Calculating Radial Acceleration in Olympic Hammer Throw

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SUMMARY

The discussion focuses on calculating the radial acceleration of a hammer in the Olympic hammer throw, specifically a 7.3 kg ball attached to a 1.2-meter cable released at a 21° angle from a height of 1.3 meters. The participant calculated a velocity of 103.69 m/s using the equation a = v² / r, leading to an unexpectedly high radial acceleration. The conversation emphasizes the importance of separating x and y motion equations to accurately determine the initial velocity and centripetal acceleration just before release.

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physic1GUY
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Homework Statement


confession Olympic hammer competition, competitor turns mass 7.3 kg ball at the end of a metal cable length 1.2 meters. Throw the hammer was released some 1.3 meters high 21 ° angle above horizon.
Last horizontal distance was 83 m hammer, what is the radial acceleration of the hammer just before liberation?


Homework Equations


Path equation - y=xtanTHETA - g / (2v0^2cos^2THETA) x x^2

The Attempt at a Solution


i set the details i know at the equation and i got a huge velocity 103.69 m/s
i put it in the equation a=v^2 / r but this is an astronom acceleration
what did i do wrong , i will glad for help and explanation...
thanks!
 
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physic1GUY said:

Homework Statement


confession Olympic hammer competition, competitor turns mass 7.3 kg ball at the end of a metal cable length 1.2 meters. Throw the hammer was released some 1.3 meters high 21 ° angle above horizon.
Last horizontal distance was 83 m hammer, what is the radial acceleration of the hammer just before liberation?


Homework Equations


Path equation - y=xtanTHETA - g / (2v0^2cos^2THETA) x x^2

The Attempt at a Solution


i set the details i know at the equation and i got a huge velocity 103.69 m/s
i put it in the equation a=v^2 / r but this is an astronom acceleration
what did i do wrong , i will glad for help and explanation...
thanks!

It would be good to see more details of your calculations. Generally in problems like this, you write one equation for the x motion as a function of time and one for the y motion as a function of time, and solve for the time where the object hits the ground. This let's you get back to the initial velocity, which will give you the centripetal acceleration in the circular throwing motion just before release...
 
i will post mu calculiting :
i talk the Y as the sin21 x 1.3
sin21x1.3= 83 tan21 - ( ( 9.8 x 83^2) /(2Vo^2-cos^2 21)
0.47 = 31.9- ( 67512.2)/(2Vo^2-0.88)
0.94Vo^2= 0.141+(-28072)-67512.2
-62.86Vo^2 = -28.072 - 67512.2
-62.86Vo^2 =-67540.2
V0=103/69 [m/s]
Ar=v^2/r
------------------
i didn't reference to any mass when do i need to refer it ?
thanks
 

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