angular velocity

Question

The hammer throwers arm is 0.8m long and the hammer extends an additional 1.2m. It takes 0.75 seconds to complete 1 revolution (angular displacement of 6.28 rad) before the thrower releases the hammer. Assume the thrower is at rest when t=0s, and that the path of the arm is circular.

a) find the angular velocity of the hammer just before it is released

b) find the magnitude of the angular acceleration immediately before release of the hammer.

c) find the tangential velocity of end of the hammer immediately before release of the hammer, if the axis of rotation is the throwers torso (where his arm attaches to his body).

Answer 1

a) So knowns are as follows: ω_o = 0, t = 0.75 s, and θ = 6.28 rad. This is enough to find final angular velocity ω.

The semi-secret kinematics formula where acceleration is unknown but not asked for is d = 1/2 (v + v_o) t. Its angular analog is θ = 1/2 (ω + ω_o) t. Rearranging and solving for ω, ω = (2θ / t) - ω_o. Plugging in, ω = 2 * (6.28 rad)/(0.75s) - 0, which gives us 16.75 rad/s.

b) Are we assuming angular acceleration is constant? If not, this is unknowable based on the information given. Otherwise, the acceleration at the end is the same as it is during the entire motion.

α = (ω - ω_o)/t is the relevant formula. α = (16.75 rad/s - 0) / (0.75 s) gives me 22.33 rad/s².

c) So since the hammer is connected to the thrower at this instant, we know its angular velocity and tangential velocity are linked.

v = ωr is what we want. v = (16.75 rad//s) (2 m) gives us 33.5 m/s.