Ahammer 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=/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=

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

1) a=5.98 rad/sec² 2) a tan= 8.97 m/s² 3) a rad= 450.7 m/s²

4) F= 3,560.5 N 5) theta = 180º

Explanation:

1) By definition, angular acceleration is equal to the change in angular velocity over time.

Assuming an constant angular acceleration, we can use one of the equivalent kinematic equations for circular movement, as follows:

ωf² - ω₀² = 2. Δθ.γ

and we know also that ω = v/r = 26.0 m/s / 1.50 m = 17.3 rad/s

As the hammer thrower accelerated from rest, ω₀ = 0, so replacing by the values, we get the angular acceleration, γ, as follows:

γ = ωf² / 2. Δθ = (17.3)² rad²/s² / 2. 8 π rad = 5.98 rad/s².

2) Tangential acceleration, has the same relationship with radius that angular velocity, so we can write the following:

at = γ . r = 5.98 rad/sec². 1.50 m = 8.97 m/s²

3) The centripetal acceleration, by definition, is the change in direction of the linear velocity vector, over time, is always directed towards the center of the circle, and her magnitude is as follows:

ac = v² / r

Just before release, the velocity has a magnitude of 26.0 m/s, so ac is as follows:

ac = (26.0)² m²/s² / 1.50 m = 450.7 m/s²

4) Ignoring gravity, the only force acting on the hammer, is the one exerted by the thrower, and this force is just the centripetal force, which is the product of the hammer mass times the centripetal acceleration, as follows:

Fc = m . ac = 7.9 Kg. 450. 7 m/s² = 3,560.5 N

5) Ignoring gravity, as the force exerted by the thrower is always along the radius of the circle, towards the centre, if we represent the radius as a vector with origin in the center, the force is always anti-parallel to it, so the angle of the force with respect to the radius of the circular motion is 180°.

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