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One type of slingshot can be made from a length of rope and
aleather pocket for holding the stone. The stone can be thrown
bywhirling it rapidly in a horizontal circle and releasing it at
theright moment. Such a slingshot is used to throw a stone from
theedge of a cliff, the point of release being 28.0 m above the
base of the cliff. The stonelands on the ground below the cliff at
a point X. Thehorizontal distance of point X from the
base of the cliff(directly beneath the point of release) is thirty
times the radiusof the circle on which the stone is whirled.
Determine the angularspeed of the stone at the moment of
release.

Answer :

Answer:

Angular speed of the stone at the moment of  release, w = 12.55 rad/s

Explanation:

  • The vertical distance (or Height) from the point where stone lands to the point of release = 28 m
  • Radius of the circle on which the stone is whirled = R
  • The horizontal distance from the point where stone lands to the point of release, d = 30*R
  • Linear velocity is the product of radius of circle and the angular velocity of the rotation

Linear Velocity, v = R*w,  

where w is the angular speed

Using the constant acceleration equation,

[tex]S = u*t + (1/2)*a*t^2[/tex],  where

  • S is the vertical distance (in our case 28 m),
  • u is the initial velocity (in our case 0 m/s),
  • t is the time taken by stone to reach point X,
  • a is the acceleration (in our case it will be g = 9.8 m/s)

Putting the values in our above constant acceleration equation we can find the time ,

(28) = (1/2)*(9.8)*t^2

t = 2.39 s

v = d/t

R*w = (30*R)/t

R*w*t = 30*R

w = 30/t

w = 30/2.39

w =  12.55 rad/s

${teks-lihat-gambar} syedabulhassan19

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