The hammer throw event in track and field involves spinning a heavy ball attached to a strong wire in a circle, then releasing it at just the right moment in the throw, and at the right angle, to maximize its trajectory away from the starting point. If the hammer thrower shown below is spinning at an angular rate of 3.31 revolutions per second, and the ball is 1.14 m from the center axis of the spin, what is the tangential velocity of ball?

When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the ball at an angle of 42.4 degrees above horizontal, and the ball travels a total distance of 32.6 m. What must have her angular rate been (in radians/s), assuming a chain length of 1.0 meter?

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cjmejiab cjmejiab · Answer 1 · 2019-11-11T04:19:36+01:00

To solve the problem we resort to the mechanics equations, our values are,

[tex]\nu=3.31rps[/tex]

[tex]r_1 = 1.14m[/tex]

By definition we know that the angular velocity is given by

[tex]\omega= 2\pi \nu[/tex]

[tex]\omega = 2\pi( 3.31)[/tex]

[tex]\omega = 20.8rad/s[/tex]

The relation of the tangential velocity with the angular velocity is given by the radius, thus

[tex]\omega = \frac{V}{r}[/tex]

[tex]V=\omega r[/tex]

[tex]V=20.8*1.14[/tex]

[tex]V=23.712m/s[/tex]

Part B)

In the case of an ideal speed at a certain angle, we have the values,

[tex]\theta = 42.4\°[/tex]

[tex]x=32.6m[/tex]

[tex]r_2 = 1m[/tex]

By definition we have to

[tex]x= \frac{V^2 sin\theta}{g}[/tex]

[tex]V= \sqrt{(\frac{g}{sin2\theta})}[/tex]

[tex]V=\frac{32.6*9.8}{sin(2*42.4)}[/tex]

[tex]V=17.81m/s[/tex]

The angular rate must be

[tex]\omega_2 = \frac{V}{r_2} = \frac{17.91}{1} = 17.91rad/s[/tex]