Calculating Radial Acceleration in Olympic Hammer Throw

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The discussion focuses on calculating the radial acceleration of a hammer in an Olympic throw scenario, where a 7.3 kg ball is swung on a 1.2-meter cable and released at a height of 1.3 meters and a 21° angle. The user initially calculated a velocity of 103.69 m/s but found this resulted in an unrealistic radial acceleration. Suggestions were made to separate the x and y motion equations to determine the time of flight and initial velocity more accurately. The user also questioned the need to reference mass in their calculations. Clarifying these calculations and the role of mass in radial acceleration is essential for solving the problem correctly.
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 centripital 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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