The hammer throw is a track-and-field event in which a 7.30-kg ball (the hammer)

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 2.97 m and that its velocity at the instant of release is directed 37.2 ° above the horizontal. Find the magnitude of the centripetal force acting on the ball just prior to the moment of release.

Explanation / Answer

Here ,

let the initail speed of the hammer is u

theta = 37.2 degree

mass , m = 7.3 Kg

horizontal distance , x = 86.75 m

for the equation of trajctory

y = x * tan(theta) - g * x^2/(2 * (v * cos(theta))^2)

0 = 86.75 * tan(37.2) - 9.8 * 86.75 /(2 * (v * cos(37.2))^2)

solving for v

v = 29.7 m/s

the initial speed of the ball is 29.7 m/s

the centripetal force is

F = m * v^2/R

F = 7.3 * 29.7^2/(2.97)

F = 2168 N

the magniude of centripetal force is 2168 N

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