how does this shot put compare?

Question

i can throw a 14 lb. shot put about 32 ft. i was wondering what that distance would become, or compare to a 12 lb shot put throw. i was just wondering if someone had a math of physics equation i could use. thank you.

Answer 1

This is a very interesting question! There is a "world class" high school shot putter living here in the Dayton, Ohio area, so I was much intrigued by your question.

In the article I cite below (it's in PDF format so you can download and print it), it is shown that the distance a put goes is proportional to the square of the launch velocity. There are other important factors, such as release angle and height at release, but velocity is the most important. So, with a lighter shot, you should be able to launch it faster. What I don't know, and cannot tell you, is how much faster.

There is obviously a 2/14 or about 14% reduction in mass of a 12 lb. shot versus a 14 lb. shot, so the "impulse" (force versus time) you apply to the shot should be greater by that factor at least, in other words, about 14% greater. Does that translate into a 14% increase in release velocity? And if so, does that mean a 30% increase in distance? I doubt it, but who knows?

If you are training with the lighter shot, and want to compare those distance results with the expected competition shot distances, I suggest you just do the exercise and measure the distances with 14 lb. and 12 lb. shots on alternate days for a period of one week. Whatever that ratio turns out to be, treat it as gospel for the rest of your training with the 12 lb. shot. But go back to training with the 14 lb. shot no later than the final two weeks of training before putting for competition. It will take at least that long for your muscles and coordination to become accommodated to the increased weight.

Good luck!

Answer 2

let the 14 lb shotput by by m1 = 14lb/32 ft/s^2, g=32 ft/s^2

Assume that the launch angle is q0, the launch height is y0, and that your force is a constant F regardless of the weight of the shot put. Also assume that you accelerate the shot over a distance d before you release it.

Ok - the release speed is found using

d = (Vr^2 - V0^2)/(2*a) where V0 = 0 (starting from rest) and

a =F/m. For mass1 (m1), a=a1=F/m1. So release speed is

Vr = sqrrt(2*a*d)

Now the velocity parallel to the ground (x direction) is:

Vx =Vr*cos(q)

and the velocity in the vertical direction (y) is:

Vy = Vr*sin(q)

The shot travels from the launch height to the ground (y=0) in an arc. So you can write

y = 0 = -1/2*g*t^2+Vy*t+ y0 and solve for t

t = Vy/g +/- sqrt{(Vy/g)^2 + 2Vy/g}

The distance the shot travels in x is

x = Vx*t where t is found from above

If you do a lot of algebra you could arrive at an expression that contains the mass of the shotput, but it will be somewhat messy. Also F, d, y0, and q are arbitrary and you could come up with a number of combinations that would achieve the 14 lb shot going 32 ft.

So I guess I don't have a simple answer to your question but this is one way to get to an answer.

Answer 3

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Answer 4

it should be directly proportional.

14/12 = 32/x