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 28.0 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 25.7 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

ω = 10.75 rad/s

Step-by-step explanation:

The stone leaves the circular path with a horizontal speed

v₀ = vt = R*ω

So

ω = v₀ / R

we are given that R = x / 25.7 so ω = 25.7*v₀ / x

Kinematics gives x = v₀*t

With this substitution for x the expression for ω becomes ω = 25.7 / t

Kinematics also gives for the vertical displacement y that

y = v₀y*t + 0.5*ay*t²

we know that v₀y = 0 m/s

since the stone is launched horizontally, so that

y = 0.5*ay*t² ⇒ t = √(2*y / ay)

Using this result for t in the expression for ω and assuming that upward is positive, we get

ω = 25.7 / √(2*y / ay)

⇒ ω = 25.7 / √(2*(-28) / (-9.8))

⇒ ω = 10.75 rad/s