What is the total momentum of a high school discus thrower's throw?

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SUMMARY

The total momentum of a high school discus thrower's throw can be calculated using the final velocity of the discus upon landing. The discus, weighing 2 kg, is released at an initial velocity of 16.99 m/s and reaches a final vertical velocity of 10.29 m/s, resulting in a final resultant velocity of 17.95 m/s. The final momentum is determined using the formula P = mv, yielding a total momentum of 35.90 kg·m/s upon landing. The calculations involve understanding projectile motion and conservation of energy principles.

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leftdalltheway
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Hi, I'm new here. I have a question with one problem and i was wondering if any of you could help me out!

So um here's the problem..
A high school discus thrower rotates at an average angular acceleration of 895 degrees/sec^2 for 1.25 sec. The 2kg discus is realeased 0.87 meters away from the axis of rotation at his spine, at an angle of 30 degrees above the horizontal. The release point is 0.32 meters in front of the throwing circle, and the release height is 1.72 meters above the ground. THe throw is measued from the front of the throwing circle to the mark the discus makes in the ground at landing.

So I figured out the total distance of the throw which is 28.49
But i need to figure out the total momentum of the discus on landing.
Now I know P=mv, so i calculated 2kg * 16.99m/s (that'S the velocity i got). But I know this isn't the asnwer because I have to find out what the velocity is from the peak of the throw to the landing...I don't know how to do this! please help me!
 
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Originally posted by leftdalltheway
So I figured out the total distance of the throw which is 28.49
But i need to figure out the total momentum of the discus on landing.
Now I know P=mv, so i calculated 2kg * 16.99m/s (that'S the velocity i got). But I know this isn't the asnwer because I have to find out what the velocity is from the peak of the throw to the landing...I don't know how to do this! please help me!
Think conservation of energy. You have the initial speed (when the discus was released), so what's the initial KE and PE? Compare that to the KE and PE just as it lands.
 
Hmmm I don't understand how PE and KE comes into this...
I practically know nothing about physics so...haha.
Well i know that the initial momentum is 33.98 because i just
multiplied the weight of the discus (2kg) times the velocity
it was going (16.99m/s). But then, how do i figure out the velocity
from the peak of the projectile all the way down to the surface?
 
There are many ways to skin this cat.

If you wish, you could treat this as a projectile motion problem. You have the initial speed and direction (and height). So, write down the equations for the vertical and horizontal motions. Can you?

But realize that using conservation of energy can solve this problem much quicker. (Learn to do it both ways!)
 
hmm I'm sooo lost.
initial speed is 16.99.
direction and height?

and how do i find the vertical and horizontal motion?
i know that the initial vertical velocity is 8.49 (16.99sin30)
and horizontal velocity is 14.71 (16.99cos30)

what do i do after that?

this is what I've done so far because there was another problem where i had to find out the distance of the throw.

so,, angular acc 895 degrees/s2 so i changed that to rad/s2.
895 * (pi/180)=15.62 rad/s2.
to find the angular velocity, i used the angular acceleration equation, transposed it find the angular velocity.
w=at so 15.62 * 1.25 sec = 19.53 rad/s
and to find the intial velocity, i used the tangential velocity equation.
V tangential = 19.53 * 0.87m = 16.99m/s

V vertical = 8.49 V horizontal = 14.71
t(starting point to the peak) = -8.49/-9.81 = 0.87s
d(starting point to the peak) = -(8.49)^2/2(-9.81) = 3.67m
d (peak to the ground) = -5.4m
t(peak to the ground) = (-5.4/0.5*-9.81)^0.5 = 1.05s
t air = 0.87 + 1.05 = 1.92
range = 14.71 * 1.92 = 28.24m
distance = 28.24 + 0.32 = 28.56m
 
Well, I don't know why you did all those calculations regarding the peak, etc, unless there were other parts to the question that you didn't mention.

Anyway, your figures so far seem correct, except I would put the range at 28.2 m. I think you got 28.24 because you rounded the flight time up, and then used the rounded time to do that last calculation.

Now you know the horizontal component of the final velocity, but if you want to solve for the momentum this way, you still have to get the vertical component as well. (Easy enough, now that you know the initial vertical velocity and the total time in the air, but keep one or two extra decimal places in your intermediate results to reduce rounding errors). After you find the vertical component of the final velocity, you have to find the resultant velocity in order to compute the momentum. You know how to do that, right?
 
well the questions are:
a) what is the distance of the throw
b) what is the total momentum of the discus on landing?
c) create a type 1 scatterplot to show the relationship between the
projection angles, 0 degrees to 90 degrees, and the throw distance
d) what is the optimal angle of projection given this situation and these constants?


um..so the horizontal component of the final velocity is 14.71? (same as the intial horizontal velocity?) or is it 16.99?

i don't know how to calculate the final vertical component...
could you point me in the right direction?
 
The horizontal component of the velocity is constant, unless some force is applied in the horizontal direction after the discus is released. So, ignoring air resistance, it will be constant until it hits the ground. I thought you knew that. (You had to use that fact in order to solve for the range).

So, did you learn how to add two vectors? Here, you have one vector in the horizontal direction, and the other in the vertical (downward). So think of them as the two legs of a right triangle. The magnitude of the resultant final velocity is equal to the length of the hypotenuse. Does that ring a bell?
 
Oh, for the final vertical velocity:

That's just a simple free-fall problem. Ignore the horizontal component. Think of an object thrown vertically upwards; you know the initial vertical component. And you know the initial distance above ground when it was released.

vf2 = vi2 + 2a(yf-yi)
 
  • #10
okay, i think i got the ansewr.

So i used vf^2 = vi^2 + 2a(yf-yi)

vf2 = (8.49)^2 + 2 (-9.81(0-1.72))
vf = 10.29

so,

resultant = ((10.29)^2 + (14.71)^2)^0.5
= 17.95 so this is the final velocity (?)

then total momentum would be

(2kg * 16.99m/s) + (2kg * 17. 95)
= 69.88

is this right?
 
  • #11
Originally posted by leftdalltheway
okay, i think i got the ansewr.

So i used vf^2 = vi^2 + 2a(yf-yi)

vf2 = (8.49)^2 + 2 (-9.81(0-1.72))
vf = 10.29

so,

resultant = ((10.29)^2 + (14.71)^2)^0.5
= 17.95 so this is the final velocity (?)
Yes!

then total momentum would be

(2kg * 16.99m/s) + (2kg * 17. 95)
= 69.88

is this right?
I don't know what you mean by "total" momentum. There's an initial momentum and a final momentum. (Don't add them!) The final momentum is: M*Vf = 2*17.95
 
  • #12
Well the question says,
find the TOTAL momentum.
I'll ask my teacher about it.

Thanks for your help!
 

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