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******10points***** physics help?
A hammer thrower accelerates the hammer (mass = 7.30 kg) from rest within four full turns (revolutions) and releases it at a speed of 28.0 m/s.
a) Assuming a uniform rate of increase in angular velocity and a horizontal circular path of radius 1.20 m, 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.
1 Answer
- ?Lv 79 years agoFavorite Answer
At release, ω = v/r = 28m/s / 1.2m = 23.3 rad/s
Also, Θ = ½αt²
"four full turns" = 4 * 2π = ½αt²
and ω = αt = 23.3 rad/s = αt
so α = 23.3rad/s / t
sub into Θ eqn:
8π = ½(23.3rad/s / t)t²
t = 2.15 s
α = 23.3rad/s / 2.15s = 10.83 rad/s² ← (a)
b) a = α r = 10.83rad/s² * 1.2m = 13 m/s²
c) a = v² / r = (28m/s)² / 1.2m = 653 m/s²
d) The gravity component is trivial here, so F = ma = 7.3kg * 653m/s² = 4769 N
e) OK, I guess they don't think gravity is trivial.
Fg = 7.3kg * 9.8m/s² = 71.5 N
True net force Fnet = √(4769² + 71.5²) N = 4769.5 N (told ya)
Θ = arctan(71.5/4769) = 0.89º
