The hammer throw is a track-and-field event in which a 7.30-kg ball (the hammer)
ID: 1533641 • Letter: T
Question
The hammer throw is a track-and-field event in which a 7.30-kg ball (the hammer) is whirled around in a circle several times and released. It then moves upward on the familiar curved path of projectile motion and eventually returns to the ground some distance away. The world record for the horizontal distance is 86.75 m, achieved in 1986 by Yuriy Sedykh. Ignore air resistance and the fact that the ball was released above the ground rather than at ground level. Furthermore, assume that the ball is whirled around a circle that has a radius of 1.71 m and that its velocity at the instant of release is directed 24.0 ° above the horizontal. Find the magnitude of the centripetal force acting on the ball just prior to the moment of release.
Explanation / Answer
Okay, I think the hard part is finding the height of release.
Assuming the bottom of the circle is essentially on the ground.
Then, if you are familiar with calculus, since the point of release is at an angle of 55.8 above the horizontal,
the derivative of the equation of the circle must be equal to tan(55.8)x at that point.
The equation of the lower half of a circle is y=-sqrt(r^2-x^2) so y=-(4-x^2)^(1/2)
derivative of that is (-x)/((4-x^2)^(1/2)) set that equal to tan(55.8)x, and x = 1.8612.
Take this back to the circle equation, y = -(4-x^2)^(1/2) and y = .731889.
That is the initial height of the object.
From here, you can proceed like a kinematics problem.
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