The hammer throw is a track-and-field event in which a 7.3-kg ball (the \'\'hamm

Question

The hammer throw is a track-and-field event in which a 7.3-kg ball (the ''hammer'') is whirled around in a circle several times and released. It then moves upward on the familiar curving path of projectile motion and eventually returns to earth some distance away. The world record for this distance is 86.75 m, achieved in 1986 by Yuriy Sedykh. Ignore air resistance and the fact that the ball is released above the ground rather than at ground level. Furthermore, assume that the ball is whirled on a circle that has a radius of 1.85 m and that its velocity at the instant of release is directed 39.0

Explanation / Answer

You can use kinematic eqs. to find the velocity V(o) at release, and then calculate the centripetal force. The X and Y position of a projectile ,projected at angle "a" with velocity V(o) are; X = [V(o)Cos(a)]t Y = [V(o)Sin(a)]t - (1/2)g(t^2) Now when X = 86.75 m , Y = 0. If we put Y=0 and cancel a "t" this will give us two eqs. in time "t"; X = [V(o)Cos(a)]t 0 = [V(o)Sin(a)] - (1/2)gt Now elliminate "t" 2(V(o)^2)[Sin(a)Cos(a)] = gX (V(o))^2 = gX/2Sin(a)Cos(a) = gX/Sin(2a) BTW, if you're unknown was X you could solve for X in terms of the initial velocity and projected angle. This is known as the "range" equation. But we need V(o)^2 , so ; V(o)^2 = (9,8)(86.75)/Sin(50) = 1109.8 (m/s)^2 Now the centripetal force associated with this velocity is; F = m(V(o)^2)/R = (7.3)(1109.8)/(1.82) = 4451.4 N

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