Hammer thrower - Angular acceleration

[physicsss]

A hammer thrower accelerates the hammer (mass = 7.3 kg) from rest within four full turns (revolutions) and releases it at a speed of 27.2 m/s. Assuming a uniform rate of increase in angular velocity and a radius of 2.0 m, calculate (a) the angular acceleration, (b) the (linear) tangential acceleration, (c) the centripetal acceleration just before release, (d) the net force exerted on the hammer by the athlete just before release, and (e) the angle of this force with respect to the radius of the circular motion. Neglect the effect of gravity.

I know that angular acceleration is w(omega)/t, but I can't figure out how to find t...

 

[Nate810]

a) Angular acceleration = (27.2 m/s)/(4 revolutions) = 6.8 m/s2/revolutionb) Tangential acceleration = angular acceleration x radius = 6.8 m/s2 x 2.0 m = 13.6 m/s2c) Centripetal acceleration = tangential acceleration/radius = 13.6 m/s2/2.0 m = 6.8 m/s2d) Net force exerted on the hammer by the athlete just before release = mass x centripetal acceleration = 7.3 kg x 6.8 m/s2 = 49.84 Ne) Angle of this force with respect to the radius of the circular motion = 90 degrees

 

[wais]

(a) To find the angular acceleration, we can use the formula for angular velocity (w) which is w = ∆θ/∆t. In this case, the hammer thrower completes four full turns, which is equivalent to 8π radians (2π radians per turn x 4 turns). The final angular velocity is 27.2 m/s, and the initial angular velocity is 0 since the hammer starts from rest. So, we can write the equation as:

27.2 m/s = (8π radians)/∆t

Solving for ∆t, we get ∆t = (8π radians)/27.2 m/s = 0.929 seconds

Now, we can use the formula for angular acceleration: α = ∆w/∆t = (27.2 m/s)/0.929 s = 29.3 rad/s^2

Therefore, the angular acceleration of the hammer is 29.3 rad/s^2.

(b) The tangential acceleration (at) is the linear acceleration of a point on the edge of a rotating object. We can find it using the formula: at = rα, where r is the radius of the circular motion and α is the angular acceleration.

So, at = (2.0 m)(29.3 rad/s^2) = 58.6 m/s^2

Therefore, the tangential acceleration is 58.6 m/s^2.

(c) The centripetal acceleration (ac) is the acceleration towards the center of the circular motion. We can find it using the formula: ac = v^2/r, where v is the tangential velocity and r is the radius of the circular motion.

We know that the tangential velocity at release is 27.2 m/s and the radius is 2.0 m. So, ac = (27.2 m/s)^2/2.0 m = 369.44 m/s^2

Therefore, the centripetal acceleration just before release is 369.44 m/s^2.

(d) The net force exerted on the hammer by the athlete is equal to the mass of the hammer (7.3 kg) multiplied by the tangential acceleration.

So, Fnet = (7.3 kg)(58.6 m/s^2) = 427.18 N

Therefore, the net force