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

1) You have to assume it's a constant angular acceleration in order to calculate this - it should be stated in the problem. The appropriate equation given Δθ (2πi rads), t (.75 sec), and ω₀ = 0 in order to find ω(t = .75) is

Δθ = (1/2)(ω + ω₀)t or ω = 2Δθ/t

2) The constant angular acceleration is defined as α = (ω - ω₀)/t

3) v = (2 meters)ω

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