An athlete whirls a 9.0-kg hammer six or seven times around and then releases it

Question

An athlete whirls a 9.0-kg hammer six or seven times around and then releases it. Although the purpose of whirling it around several times is to increase the hammer's speed, assume that just before the hammer is released, it moves at constant speed along a circular arc of radius 1.4 m. At the instant she releases the hammer, it is 3.8 m above the ground and its velocity is directed 7.0o above the horizontal. The hammer lands a horizontal distance of 74 m away. What force does the athlete apply to the grip just before she releases it? Ignore air resistance.

Explanation / Answer

if the initial velocity of the projectile is u

constant horizontal speed = ucos70°

initial vertical speed = - usin70° (upwards)

equation is:

y = -xtan70° + ½gx²/[u²cos²70°]

y = 3.8 m, x = 74 m

=> 3.8 = -74(2.747) + 4.905(74)² / [u²cos²70°]

=> [u²cos²70°] = [4.905(74)²] / 207.07

=> u² = 129.713/cos²70° => u = 33.29 m/s

centripetal force = mu²/r

if F is the force required:

F + mgcos60° = mu²/r

=> F = -4.5(9.81) + 9(1108.22)/1.4 ~= 7080 N

[please note: if we ignore gravity F ~ 7100 N

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