A hammer thrower accelerates the hammer (mass = 7.30kg ) from rest within four f

Question

A hammer thrower accelerates the hammer (mass = 7.30kg ) from rest within four full turns (revolutions) and releases it at a speed of 27.3m/s . a.Assuming a uniform rate of increase in angular velocity and a horizontal circular path of radius 1.24m , calculate the angular acceleration. b.Calculate the (linear) tangential acceleration. c.Calculate the centripetal acceleration just before release. d.Calculate the net force being exerted on the hammer by the athlete just before release. e.Calculate the angle of this force with respect to the radius of the circular motion.

Explanation / Answer

The hammer is moving tangent to the circle. So, 28.5 m/s is the speed of an object that is moving around the circumference of the circle. 4 revolutions means the object moved a distance of 4 circumferences. Distance = 4 * 2 * p * 1.3 = 32.67 m Use the following equation to determine the tangential acceleration. vf^2 = vi^2 + 2 * a * d vi = 0 m/s, vf = 28.5 m/s, d = 32.67 m 28.5^2 = 2 * a * 32.67 tangential acceleration = 28.5^2 ÷ 65.34 = 12.43 m/s^2 a) Assuming a uniform rate of increase in angular velocity and a horizontal circular path of radius 1.30 m, calculate the angular acceleration. Angular acceleration = tangential acceleration ÷ radius = 12.43 ÷ 1.3 = 9.56 rad.s^2 OR ?f^2 = ?i^2+ 2 * a * ? ?f = vf ÷ r = 28.5 ÷ 1.3 = 21.92 rad/s ? = 4 revolutions = 4 * 2 * p radians = 8 * p radians 21.92^2 = 2 * a * 8 * p a = 21.92^2 ÷ (16 * p) = 9.56 m/s^2

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