Shotput acceleration and speed calculation

  • Thread starter jarny
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In summary, the shot's speed at the end of the acceleration phase is decreased by a percentage depending on how much the angle between the path and the horizontal changes.
  • #1
jarny
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Homework Statement



In shot putting, many athletes elect to launch the shot at an angle that is smaller than the theoretical one (about 42°) at which the distance of a projected ball at the same speed and height is greatest. One reason has to do with the speed the athlete can give the shot during the acceleration phase of the throw. Assume that a 7.42 kg shot is accelerated along a straight path of length 1.65 m by a constant applied force of magnitude 410 N, starting with an initial speed of 2.5 m/s (due to the athlete's preliminary motion). What is the shot's speed at the end of the acceleration phase if the angle between the path and the horizontal is (a) 27° and (b) 42°? (Hint: Treat the motion as though it were along a ramp at the given angle.) (c) By what percent is the launch speed decreased if the athlete increases the angle from 27° to 42°?

Homework Equations



Fnet=m*a->

The Attempt at a Solution



Ok, as of now I think I have the hang of it, I just have a question and the rest makes sense. What I am going to find the acceleration vector and use that with v_0 to find v in v=v_0+at. I am going to find the acceleration vector via pythagorean theorem from the Net forces of the x and y components of the ball. The y is simple (i think) and I will find it by the sum of the F_g (negative) and the y component of the 410N vector at the degrees specified. I am stumped with the x though, Am I just summing the x component of fapp vector with the force of the ball (directed along the ground)? Thanks
 
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  • #2


You seem to be on the right track. The x component of your acceleration should just be

[tex] a_x = \frac{f_x}{m} [/tex]
 
  • #3


so only the x component of the fapp is all I need?
 
  • #4


jarny said:
so only the x component of the fapp is all I need?

Yes, because gravity acts in the vertical direction and I'm sure you aren't including wind resistance and I can't think of any other force that would act in the horizontal direction.
 
  • #5


Well I thought since the ball is at an angle the sum of the forces would be -f_gsintheta+fapp*costheta
 
  • #6


jarny said:
Well I thought since the ball is at an angle the sum of the forces would be -f_gsintheta+fapp*costheta

Oh, no! That's scalar addition. To add those two components to get a vector, you have multiply them by the appropriate unit vectors along the x and y axes. You find the magnitude of this vector sum by using the Pythagorean theorem.
 
  • #7


I am not following you here, sorry. can you put it in terms of the equations? (don't do work but is it possible just to say it a different way? thanks and sorry
 
  • #8


jarny said:

Homework Statement



In shot putting, many athletes elect to launch the shot at an angle that is smaller than the theoretical one (about 42°) at which the distance of a projected ball at the same speed and height is greatest. One reason has to do with the speed the athlete can give the shot during the acceleration phase of the throw. Assume that a 7.42 kg shot is accelerated along a straight path of length 1.65 m by a constant applied force of magnitude 410 N, starting with an initial speed of 2.5 m/s (due to the athlete's preliminary motion). What is the shot's speed at the end of the acceleration phase if the angle between the path and the horizontal is (a) 27° and (b) 42°? (Hint: Treat the motion as though it were along a ramp at the given angle.) (c) By what percent is the launch speed decreased if the athlete increases the angle from 27° to 42°?

Okay, let's start from the top. You start out with a constant force applied by the athlete of 410 N along a straight path. Since

[tex] \mathbf{F} = m \mathbf{a} [/tex]

you have can figure out the acceleration by dividing by m if you figure out all the forces acting and add them up vectorially.

The forces are (1) the given force and (2) gravity acting down.

In vector form, the given force is

[tex] \mathbf{F} = 410 cos(\theta) \mathbf{i} + 410 sin(\theta) \mathbf{j} [/tex]

and gravity is

[tex] \mathbf{F}_g = -mg \mathbf{j} [/tex]

the total force in vector form is

[tex] \mathbf{F}_T = 410 cos(\theta) \mathbf{i} + (410 sin(\theta) - mg) \mathbf{j} [/tex]

Divide through by m and you can then pick off the acceleration in the x and y directions. That will allow you to answer your problem.

Good luck. I must leave now.
 

1. How does the mass of the shotput affect its distance?

The mass of the shotput does not directly affect the distance it can be thrown. However, a heavier shotput may require more force to be thrown the same distance as a lighter shotput due to its inertia.

2. What is the optimal angle for throwing a shotput?

The optimal angle for throwing a shotput is around 45 degrees. This angle allows for the maximum horizontal distance to be achieved while also taking into account air resistance.

3. How does air resistance affect the trajectory of a shotput?

Air resistance can significantly decrease the distance a shotput can be thrown by slowing down its forward motion. This effect is more noticeable at higher speeds and can be reduced by using a more aerodynamic shotput or throwing at a lower angle.

4. What role does the release angle play in shotput throwing?

The release angle, or the angle at which the shotput is released, affects the trajectory and distance of the throw. A lower release angle will result in a longer distance, while a higher release angle will result in a shorter distance but a higher arc.

5. How does the height of a shotput thrower affect the distance of the throw?

The height of the thrower does not directly affect the distance of the throw. However, taller throwers may have an advantage due to their longer arms allowing for a longer release distance and potentially more force being applied to the shotput.

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