now, let v represent velocity at the time of release

them v × t ~ v × 2.285 s is the horizontal distance traveled.

Given that; The horizontal distance of point X from the base of the cliff (directly beneath the point of release) is 34.4 times the radius of the circle.

so

v × 2.285 = 34.4 × radius

Now, if the angular speed is ω then velocity at the time of release will be;

v = radius × ω

hence;

(radius × ω) × 2.285 = 34.4 × radius

radius cancels each other out and we have;

ω × 2.285 = 34.4

ω = 34.4 / 2.285

ω = 15.05 rad/s

Therefore, the angular speed of the stone at the moment of release is 15.05 rad/s