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.