A hammer thrower accelerates the hammer (mass = 7.90 kg ) from rest within four

Question

A hammer thrower accelerates the hammer (mass = 7.90 kg ) from rest within four full turns (revolutions) and releases it at a speed of 26.0 m/s .

1. Assuming a uniform rate of increase in angular velocity and a horizontal circular path of radius 1.50 m/s , calculate the angular acceleration. Ignore gravity.
a=_________________ rad/s^2

2. Assuming a uniform rate of increase in angular velocity and a horizontal circular path of radius 1.50 m , calculate the (linear) tangential acceleration. Ignore gravity

a tan=______________ m/s^2

3. Assuming a uniform rate of increase in angular velocity and a horizontal circular path of radius 1.50 m, calculate the centripetal acceleration just before release. Ignore gravity.
a rad=__________________m/s^2

4. Assuming a uniform rate of increase in angular velocity and a horizontal circular path of radius 1.50m , calculate the net force being exerted on the hammer by the athlete just before release. Ignore gravity.
F=______________N

5. Assuming a uniform rate of increase in angular velocity and a horizontal circular path of radius 1.50m , calculate the angle of this force with respect to the radius of the circular motion. Ignore gravity.
theta=______________ degrees

Explanation / Answer

Mass of the hammer M = 7.9 Kg Angular displacement = number of revolutions * 2                                      = 4 * 2                                      = 25.132 rad Linear speed v = 26 m/s Radius of the circle r = 1.5 m 1) Initial angular speed 1 = 0 Final angular speed 2 = v/r = (26 m/s)/(1.5 m)                                             = 17.33 rad/s Angular acceleration = [2^2 - 1^2]/2                                    = [(17.33 rad/s)^2 - (0)^2]/(2*25.132 rad)                                    = 5.977 rad/s^2 2) Tangential acceleration a tan = r = (1.5 m)(5.977 rad/s^2)                                                    = 8.965 m/s^2 3) Radial acceleration a rad = v^2/r = (26 m/s)^2/(1.5 m)                                                    = 450.66 m/s^2 4) Net acceleration a = Sqrt[(a tan)^2 + (a rad)^2]                              = Sqrt[(8.965 m/s^2)^2 + (450.66 m/s^2)^2]                              = 450.749 m/s^2 Net force acting on the hammer F = ma                                                     = ( 7.9 Kg)(450.749 m/s^2)                                                     = 3560.9 N 5) Direction = Tan^-1[8.965 / 450.66 ]                   = 1.140 with radius of the circle.           

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