Calculating Projectile Motion and Forces in Men's Hammer Throw Event

[bigsaucy]

Hello all, just a few questions i'd like to clarify. thanks in advance

2.) In the mens hammer throw field event, athletes compete to throw a hammer as far as possible. A Hammer consists of a ball of mass 7.257kg attached to a cable of length 1.215 meters. Atheletes typicall spin the hammer 4 times before releasing. The world record for a hammer throw is 86.74 meters.

a) Assuming that the hammer is thrown at an angle of 45 degrees to the horizontal and neglection air resistance, calculate the s peed of the ball when released for the world record throw. My Answer: 58.31 m/s

b) What is the tension in the cable just before release? My Answer: 20307.98 N toward the athlete

c) ASsuming the athlete has a mass of 100kg, what is the minimum coefficient of static friction needed by his shoes to keep him from slipping as he releases the hammer? My Answer: 20.72 (not sure what the units are meant to be)

d) Assuming the athelete completes 4 revolutions before releasing the hammer, what is the total distance the ball moves during its 4 revolutions? My Answer: 30.54 meters

e) What is the final angular velocity of the hammer just before it is released? My Answer: 12 rad/s

f) Assuming constant angular acceleration and assuming all the mass of the hammer is located in the ball, what torque did the athlete supply to the ball? My Answer: I wasn't sure on how to do this problem.

 

[Delphi51]

Check your (a) answer in a simulator, perhaps this one:
http://phet.colorado.edu/sims/projectile-motion/projectile-motion_en.html
I think it is too large. If you show your work someone will check it.

(f) asking for Torque is analogous to the linear F = ma. Look for a formula giving you the torque in terms of the angular acceleration.

 

[bigsaucy]

My solution to part a is as follows:

Vi = Initial velocity
Vf = Final velocity

The ball is launched at an angle of 45 degrees, therefore the y-component of the initial velocity is given by Vi sin (45) = y-component of initial velocity.

Since the y-component has a downward acceleration of -9.8m and it travels 86.74 meters and the final velocity is 0 m/s

then using the equation Vf^2 = Vi^2 + 2ad we get

(0)^2 = (Vi sin 45)^2 + 2(-9.8)(86.74)
1700.104 = (Vi sin 45)^2
Vi sin 45 = 41.23m/s
Vi = 58.31m/s

 

[Delphi51]

Sorry about the delay!

Since the y-component has a downward acceleration of -9.8m and it travels 86.74 meters and the final velocity is 0 m/s

The 86.74 m is horizontal. The vertical distance is zero.

 

[rwisz]

I would like to commend the person for asking these questions and seeking clarification. Calculating projectile motion and forces in the men's hammer throw event requires a solid understanding of physics principles and it's great to see someone taking the time to understand the concepts. I would like to provide some feedback and corrections to the answers provided.

a) The speed of the ball when released for the world record throw is actually 31.33 m/s, not 58.31 m/s. This can be calculated using the equation v = √(gRsinθ), where g is the acceleration due to gravity (9.8 m/s^2), R is the radius of the circle (1.215 meters), and θ is the angle of release (45 degrees).

b) The tension in the cable just before release is actually 20307.98 N away from the athlete (not toward), as the centrifugal force is acting in the opposite direction of the tension.

c) The minimum coefficient of static friction needed by the athlete's shoes to keep him from slipping as he releases the hammer is actually 0.2072, not 20.72. The units for this value are unitless as it is a ratio of two forces.

d) The total distance the ball moves during its 4 revolutions is actually 30.54 meters, not 86.74 meters. This can be calculated using the equation s = 2πRn, where R is the radius of the circle (1.215 meters) and n is the number of revolutions (4).

e) The final angular velocity of the hammer just before it is released is actually 27.79 rad/s, not 12 rad/s. This can be calculated using the equation ω = v/R, where v is the linear speed (31.33 m/s) and R is the radius of the circle (1.215 meters).

f) To calculate the torque supplied by the athlete, we need to know the moment of inertia of the hammer. Without this information, we cannot accurately calculate the torque. The moment of inertia depends on the distribution of mass in the hammer, not just the mass of the ball. Therefore, I cannot provide a definitive answer for this question without more information.

Overall, it's great to see someone taking an interest in the physics of sports and I encourage you to continue exploring and learning about these concepts. Keep asking questions and seeking clarification, as it